Class X Chapter 4 Triangles Maths - kopykitab.com
Transcript of Class X Chapter 4 Triangles Maths - kopykitab.com
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Exercise β 4A
1. D and E are points on the sides AB and AC respectively of a βABC such that DEβBC.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If π΄π·
π·π΅ =
4
7 and AC = 6.6cm, find AE.
(iv) If π΄π·
π΄π΅ =
8
15 and EC = 3.5cm, find AE.
Sol:
(i) In β ABC, it is given that DE β₯ BC.
Applying Thalesβ theorem, we get: π΄π·
π·π΅=
π΄πΈ
πΈπΆ
β΅ AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
β΄ DB = 10 β 3.6 = 6.4cm
Or, 3.6
6.4=
4.5
πΈπΆ
Or, EC = 6.4Γ4.5
3.6
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) In β ABC, it is given that DE || BC.
Applying Thalesβ Theorem, we get : π΄π·
π·π΅=
π΄πΈ
πΈπΆ
Adding 1 to both sides, we get : π΄π·
π·π΅+ 1 =
π΄πΈ
πΈπΆ+1
βΉ π΄π΅
π·π΅=
π΄πΆ
πΈπΆ
βΉ13.3
π·π΅=
11.9
5.1
βΉDB = 13.3Γ5.1
11.9 = 5.7 cm
Therefore, AD=AB-DB=13.5-5.7=7.6 cm
(iii) In β ABC, it is given that DE || BC.
Applying Thalesβ theorem, we get :
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
βΉ 4
7=
π΄πΈ
πΈπΆ
Adding 1 to both the sides, we get :
11
7=
π΄πΆ
πΈπΆ
βΉ EC = 6.6Γ7
11 = 4.2 cm
Therefore,
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
AE = AC β EC = 6.6 β 4.2 = 2.4 cm
(iv) In β ABC, it is given that DE β BC.
Applying Thalesβ theorem, we get:
π΄π·
π΄π΅=
π΄πΈ
π΄πΆ
βΉ 8
15=
π΄πΈ
π΄πΈ+ πΈπΆ
βΉ 8
15=
π΄πΈ
π΄πΈ+ 3.5
βΉ 8AE + 28 = 15AE
βΉ 7AE = 28
βΉ AE = 4cm
2. D and E are points on the sides AB and AC respectively of a βABC such that DEβBC. Find
the value of x, when
(i) AD = x cm, DB = (x β 2) cm, AE = (x + 2) cm and EC = (x β 1) cm.
(ii) AD = 4cm, DB = (x β 4) cm, AE = 8cm and EC = (3x β 19) cm.
(iii) AD = (7x β 4) cm, AE = (5x β 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Sol:
(i) In β ABC, it is given that DE || BC.
Applying Thalesβ theorem, we have : π΄π·
π·π΅=
π΄πΈ
πΈπΆ
βΉ π₯
π₯β2=
π₯+2
π₯β1
βΉX(x-1) = (x-2) (x+2)
βΉπ₯2 β x = π₯2 β 4
βΉx= 4 cm
(ii) In β ABC, it is given that DE β BC.
Applying Thalesβ theorem, we have :
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
βΉ 4
π₯β4 =
8
3π₯β19
βΉ 4 (3x-19) = 8 (x-4)
βΉ12x β 76 = 8x β 32
βΉ4x = 44
βΉ x = 11 cm
(iii) In β ABC, it is given that DE || BC.
Applying Thalesβ theorem, we have : π΄π·
π·π΅=
π΄πΈ
πΈπΆ
βΉ 7π₯β4
3π₯+4=
5π₯β2
3π₯
βΉ3x (7x-4) = (5x-2) (3x+4)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ21π₯2 β 12x = 15π₯2 + 14x-8
βΉ6π₯2 β 26x + 8 = 0
βΉ(x-4) (6x-2) =0
βΉ x = 4, 1
3
β΅ x β 1
3 (as if x =
1
3 π‘βππ π΄πΈ π€πππ ππππππ πππππ‘ππ£π)
β΄ x = 4 cm
3. D and E are points on the sides AB and AC respectively of a βABC. In each of the following
cases, determine whether DEβBC or not.
(i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm.
(ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm.
(iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.
(iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
Sol:
(i) We have:
π΄π·
π·πΈ=
5.7
9.5= 0.6 ππ
π΄πΈ
πΈπΆ=
4.8
8= 0.6 ππ
Hence, π΄π·
π·π΅=
π΄πΈ
πΈπΆ
Applying the converse of Thalesβ theorem,
We conclude that DE || BC.
(ii) We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 -6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 β 4.2 = 7 cm
Now,
π΄π·
π·π΅=
5.2
6.5=
4
5
π΄πΈ
πΈπΆ=
4.2
7
Thus, π΄π·
π·π΅β
π΄πΈ
πΈπΆ
Applying the converse of Thalesβ theorem,
We conclude that DE is not parallel to BC.
(iii) We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
DB = 10.8 β 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4cm
Therefore,
AE = 9.6 β 4 = 5.6 cm
Now,
π΄π·
π·π΅=
6.3
4.5=
7
5
π΄πΈ
πΈπΆ=
5.6
4=
7
5
βΉπ΄π·
π·π΅=
π΄πΈ
πΈπΆ
Applying the converse of Thalesβ theorem,
We conclude that DE β BC.
(iv) We have :
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 β 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 β 6.4 = 3.6 cm
Now,
π΄π·
π·π΅=
7.2
4.8=
3
2
π΄πΈ
πΈπΆ=
6.4
3.6=
16
9
This, π΄π·
π·π΅β
π΄πΈ
πΈπΆ
Applying the converse of Thalesβ theorem,
We conclude that DE is not parallel to BC.
4. In a βABC, AD is the bisector of β A.
(i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC.
(ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC.
(iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC.
(iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC.
Sol:
(i) It is give that AD bisects β A.
Applying angle β bisector theorem in β ABC, we get:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
βΉ5.6
π·πΆ=
6.4
8
βΉDC = 8Γ5.6
6.4= 7 ππ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
(ii) It is given that AD bisects β π΄.
Applying angle β bisector theorem in β ABC, we get:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
Let BD be x cm.
Therefore, DC = (6- x) cm
βΉπ₯
6βπ₯=
10
14
βΉ14x = 60-10x
βΉ24x = 60
βΉx = 2.5 cm
Thus, BD = 2.5 cm
DC = 6-2.5 = 3.5 cm
(iii) It is given that AD bisector β π΄.
Applying angle β bisector theorem in β ABC, we get:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
BD = 3.2 cm, BC = 6 cm
Therefore, DC = 6- 3.2 = 2.8 cm
βΉ3.2
2.8=
5.6
π΄πΆ
βΉAC = 5.6Γ2.8
3.2= 4.9 ππ
(iv) It is given that AD bisects β π΄.
Applying angle β bisector theorem in β ABC, we get:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
βΉπ΅π·
3=
5.6
4
βΉBD = 5.6Γ3
4
βΉBD = 4.2 cm
Hence, BC = 3+ 4.2 = 7.2 cm
5. M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB
produced at N. Prove that
(i) DM DC
MN BN (ii)
DN AN
DM DC
Sol:
(i) Given: ABCD is a parallelogram
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
To prove:
(i) π·π
ππ=
π·πΆ
π΅π
(ii) π·π
π·π=
π΄π
π·πΆ
Proof: In β DMC and β NMB
β DMC = β NMB (Vertically opposite angle)
β DCM = β NBM (Alternate angles)
By AAA- Similarity
βDMC ~ βNMB
β΄ π·π
ππ=
π·πΆ
π΅π
NOW, ππ
π·π=
π΅π
π·πΆ
Adding 1 to both sides, we get
ππ
π·π+ 1 =
π΅π
π·πΆ+ 1
βΉ ππ+π·π
π·π=
π΅π+π·πΆ
π·πΆ
βΉππ+π·π
π·π =
π΅π+π΄π΅
π·πΆ [β΅ ABCD is a parallelogram]
βΉπ·π
π·π=
π΄π
π·πΆ
6. Show that the line segment which joins the midpoints of the oblique sides of a trapezium is
parallel sides
Sol:
(i)
Let the trapezium be ABCD with E and F as the mid Points of AD and BC,
Respectively Produce AD and BC to Meet at P.
In β PAB, DC || AB.
Applying Thalesβ theorem, we get
ππ·
π·π΄=
ππΆ
πΆπ΅
Now, E and F are the midpoints of AD and BC, respectively.
βΉππ·
2π·πΈ=
ππΆ
2πΆπΉ
βΉππ·
π·πΈ=
ππΆ
πΆπΉ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Applying the converse of Thalesβ theorem in β PEF, we get that DC
Hence, EF || AB.
Thus. EF is parallel to both AB and DC.
This completes the proof.
7. In the given figure, ABCD is a trapezium in which ABβDC and its diagonals intersect at O.
If AO = (5x β 7), OC = (2x + 1) , BO = (7x β 5) and OD = (7x + 1), find the value of x.
Sol:
In trapezium ABCD, AB β CD and the diagonals AC and BD intersect at O.
Therefore,
π΄π
ππΆ=
π΅π
ππ·
βΉ 5π₯β7
2π₯+1=
7π₯β5
7π₯+1
βΉ (5x-7) (7x+1) = (7x-5) (2x+1)
βΉ35π₯2 + 5x β 49x β 7 = 14π₯2 β 10x + 7x β 5
βΉ21π₯2 β 41x β 2 = 0
βΉ21π₯2 β 42x + x -2 =0
βΉ21x (x-2) +1 (x-2) =0
βΉ (x-2) (21x + 1) =0
βΉ x = 2, - 1
21
β΅ x β - 1
21
β΄ x = 2
8. In ABC , M and N are points on the sides AB and AC respectively such that BM= CN. If
B C then show that MN||BC
Sol:
In βABC, β B = β πΆ
β΄ AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
AB β BM = AC β BM
βΉAB β BM = AC β CN (β΅BM =CN)
βΉAM =AN
β΄β AMN =β ANM (Angles opposite to equal sides are equal)
Now, in βABC,
β π΄+ β B + β C =1800 ----(1)
(Angle Sum Property of triangle)
Again In In βAMN,
β A + β AMN + β ANM =1800 ----(2)
(Angle Sum Property of triangle)
From (1) and (2), we get
β B + β C = β AMN + β ANM
βΉ 2β B = 2β AMN
βΉβ B = β AMN
Since, β B and β AMN are corresponding angles.
β΄ MN β BC.
9. ABC and DBC lie on the same side of BC, as shown in the figure. From a point P on BC,
PQ||AB and PR||BD are drawn, meeting AC at Q and CD at R respectively. Prove that
QR||AD.
Sol:
In β CAB, PQ || AB.
Applying Thales' theorem, we get: πΆπ
ππ΅=
πΆπ
ππ΄ β¦(1)
Similarly, applying Thales theorem in BDC , Where PR||DM we get:
πΆπ
ππ΅=
πΆπ
π π· β¦(2)
Hence, from (1) and (2), we have :
πΆπ
ππ΄=
πΆπ
π π·
Applying the converse of Thalesβ theorem, we conclude that QR β AD in β ADC.
This completes the proof.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
10. In the given figure, side BC of a βABC is bisected at D
and O is any point on AD. BO and CO produced meet
AC and AB at E and F respectively, and AD is
produced to X so that D is the midpoint of OX.
Prove that AO : AX = AF : AB and show that EFβBC.
Sol:
It is give that BC is bisected at D.
β΄ BD = DC
It is also given that OD =OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram.
β΄ BO || CX and BX || CO
BX || CF and CX || BE
BX || OF and CX || OE
Applying Thalesβ theorem in β ABX, we get:
π΄π
π΄π=
π΄πΉ
π΄π΅ β¦(1)
Also, in β ACX, CX || OE.
Therefore by Thalesβ theorem, we get:
π΄π
π΄π=
π΄πΈ
π΄πΆ β¦(2)
From (1) and (2), we have:
π΄π
π΄π=
π΄πΈ
π΄πΆ
Applying the converse of Theorem in β ABC, EF || CB.
This completes the proof.
11. ABCD is a parallelogram in which P is the midpoint of DC and Q is a point
on AC such that CQ = 1
4 AC. If PQ produced meets BC at R, prove that R
is the midpoint of BC.
Sol:
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = 1
2 AC β¦(i)
Also, it is given that CQ = 1
4 AC β¦(ii)
Dividing equation (ii) by (i), we get:
πΆπ
πΆπ=
1
4 π΄πΆ
1
2 π΄πΆ
Or, CQ = 1
2 πΆπ
Hence, Q is the midpoint of CS.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Therefore, according to midpoint theorem in βCSD
PQ || DS
If PQ || DS, we can say that QR || SB
In β CSB, Q is midpoint of CS and QR β SB.
Applying converse of midpoint theorem , we conclude that R is the midpoint of CB.
This completes the proof.
12. In the adjoining figure, ABC is a triangle in which AB = AC. IF D and E are
points on AB and AC respectively such that AD = AE, show that the points B,
C, E and D are concyclic.
Sol:
Given:
AD = AE β¦(i)
AB = AC β¦(ii)
Subtracting AD from both sides, we get:
βΉ AB β AD = AC β AD
βΉ AB β AD = AC - AE (Since, AD = AE)
βΉ BD = EC β¦(iii)
Dividing equation (i) by equation (iii), we get:
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
Applying the converse of Thalesβ theorem, DEβBC
βΉ β DEC + β ECB = 1800 (Sum of interior angles on the same side of a
Transversal Line is 00 .)
βΉ β DEC + β CBD = 1800 (Since, AB = AC βΉ β B = β C)
Hence, quadrilateral BCED is cyclic.
Therefore, B,C,E and D are concylic points.
13. In βABC, the bisector of β B meets AC at D. A line OQβAC meets AB, BC and BD at O, Q
and R respectively. Show that BP Γ QR = BQ Γ PR
Sol:
In triangle BQO, BR bisects angle B.
Applying angle bisector theorem, we get:
ππ
ππ =
π΅π
π΅π
βΉBP Γ QR = BQ Γ PR
This completes the proof.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Exercise β 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
β BAC = β PQR = 500
β ABC = β QPR = 600
β ACB = β PRQ = 700
Therefore, by AAA similarity theorem, β ABC β QPR
(ii)
We have:
π΄π΅
π·πΉ=
3
6=
1
2 πππ
π΅πΆ
π·πΈ=
4.5
9=
1
2
But , β ABC β β EDF (Included angles are not equal)
Thus, this triangles are not similar.
(iii)
We have:
πΆπ΄
ππ =
8
6=
4
3 πππ
πΆπ΅
ππ=
6
4.5=
4
3
βΉ πΆπ΄
ππ =
πΆπ΅
ππ
Also, β ACB = β PQR = 800
Therefore, by SAS similarity theorem, β ACB - β RQP.
(iv)
We have
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π·πΈ
ππ =
2.5
5=
1
2
πΈπΉ
ππ=
2
4=
1
2
π·πΉ
ππ =
3
6=
1
2
βΉ π·πΈ
ππ =
πΈπΉ
ππ=
π·πΉ
ππ
Therefore, by SSS similarity theorem, β FED- β PQR
(v)
In β ABC
β A + β π΅ + β πΆ = 1800 (Angle Sum Property)
βΉ800 + β π΅ + 700 = 1800
βΉ β π΅ = 300
β π΄ = β π πππ β π΅ = β π
Therefore, by AA similarity , β ABC - β MNR
2. In the given figure, βODC~βOBA, β BOC = 1150 and β CDO = 700.
Find (i) β DCO (ii) β DCO (iii) β OAB (iv) β OBA.
Sol:
(i)
It is given that DB is a straight line.
Therefore,
β π·ππΆ + β πΆππ΅ = 1800
β π·ππΆ = 1800 β 1150 = 650
(ii)
In β DOC, we have:
β ππ·πΆ + β π·πΆπ + β π·ππΆ = 1800
Therefore,
700 + β π·πΆπ + 650 = 1800
βΉ β π·πΆπ = 180 β 70 β 65 = 450
(iii)
It is given that β ODC - β OBA
Therefore,
β ππ΄π΅ = β ππΆπ· = 450
(iv)
Again, β ODC- β OBA
Therefore,
β ππ΅π΄ = β ππ·πΆ = 700
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
3. In the given figure, βOAB ~ βOCD. If AB = 8cm, BO = 6.4cm, OC = 3.5cm and
CD = 5cm, find (i) OA (ii) DO.
Sol:
(i) Let OA be X cm.
β΅ β OAB - β OCD
β΄ ππ΄
ππΆ=
π΄π΅
πΆπ·
βΉ π₯
3.5=
8
5
βΉ π₯ =8Γ3.5
5= 5.6
Hence, OA = 5.6 cm
(ii) Let OD be Y cm
β΅ β OAB - β OCD
β΄ π΄π΅
πΆπ·=
ππ΅
ππ·
βΉ 8
5=
6.4
π¦
βΉ π¦ =6.4 Γ5
8= 4
Hence, DO = 4 cm
4. In the given figure, if β ADE = β B, show that βADE ~ βABC. If AD = 3.8cm, AE = 3.6cm,
BE = 2.1cm and BC = 4.2cm, find DE.
Sol:
Given :
β π΄π·πΈ = β π΄π΅πΆ πππ β π΄ = β π΄
Let DE be X cm
Therefore, by AA similarity theorem, β ADE - β ABC
βΉ π΄π·
π΄π΅=
π·πΈ
π΅πΆ
βΉ 3.8
3.6+2.1=
π₯
4.2
βΉ π₯ =3.8 Γ4.2
5.7= 2.8
Hence, DE = 2.8 cm
5. The perimeter of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ
= 12cm, find AB.
Sol:
It is given that triangles ABC and PQR are similar.
Therefore,
πππππππ‘ππ (βπ΄π΅πΆ)
πππππππ‘ππ (βπππ )=
π΄π΅
ππ
βΉ32
24=
π΄π΅
12
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ π΄π΅ =32Γ12
24= 16 ππ
6. The corresponding sides of two similar triangles ABC and DEF are BC = 9.1cm and EF =
6.5cm. If the perimeter of βDEF is 25cm, find the perimeter of βABC.
Sol:
It is given that β ABC - β DEF.
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their
corresponding sides.
βΉ πππππππ‘ππ ππ βπ΄π΅πΆ
πππππππ‘ππ ππ βπ·πΈπΉ=
π΅πΆ
πΈπΉ
Let the perimeter of βABC be X cm
Therefore,
π₯
25=
9.1
6.5
βΉ π₯ =9.1Γ25
6.5= 35
Thus, the perimeter of βABC is 35 cm.
7. In the given figure, β CAB = 900 and ADβ₯BC. Show that βBDA ~ βBAC. If AC = 75cm,
AB = 1m and BC = 1.25m, find AD.
Sol:
In β BDA and β BAC, we have :
β π΅π·π΄ = β π΅π΄πΆ = 900
β π·π΅π΄ = β πΆπ΅π΄ (πΆπππππ)
Therefore, by AA similarity theorem, β BDA - β BAC
βΉ π΄π·
π΄πΆ=
π΄π΅
π΅πΆ
βΉ π΄π·
0.75=
1
1.25
βΉ AD = 0.75
1.25
= 0.6 m or 60 cm
8. In the given figure, β ABC = 900 and BDβ₯AC.
If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm,
find BC.
Sol:
It is given that ABC is a right angled triangle and BD is the altitude drawn from the right
angle to the hypotenuse.
In β BDC and β ABC, we have :
β π΄π΅πΆ = β π΅π΅πΆ = 900 (πππ£ππ)
β πΆ = β πΆ (ππππππ)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
By AA similarity theorem, we get :
β BDC- β ABC
π΄π΅
π΅π·=
π΅πΆ
π·πΆ
βΉ 5.7
3.8=
π΅πΆ
5.4
βΉ π΅πΆ =5.7
3.8 Γ5.4
= 8.1
Hence, BC = 8.1 cm
9. In the given figure, β ABC = 900 and BDβ₯AC.
If BD = 8cm, AD = 4cm, find CD.
Sol:
It is given that ABC is a right angled triangle
and BD is the altitude drawn from the right angle to the hypotenuse.
In β DBA and β DCB, we have :
β π΅π·π΄ = β πΆπ·π΅
β π·π΅π΄ = β π·πΆπ΅ = 900
Therefore, by AA similarity theorem, we get :
βDBA - β DCB
βΉ π΅π·
πΆπ·=
π΄π·
π΅π·
βΉ πΆπ· =π΅π·2
π΄π·
CD = 8Γ8
4= 16 ππ
10. P and Q are points on the sides AB and AC respectively of a βABC. If AP = 2cm, PB = 4cm,
AQ = 3cm and QC = 6cm, show that BC = 3PQ.
Sol:
We have :
π΄π
π΄π΅=
2
6=
1
3 πππ
π΄π
π΄πΆ=
3
9=
1
3
βΉ π΄π
π΄π΅=
π΄π
π΄πΆ
In β APQ and β ABC, we have:
π΄π
π΄π΅=
π΄π
π΄πΆ
β π΄ = β π΄
Therefore, by AA similarity theorem, we get:
β APQ - β ABC
Hence, ππ
π΅πΆ=
π΄π
π΄πΆ=
1
3
βΉ ππ
π΅πΆ=
1
3
βΉ BC = 3PQ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
This completes the proof.
11. ABCD is parallelogram and E is a point on BC.
If the diagonal BD intersects AE at F, prove that
AF Γ FB = EF Γ FD.
Sol:
We have:
β π΄πΉπ· = β πΈπΉπ΅ (ππππ‘ππππππ¦ πππππ ππ‘π ππππππ )
β΅ DA || BC
β΄ β π·π΄πΉ = β π΅πΈπΉ (π΄ππ‘πππππ‘π ππππππ )
β DAF ~ β BEF (AA similarity theorem)
βΉ π΄πΉ
πΈπΉ=
πΉπ·
πΉπ΅
Or, AF Γ FB = FD Γ EF
This completes the proof.
12. In the given figure, DBβ₯BC, DEβ₯AB and ACβ₯BC.
Prove that π΅πΈ
π·πΈ=
π΄πΆ
π΅πΆ.
Sol:
In βBED and βACB, we have:
β π΅πΈπ· = β π΄πΆπ΅ = 900
β΅ β π΅ + β πΆ = 1800
β΄ BD || AC
β πΈπ΅π· = β πΆπ΄π΅ (π΄ππ‘πππππ‘π ππππππ )
Therefore, by AA similarity theorem, we get :
β BED ~ β ACB
βΉ π΅πΈ
π΄πΆ=
π·πΈ
π΅πΆ
βΉ π΅πΈ
π·πΈ=
π΄πΆ
π΅πΆ
This completes the proof.
13. A vertical pole of length 7.5cm casts a shadow 5m long on the ground and at the same time
a tower casts a shadow 24m long. Find the height of the tower.
Sol:
Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.
In β ABC and β PQR, we have:
β π΄π΅πΆ = β πππ = 900
β π΄πΆπ΅ = β ππ π (π΄πππ’πππ ππππ£ππ‘πππ ππ π‘βπ ππ’π ππ‘ π‘βπ π πππ π‘πππ)
Therefore, by AA similarity theorem , we get :
β ABC ~ β PQR
βΉ π΄π΅
π΅πΆ=
ππ
ππ
βΉ 7.5
5=
π₯
24
7.524 36
5x cm
Therefore, PQ = 36 m
Hence, the height of the tower is 36 m.
14. In an isosceles βABC, the base AB is produced both ways in P and Q such that
AP Γ BQ = AC2.
Prove that βACP~βBCQ.
Sol:
Disclaimer: It should be βAPC ~ βBCQ instead of βACP ~
βBCQ
It is given that βABC is an isosceles triangle.
Therefore,
CA = CB
βΉ β πΆπ΄π΅ = β πΆπ΅π΄
βΉ 1800 β β πΆπ΄π΅ = 1800 β β πΆπ΅π΄
βΉ β πΆπ΄π = β πΆπ΅π
Also,
AP Γ BQ = π΄πΆ2
βΉ π΄π
π΄πΆ=
π΄πΆ
π΅π
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ π΄π
π΄πΆ=
π΅πΆ
π΅π (β΅ π΄πΆ = π΅πΆ )
Thus, by SAS similarity theorem, we get
βAPC ~ βBCQ
This completes the proof.
15. In the given figure, β 1 = β 2 and π΄πΆ
π΅π·=
πΆπ΅
πΆπΈ.
Prove that β ACB ~ β DCE.
Sol:
We have :
π΄πΆ
π΅π·=
πΆπ΅
πΆπΈ
βΉ π΄πΆ
πΆπ΅=
π΅π·
πΆπΈ
βΉ π΄πΆ
πΆπ΅=
πΆπ·
πΆπΈ (πππππ, π΅π· = π·πΆ ππ β 1 = β 2 )
Also, β 1 = β 2
i.e, β π·π΅πΆ = β π΄πΆπ΅
Therefore, by SAS similarity theorem, we get :
β ACB - β DCE
16. ABCD is a quadrilateral in which AD = BC.
If P, Q, R, S be the midpoints of AB, AC, CD
and BD respectively, show that PQRS
is a rhombus.
Sol:
In β ABC, P and Q are mid points of AB and AC respectively.
So, PQ || BC, and PQ = 1
2 π΅πΆ β¦(1)
Similarly, in βADC, β¦(2)
Now, in βBCD, SR =1
2 π΅πΆ β¦(3)
Similarly, in βABD, PS = 1
2 π΄π· =
1
2 π΅πΆ β¦(4)
Using (1), (2), (3), and (4).
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
17. In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that
(a) PAC PDB (b) PA. PB= PC.PD
Sol:
Given : AB and CD are two chords
To Prove:
(a) β PAC ~ βPDB
(b) PA.PB = PC.PD
Proof: In β PAC and β PDB
β π΄ππΆ = β π·ππ΅ (ππππ‘ππππππ¦ πππππ ππ‘π ππππππ )
β πΆπ΄π = β π΅π·π (π΄πππππ ππ π‘βπ π πππ π ππππππ‘ πππ πππ’ππ)
ππ¦ π΄π΄ π ππππππππ‘π¦ ππππ‘πππππ βππ΄πΆ ~ππ·π΅
When two triangles are similar, then the ratios of lengths of their corresponding sides
are proportional.
β΄ ππ΄
ππ·=
ππΆ
ππ΅
βΉ PA. PB = PC. PD
18. Two chords AB and CD of a circle intersect at a point P outside the circle.
Prove that: (i) β PAC ~ β PDB
(ii) PA. PB = PC.PD
Sol:
Given : AB and CD are two chords
To Prove:
(a) β PAC - β PDB
(b) PA. PB = PC.PD
Proof: β π΄π΅π· + β π΄πΆπ· = 1800 β¦(1) (Opposite angles of a cyclic quadrilateral are
supplementary)
β ππΆπ΄ + β π΄πΆπ· = 1800 β¦(2) (Linear Pair Angles )
Using (1) and (2), we get
β π΄π΅π· = β ππΆπ΄
β π΄ = β π΄ (Common)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
By AA similarity-criterion β PAC - β PDB
When two triangles are similar, then the rations of the lengths of their corresponding
sides are proportional.
β΄ ππ΄
ππ·=
ππΆ
ππ΅
βΉ PA.PB = PC.PD
19. In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD AC ,
if DP AB and DQ BC then prove that
(a) 2 .DQ Dp QC (b) 2 .DP DQ AP 2
Sol:
We know that if a perpendicular is drawn from the vertex of the right angle of a right
triangle to the hypotenuse then the triangles on the both sides of the perpendicular are
similar to the whole triangle and also to each other.
(a) Now using the same property in In βBDC, we get
βCQD ~ βDQB
πΆπ
π·π=
π·π
ππ΅
βΉ π·π2 = ππ΅. πΆπ
Now. Since all the angles in quadrilateral BQDP are right angles.
Hence, BQDP is a rectangle.
So, QB = DP and DQ = PB
β΄ π·π2 = π·π. πΆπ
(b)
Similarly, βAPD ~ βDPB
π΄π
π·π=
ππ·
ππ΅
βΉ π·π2 = π΄π. ππ΅
βΉ π·π2 = π΄π. π·π [β΅ π·π = ππ΅]
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Exercise β 4C
1. βABC~βDEF and their areas are respectively 64 cm2 and 121cm2. If EF = 15.4cm, find BC.
Sol:
It is given that β ABC ~ β DEF.
Therefore, ratio of the areas of these triangles will be equal to the ration of squares of their
corresponding sides.
ππ (βπ΄π΅πΆ)
ππ (βπ·πΈπΉ)=
π΅πΆ2
πΈπΉ2
Let BC be X cm.
βΉ 64
121=
π₯2
(15.4)2
βΉ π₯2 =64Γ15.4 Γ15.4
121
βΉ π₯ =(64Γ15.4 Γ 15.4)
121
= 8Γ15.4
11
= 11.2
Hence, BC = 11.2 cm
2. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5cm, find
the length of QR.
Sol:
It is given that β ABC ~ β PQR
Therefore, the ration of the areas of triangles will be equal to the ratio of squares of their
corresponding sides.
ππ (βπ΄π΅πΆ)
ππ (βπππ )=
π΅πΆ2
ππ 2
βΉ 9
16=
42
ππ 2
βΉ ππ 2 =4.5 Γ4.5 Γ16
9
βΉ ππ =(4.5 Γ4.5 Γ16)
9
= 4.5 Γ4
3
= 6 cm
Hence, QR = 6 cm
3. βABC~βPQR and ar(βABC) = 4, ar(βPQR) . If BC = 12cm, find QR.
Sol:
Given : ππ ( β π΄π΅πΆ ) = 4ππ (β πππ )
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
ππ(βπ΄π΅πΆ)
ππ(βπππ )=
4
1
β΅ βABC ~ βPQR
β΄ ππ(βπ΄π΅πΆ)
ππ(βπππ )=
π΅πΆ2
ππ 2
β΄ π΅πΆ2
ππ 2=
4
1
βΉ ππ 2 =122
4
βΉ ππ 2 = 36
βΉ ππ = 6 ππ
Hence, QR = 6 cm
4. The areas of two similar triangles are 169cm2 and 121cm2 respectively. If the longest side of
the larger triangle is 26cm, find the longest side of the smaller triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Let the longest side of smaller triangle be X cm.
ππ (πΏπππππ π‘πππππππ)
ππ (πππππππ π‘πππππππ)=
(πΏπππππ π‘ π πππ ππ ππππππ π‘πππππππ)2
( πΏπππππ π‘ π πππ ππ π ππππππ π‘πππππππ)2
βΉ 169
121=
262
π₯2
βΉ π₯ = β26 Γ26 Γ121
169
= 22
Hence, the longest side of the smaller triangle is 22 cm.
5. βABC ~ βDEF and their areas are respectively 100cm2 and 49cm2. If the altitude of βABC
is 5cm, find the corresponding altitude of βDEF.
Sol:
It is given that βABC ~ βDEF.
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of
their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their
corresponding altitudes.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Let the altitude of βABC be AP, drawn from A to BC to meet BC at P and the altitude of
βDEF be DQ, drawn from D to meet EF at Q.
Then,
ππ(βπ΄π΅πΆ)
ππ(βπ·πΈπΉ)=
π΄π2
π·π2
βΉ 100
49=
52
π·π2
βΉ 100
49=
25
π·π2
βΉ π·π2 =49 Γ25
100
βΉ π·π = β49 Γ25
100
βΉ π·π = 3.5 ππ
Hence, the altitude of βDEF is 3.5 cm
6. The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the
ratio of their areas.
Sol:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
It is given that β ABC ~ β DEF.
We know that the ration of areas of two similar triangles is equal to the ratio of squares of
their corresponding altitudes.
β΄ ππ(βπ΄π΅πΆ)
ππ(βπ·πΈπΉ)=
(π΄π)2
(π·π)2
βΉ ππ(βπ΄π΅πΆ)
ππ(π·πΈπΉ)=
62
92
= 36
81
= 4
9
Hence, the ratio of their areas is 4 : 9
7. The areas of two similar triangles are 81cm2 and 49cm2 respectively. If the altitude of the
first triangle is 6.3cm, find the corresponding altitude of the other.
Sol:
It is given that the triangles are similar.
Therefore, the areas of these triangles will be equal to the ratio of squares of their
corresponding sides.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their
corresponding altitudes.
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
ππ(βπ΄π΅πΆ)
ππ(βπππ )=
π΄π2
π·π2
βΉ 81
49=
6.32
π·π
βΉ π·π2 =49
81 Γ6.32
βΉ π·π2 = β49
81 Γ6.3 Γ6.3
Hence, the altitude of the other triangle is 4.9 cm.
8. The areas of two similar triangles are 64cm2 and 100cm2 respectively. If a median of the
smaller triangle is 5.6cm, find the corresponding median of the other.
Sol:
Let the two triangles be ABC and PQR with medians AM and PN, respectively.
Therefore, the ratio of areas of two similar triangles will be equal to the ratio of squares of
their corresponding medians.
β΄ ππ(βπ΄π΅πΆ)
ππ(βπππ )=
π΄π2
ππ2
βΉ 64
100=
5.62
ππ2
βΉ ππ2 =64
100 Γ5.62
βΉ ππ2 = β100
64 Γ5.6 Γ5.6
= 7 cm
Hence, the median of the larger triangle is 7 cm.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
9. In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in
Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of βAPQ is 1
16 of the
area of βABC.
Sol:
We have :
π΄π
π΄π΅=
1
1+3=
1
4 πππ
π΄π
π΄πΆ=
1.5
1.5+4.5=
1.5
6=
1
4
βΉ π΄π
π΄π΅=
π΄π
π΄πΆ
Also, β π΄ = β π΄
By SAS similarity, we can conclude that βAPQ- βABC.
ππ(βπ΄ππ)
ππ(βπ΄π΅πΆ)=
π΄π2
π΄π΅2 =12
42 =1
16
βΉ ππ(βπ΄ππ)
ππ(βπ΄π΅πΆ)=
1
16
βΉ ππ(βπ΄ππ) =1
16 Γππ (βπ΄π΅πΆ)
Hence proved.
10. In the given figure, DEβBC. If DE = 3cm, BC = 6cm and ar(βADE) = 15cm2, find the area
of βABC.
Sol:
It is given that DE || BC
β΄ β π΄π·πΈ = β π΄π΅πΆ (πΆπππππ πππππππ ππππππ )
β π΄πΈπ· = β π΄πΆπ΅ (πΆπππππ πππππππ ππππππ )
By AA similarity, we can conclude that β ADE ~ β ABC
β΄ ππ(βπ΄π·πΈ)
ππ(βπ΄π΅πΆ)=
π·πΈ2
π΅πΆ2
βΉ 15
ππ(βπ΄π΅πΆ)=
32
62
βΉ ππ (βπ΄π΅πΆ) =15 Γ36
9
= 60 ππ2
Hence, area of triangle ABC is 60 ππ2
11. βABC is right angled at A and ADβ₯BC. If BC = 13cm and AC = 5cm, find the ratio of the
areas of βABC and βADC.
Sol:
In βABC and βADC, we have:
β π΅π΄πΆ = β π΄π·πΆ = 900
β π΄πΆπ΅ = β π΄πΆπ· (ππππππ)
By AA similarity, we can conclude that β BAC~ β ADC.
Hence, the ratio of the areas of these triangles is equal to the ratio of squares of their
corresponding sides.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
β΄ ππ(βπ΅π΄πΆ)
ππ(βπ΄π·πΆ)=
π΅πΆ2
π΄πΆ2
βΉ ππ(βπ΅π΄πΆ)
ππ(βπ΄π·πΆ)=
132
52
= 169
25
Hence, the ratio of areas of both the triangles is 169:25
12. In the given figure, DEβBC and DE: BC = 3:5. Calculate the ratio of the areas of βADE and
the trapezium BCED.
Sol:
It is given that DE || BC.
β΄ β π΄π·πΈ = β π΄π΅πΆ (πΆπππππ πππππππ ππππππ )
β π΄πΈπ· = β π΄πΆπ΅ (πΆπππππ πππππππ ππππππ )
Applying AA similarity theorem, we can conclude that β ADE ~ βABC.
β΄ ππ(βπ΄π΅πΆ)
ππ(π΄π·πΈ)=
π΅πΆ2
π·πΈ2
Subtracting 1 from both sides, we get:
ππ(βπ΄π΅πΆ)
ππ(βπ΄π·πΈ)β 1 =
52
32 β 1
βΉ ππ(βπ΄π΅πΆ)βππ(βπ΄π·πΈ)
ππ(βπ΄π·πΈ)=
25β9
9
βΉ ππ(π΅πΆπΈπ·)
ππ(βπ΄π·πΈ)=
16
9
Or, ππ(βπ΄π·πΈ)
ππ(π΅πΆπΈπ·)=
9
16
13. In βABC, D and E are the midpoints of AB and AC respectively. Find the ratio of the areas
of βADE and βABC.
Sol:
It is given that D and E are midpoints of AB and AC.
Applying midpoint theorem, we can conclude that DE β BC.
Hence, by B.P.T., we get :
π΄π·
π΄π΅=
π΄πΈ
π΄πΆ
Also, β π΄ = β π΄
Applying SAS similarity theorem, we can conclude that β ADE~ β ABC.
Therefore, the ration of areas of these triangles will be equal to the ratio of squares of their
corresponding sides.
β΄ ππ(βπ΄π·πΈ)
ππ(βπ΄π΅πΆ)=
π·πΈ2
π΅πΆ2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
= (
1
2 π΅πΆ)
2
π΅πΆ2
= 1
4
Exercise β 4D
1. The sides of certain triangles are given below. Determine which of them right triangles are.
(i) 9cm, 16cm, 18cm (ii) 7cm, 24cm, 25cm
(iii) 1.4cm, 4.8cm, 5cm (iv) 1.6cm, 3.8cm, 4cm
(v) (a β 1) cm, 2βπ cm, (a + 1) cm
Sol:
For the given triangle to be right-angled, the sum of the two sides must be equal to the
square of the third side.
Here, let the three sides of the triangle be a, b and c.
(i)
a = 9 cm, b = 16 cm and c = 18 cm
Then,
π2 + π2 = 92 + 162
= 81 + 256
= 337
π2 = 192
= 361
π2 + π2 β π2
Thus, the given triangle is not right-angled.
(ii)
A=7 cm, b = 24 cm and c = 25 cm
Then,
π2 + π2 = 72 + 242
= 49 + 576
= 625
π2 = 252
= 625
π2 + π2 = π2
Thus, the given triangle is a right-angled.
(iii)
A= 1.4 cm, b= 4.8 cm and c= 5 cm
Then,
π2 + π2 = (1.4)2 + (4.8)2
= 1.96 + 23.04
= 25
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π2 = 52
= 25
π2 + π2 = π2
Thus, the given triangle is right-angled.
(iv) A = 1.6 cm, b= 3.8 cm and c= 4 cm
Then
π2 + π2 = (1.6)2 + (3.8)2
= 2.56 + 14.44
= 16
π2 + π2 β π2
Thus, the given triangle is not right-angled.
(v)
P = (a-1) cm, q = 2 βπ ππ πππ π = (π + 1)ππ
Then,
π2 + π2 = (π β 1)2 + (2βπ)2
= π2 + 1 β 2π + 4π
= π2 + 1 + 2π
= (π + 1)2
π2 = (π + 1)2
π2 + π2 = π2
Thus, the given triangle is right-angled.
2. A man goes 80m due east and then 150m due north. How far is he from the starting point?
Sol:
Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to c.
We need to find AC.
In right- angled triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
π΄πΆ = β802 + 1502
= β6400 + 22500
= β28900
= 170 m
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Hence, the man is 170 m away from the starting point.
3. A man goes 10m due south and then 24m due west. How far is he from the starting point?
Sol:
Let the man starts from point D and goes 10 m due south at E. He then goes 24 m due west
at F.
In right βDEF, we have:
DE = 10 m, EF = 24 m
π·πΉ2 = πΈπΉ2 + π·πΈ2
π·πΉ = β102 + 242
= β100 + 576
= β676
= 26 m
Hence, the man is 26 m away from the starting point.
4. A 13m long ladder reaches a window of a building 12m above the ground. Determine the
distance of the foot of the ladder from the building.
Sol:
Let AB and AC be the ladder and height of the building.
It is given that :
AB = 13 m and AC = 12 m
We need to find distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have:
π΄π΅2 = π΄πΆ2 + π΅πΆ2
βΉ π΅πΆ = β132 β 122
= β169 β 144
= β25
= 5 m
Hence, the distance of the foot ladder from the building is 5 m
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
5. A ladder is placed in such a way that its foot is at a distance of 15m from a wall and its top
reaches a window 20m above the ground. Find the length of the ladder.
Sol:
Let the height of the window from the ground and the distance of the foot of the ladder
from the wall be AB and BC, respectively.
We have :
AB = 20 m and BC = 15 m
Applying Pythagoras theorem in right-angled ABC, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
βΉ π΄πΆ = β202 + 152
= β400 + 225
= β625
= 25 m
Hence, the length of the ladder is 25 m.
6. Two vertical poles of height 9m and 14m stand on a plane ground. If the distance between
their feet is 12m, find the distance between their tops.
Sol:
Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m and AC = 5 m
We need to find AD, the distance between their tops.
Applying Pythagoras theorem in right-angled ACD, we have:
π΄π·2 = π΄πΆ2 + π·πΆ2
π΄π·2 = 52 + 122 = 25 + 144 = 169
π΄π· = β169 = 13 π
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Hence, the distance between the tops to the two poles is 13 m.
7. A guy wire attached to a vertical pole of height 18 m is 24m long and has a stake attached to
the other end. How far from the base of the pole should the stake be driven so that the wire
will be taut?
Sol:
Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let
it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
π΄π΅2 = π΅πΆ2 + πΆπ΄2
βΉ 242 = 182 + πΆπ΄2
βΉ πΆπ΄2 = 576 β 324
βΉ πΆπ΄2 = 252
βΉ πΆπ΄ = 6β7 π
Hence, the stake should be driven 6 β7π far from the base of the pole.
8. In the given figure, O is a point inside a βPQR such that β PQR such that β POR = 900, OP
= 6cm and OR = 8cm. If PQ = 24cm and QR = 26cm, prove that βPQR is right-angled.
Sol:
Applying Pythagoras theorem in right-angled triangle POR, we
have:
ππ 2 = ππ2 + ππ 2
βΉ ππ 2 = 62 + 82 = 36 + 64 = 100
βΉ ππ = β100 = 10 ππ
IN β PQR,
ππ2 + ππ 2 = 242 + 102 = 576 + 100 = 676
And ππ 2 = 262 = 676
β΄ ππ2 + ππ 2 = ππ 2
Therefore, by applying Pythagoras theorem, we can say that βPQR is right-angled at P.
9. βABC is an isosceles triangle with AB = AC = 13cm. The length of altitude from A on BC
is 5cm. Find BC.
Sol:
It is given that β ABC is an isosceles triangle.
Also, AB = AC = 13 cm
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.
AD =5 cm
β π΄π·π΅ πππ β π΄π·πΆ are right-angled triangles.
Applying Pythagoras theorem, we have;
π΄π΅2 = π΄π·2 + π΅π·2
π΅π·2 = π΄π΅2 β π΄π·2 = 132 β 52
π΅π·2 = 169 β 25 = 144
π΅π· = β144 = 12
Hence,
BC =2(BD) = 2 Γ 12 = 24 cm
10. Find the length of altitude AD of an isosceles βABC in which AB = AC = 2a units and BC
= a units.
Sol:
In isosceles β ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.
BD = BC = π
2 π’πππ‘π
Applying Pythagoras theorem in right-angled βABD, we have:
π΄π΅2 = π΄π·2 + π΅π·2
π΄π·2 = π΄π΅2 β π΅π·2 = (2π)2 β (π
2)
2
π΄π·2 = 4π2 βπ
4
2=
15π2
4
π΄π· = β15π2
4=
πβ15
2 π’πππ‘π .
11. βABC is am equilateral triangle of side 2a units. Find each of its altitudes.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Let AD, BE and CF be the altitudes of βABC meeting BC, AC and AB at D, E and F,
respectively.
Then, D, E and F are the midpoint of BC, AC and AB, respectively.
In right-angled βABD, we have:
AB = 2a and BD = a
Applying Pythagoras theorem, we get:
π΄π΅2 = π΄π·2 + π΅π·2
π΄π·2 = π΄π΅2 β π΅π·2 = (2π)2 β π2
π΄π·2 = 4π2 β π2 = 3π2
π΄π· = β3π π’πππ‘π
Similarly,
BE = πβ3 π’πππ‘π πππ πΆπΉ = πβ3 π’πππ‘π
12. Find the height of an equilateral triangle of side 12cm.
Sol:
Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then,
D will be the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
π΄π΅2 = π΄π·2 + π΅π·2
βΉ π΄π·2 = 122 β 62 (β΅ π΅π· =1
2 π΅πΆ = 6)
βΉ π΄π·2 = 144 β 36 = 108
βΉ π΄π· = β108 = 6β3 ππ.
Hence, the height of the given triangle is 6β3 ππ.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
13. Find the length of a diagonal of a rectangle whose adjacent sides are 30cm and 16cm.
Sol:
Let ABCD be the rectangle with diagonals AC and BD meeting at O.
According to the question:
AB = CD = 30 cm and BC = AD = 16 cm
Applying Pythagoras theorem in right-angled triangle ABC, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2 = 302 + 162 = 900 + 256 = 1156
π΄πΆ = β1156 = 34 ππ
Diagonals of a rectangle are equal.
Therefore, AC = BD = 34 cm
14. Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
Sol:
Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O.
We know that the diagonals of a rhombus bisect each other at angles.
Applying Pythagoras theorem in right-angled AOB, we get:
π΄π΅2 = π΄π2 + π΅π2 = 122 + 52
π΄π΅2 = 144 + 25 = 169
π΄π΅ = β169 = 13 ππ
Hence, the length of each side of the rhombus is 13 cm.
15. In βABC, D is the midpoint of BC and AEβ₯BC. If AC>AB, show that AB2 = π΄π·2 +1
4 π΅πΆ2 β π΅πΆ. π·πΈ
Sol:
In right-angled triangle AED, applying Pythagoras theorem, we have:
π΄π΅2 = π΄πΈ2 + πΈπ·2 β¦ (π)
In right-angled triangle AED, applying Pythagoras theorem, we have:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π΄π·2 = π΄πΈ2 + πΈπ·2
βΉ π΄πΈ2 = π΄π·2 β πΈπ·2 β¦ (ππ)
Therefore,
π΄π΅2 = π΄π·2 β πΈπ·2 + πΈπ΅2 ( ππππ (π)πππ (ππ))
π΄π΅2 = π΄π·2 β πΈπ·2 + ( π΅π· β π·πΈ)2
= π΄π·2 β πΈπ·2 + (1
2 π΅πΆ β π·πΈ)
2
= π΄π·2 β π·πΈ2 +1
4 π΅πΆ2 + π·πΈ2 β π΅πΆ. π·πΈ
= π΄π·2 +1
4 π΅πΆ2 β π΅πΆ. π·πΈ
This completes the proof.
16. In the given figure, 90ACB CD AB Prove that 2
2
BC BD
AC AD
Sol:
Given: β π΄πΆπ΅ = 900 πππ πΆπ· β₯ π΄π΅
To Prove; π΅πΆ2
π΄πΆ2 =π΅π·
π΄π·
Proof: In β ACB and β CDB
β π΄πΆπ΅ = β πΆπ·π΅ = 900 (πΊππ£ππ)
β π΄π΅πΆ = β πΆπ΅π· (πΆπππππ)
By AA similarity-criterion β ACB ~ βCDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides
are proportional.
β΄ π΅πΆ
π΅π·=
π΄π΅
π΅πΆ
βΉ π΅πΆ2 = π΅π·. π΄π΅ β¦ (1)
In β ACB and β ADC
β π΄πΆπ΅ = β π΄π·πΆ = 900 (πΊππ£ππ)
β πΆπ΄π΅ = β π·π΄πΆ (πΆπππππ)
By AA similarity-criterion β ACB ~ βADC
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
When two triangles are similar, then the ratios of their corresponding sides are
proportional.
β΄ π΄πΆ
π΄π·=
π΄π΅
π΄πΆ
βΉ π΄πΆ2 = AD. AB β¦.(2)
Dividing (2) by (1), we get
π΅πΆ2
π΄πΆ2 =π΅π·
π΄π·
17. In the given figure, D is the midpoint of side BC and AEβ₯BC. If BC = a, AC = b, AB = c,
AD = p and AE = h, prove that
(i) π2 = π2 + ππ₯ +π2
π₯
(ii) π2 = π2 β ππ₯ +π2
π₯
(iii) π2 + π2 = 2π2 +π2
2
(iv) π2 β π2 = 2ππ₯
Sol:
(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:
π΄πΆ2 = π΄πΈ2 + πΈπΆ2
βΉ π2 = β2 + ( π₯ +π
2)
2
= β2 + π₯2 +π2
4+ ππ₯ β¦ (π)
In right β angled triangle AED, we have:
π΄π·2 = π΄πΈ2 + πΈπ·2
βΉ π2 = β2 + π₯2 β¦ (ππ)
Therefore,
from (i) and (ii),
π2 = π2 + ππ₯ +π2
π₯
(ii)
In right-angled triangle AEB, applying Pythagoras, we have:
π΄π΅2 = π΄πΈ2 + πΈπ΅2
βΉ π2 = β2 + (π
2β π₯)
2
( β΅ π΅π· =π
2 πππ π΅πΈ = π΅π· β π₯)
βΉ π2 = β2 + π₯2 βπ2
4 ( β΅ β2 + π₯2 = π2)
βΉ π2 = π2 β ππ₯ +π2
π₯
(iii)
Adding (i) and (ii), we get:
βΉ π2 + π2 = π2 + ππ₯ +π2
4+ π2 β ππ₯ +
π2
4
= 2π2 + ππ₯ β ππ₯ +π2+π2
4
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
= 2π2 +π2
2
(iv)
Subtracting (ii) from (i), we get:
π2 β π2 = π2 + ππ₯ +π2
4β (π2 β ππ₯ +
π2
4)
= π2 β π2 + ππ₯ + ππ₯ +π2
4β
π2
4
= 2ax
18. In βABC, AB = AC. Side BC is produced to D. Prove that π΄π·2 β π΄πΆ2 = BD. CD
Sol:
Draw AEβ₯BC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:
Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is
also the median of the isosceles triangle.
So, BE = CE
And DE+CE=DE+BE=BD
π΄π·2 = π΄πΈ2 + π·πΈ2
βΉ π΄πΈ2 = π΄π·2 β π·πΈ2 β¦(i)
In βACE,
π΄πΆ2 = π΄πΈ2 + πΈπΆ2
βΉ π΄πΈ2 = π΄πΆ2 β πΈπΆ2 β¦ (ππ)
Using (i) and (ii),
βΉ π΄π·2 β π·πΈ2 = π΄πΆ2 β πΈπΆ2
βΉ π΄π·2 β π΄πΆ2 = π·πΈ2 β πΈπΆ2
= (DE + CE) (DE β CE)
= (DE + BE) CD
= BD.CD
19. ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are
constructed on sides AC and AB. Find the ratio between the areas of βABE and βACD.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Sol:
We have, ABC as an isosceles triangle, right angled at B.
Now, AB = BC
Applying Pythagoras theorem in right-angled triangle ABC, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2 = 2π΄π΅2 (β΅ π΄π΅ = π΄πΆ) β¦ (π)
β΅ βACD ~ β ABE
We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their
corresponding sides.
β΄ ππ(βπ΄π΅πΈ)
ππ(βπ΄πΆπ·)=
π΄π΅2
π΄πΆ2 =π΄π΅2
2π΄π΅2 [ππππ (π)]
= 1
2= 1 βΆ 2
20. An aeroplane leaves an airport and flies due north at a speed of 1000km per hour. At the
same time, another aeroplane leaves the same airport and flies due west at a speed of 1200
km per hour. How far apart will be the two planes after 1
12
hours?
Sol:
Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second
aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in 1 1
2 βππ’ππ = 1000 Γ
3
2= 1500 ππ
Distance covered by plane B in 1 1
2 βππ’ππ = 1200 Γ
3
2= 1800 ππ
Now, In right triangle ABC
By using Pythagoras theorem, we have
π΄π΅2 = π΅πΆ2 + πΆπ΄2
= (1800)2 + (1500)2
= 3240000 + 2250000
= 5490000
β΄ π΄π΅2 = 5490000
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ AB = 300 β61 π
Hence, the distance between two planes after 11
2 βππ’ππ ππ 300 β61π
21. In a ABC , AD is a median and AL BC .
Prove that
(a)
2
2 2 .2
BCAC AD BC DL
(b)
2
2 2 .2
BCAB AD BC DL
(c) 2 2 2 212
2AC AB AD BC
Sol:
(a) In right triangle ALD
Using Pythagoras theorem, we have
π΄πΆ2 = π΄πΏ2 + πΏπΆ2
= π΄π·2 β π·πΏ2 + (π·πΏ + π·πΆ)2 [Using (1)]
= π΄π·2 β π·πΏ2 + ( π·πΏ +π΅πΆ
2 )
2
[β΅ AD is a median]
= π΄π·2 β π·πΏ2 + π·πΏ2 + (π΅πΆ
2)
2
+ π΅πΆ. π·πΏ
β΄ π΄πΆ2 = π΄π·2 + π΅πΆ. π·πΏ + (π΅πΆ
2)
2
β¦. (2)
(b) In right triangle ALD
Using Pythagoras theorem, we have
π΄πΏ2 = π΄π·2 β π·πΏ2 β¦ (3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
π΄π΅2 = π΄πΏ2 + πΏπ΅2
= π΄π·2 β π·πΏ2 + πΏπ΅2 [ππ πππ (3)]
= π΄π·2 β π·πΏ2 + ( π΅π· β π·πΏ)2
= π΄π·2 π·πΏ2 + (1
2 π΅πΆ β π·πΏ )
2
= π΄π·2 β π·πΏ2 + (π΅πΆ
2)
2
β π΅πΆ. π·πΏ + π·πΏ2
β΄ π΄π΅2 = π΄π·2 β π΅πΆ. π·πΏ + (π΅πΆ
2)
2
β¦ . . (4)
(c) Adding (2) and (4), we get,
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
= π΄πΆ2 + π΄π΅2 = π΄π·2 + π΅πΆ. π·πΏ + (π΅πΆ
2)
2
+ π΄π·2 β π΅πΆ. π·πΏ + (π΅πΆ
2)
2
= 2 π΄π·2 +π΅πΆ2
4+
π΅πΆ2
4
= 2 π΄π·2 +1
2 π΅πΆ2
22. Naman is doing fly-fishing in a stream. The trip fishing rod is 1.8m above the surface of the
water and the fly at the end of the string rests on the water 3.6m away from him and 2.4m
from the point directly under the tip of the rod. Assuming that the string( from the tip of his
rod to the fly) is taut, how much string does he have out (see the adjoining figure) if he pulls
in the string at the rate of 5cm per second, what will be the horizontal distance of the fly from
him after 12 seconds?
23. Sol:
Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 Γ 5 = 60 cm or 0.6 m
Now, in βBMC
By using Pythagoras theorem, we have
π΅πΆ2 = πΆπ2 + ππ΅2
= (2.4)2 + (1.8)2
= 9
β΄ BC = 3 m
Now, BCβ = BC β 0.6
= 3 β 0.6
= 2.4 m
Now, In βBCβM
By using Pythagoras theorem, we have
πΆβ²π2 = π΅πΆβ²2β ππ΅2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
= (2.4)2 β (1.8)2
= 2.52
β΄ CβM = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
CβA = CβM + MA
= 1.6 + 1.2
= 2.8 m
Exercise β 4E
1. State the two properties which are necessary for given two triangles to be similar.
Sol:
The two triangles are similar if and only if
1. The corresponding sides are in proportion.
2. The corresponding angles are equal.
2. State the basic proportionality theorem.
Sol:
If a line is draw parallel to one side of a triangle intersect the other two sides, then it
divides the other two sides in the same ratio.
3. State and converse of Thaleβs theorem.
Sol:
If a line divides any two sides of a triangle in the same ratio, then the line must be parallel
to the third side.
4. State the midpoint theorem
Sol:
The line segment connecting the midpoints of two sides of a triangle is parallel to the third
side and is equal to one half of the third side.
5. State the AAA-similarity criterion
Sol:
If the corresponding angles of two triangles are equal, then their corresponding sides are
proportional and hence the triangles are similar.
6. State the AA-similarity criterion
Sol:
If two angles are correspondingly equal to the two angles of another triangle, then the two
triangles are similar.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
7. State the SSS-similarity criterion for similarity of triangles
Sol:
If the corresponding sides of two triangles are proportional then their corresponding angles
are equal, and hence the two triangles are similar.
8. State the SAS-similarity criterion
Sol:
If one angle of a triangle is equal to one angle of the other triangle and the sides including
these angles are proportional then the two triangles are similar.
9. State Pythagoras theorem
Sol:
The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here, the hypotenuse is the longest side and itβs always opposite the right angle.
10. State the converse of Pythagoras theorem.
Sol:
If the square of one side of a triangle is equal to the sum of the squares of the other two
sides, then the triangle is a right triangle.
11. If D, E, F are the respectively the midpoints of sides BC, CA and AB of βABC. Find the
ratio of the areas of βDEF and βABC.
Sol:
By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of
those sides is parallel to the third side and is half the length of the third side.
β΄ DF || BC
And π·πΉ =1
2 π΅πΆ
βΉ DF = BE
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, in βABC and βEFD
β π΄π΅πΆ = β πΈπΉπ· (πππππ ππ‘π ππππππ ππ π πππππππππππππ)
β π΅πΆπ΄ = β πΈπ·πΉ (πππππ ππ‘π ππππππ ππ π πππππππππππππ)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
By AA similarity criterion, βABC ~ βEFD
If two triangles are similar, then the ratio of their areas is equal to the squares of their
corresponding sides.
β΄ ππππ(βπ·πΈπΉ)
ππππ(βπ΄π΅πΆ)= (
π·πΉ
π΅πΆ)
2
= (π·πΉ
2 π·πΉ)
2
=1
4
Hence, the ratio of the areas of βDEF and βABC is 1 : 4.
12. Two triangles ABC and PQR are such that AB = 3 cm, AC = 6cm, β π΄ = 70Β°, PR = 9cm
β π = 70Β° and PQ = 4.5 cm. Show that ABC PQR and state that similarity criterion.
Sol:
Now, In βABC and βPQR
β π΄ = β π = 700 (πΊππ£ππ)
π΄π΅
ππ=
π΄πΆ
ππ [β΅
3
4.5=
6
9 βΉ
1
1.5=
1
1.5 ]
By SAS similarity criterion, βABC~ βPQR
13. In βABC~βDEF such that 2AB = DE and BC = 6cm, find EF.
Sol:
When two triangles are similar, then the ratios of the lengths of their corresponding sides
are equal.
Here, βABC ~βDEF
β΄ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ
βΉ π΄π΅
2 π΄π΅=
6
πΈπΉ
βΉ EF = 12 cm
14. In the given figure, DEβBC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and
EC = (3x + 19) cm. Find the value of x.
Sol:
In βADE and βABC
β π΄π·πΈ = β π΄π΅πΆ (πΆπππππ πππππππ ππππππ ππ π·πΈ β₯ π΅πΆ)
β π΄πΈπ· = β π΄πΆπ΅ (πΆπππππ πππππππ ππππππ ππ π·πΈ β₯ π΅πΆ
By AA similarity criterion, βADE ~ βABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
β΄ π΄π·
π΄π΅=
π΄πΈ
π΄πΆ
βΉ π΄π·
π΄π·+π·π΅=
π΄πΈ
π΄πΈ+πΈπΆ
βΉ π₯
π₯+3π₯+4=
π₯+3
π₯+3+3π₯+19
βΉ π₯
4π₯+4=
π₯+3
π₯+3+3π₯+19
βΉ π₯
2π₯+2=
π₯+3
2π₯+11
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ 2π₯2 + 11π₯ = 2π₯2 + 2π₯ + 6π₯ + 6
βΉ 3x = 6
βΉ x = 2
Hence, the value of x is 2.
15. A ladder 10m long reaches the window of a house 8m above the ground. Find the distance
of the foot of the ladder from the base of the wall.
Sol:
Let AB be A ladder and B is the window at 8 m above the ground C.
Now, In right triangle ABC
By using Pythagoras theorem, we have
π΄π΅2 = π΅πΆ2 + πΆπ΄2
βΉ 102 = 82 + πΆπ΄2
βΉ πΆπ΄2 = 100 β 64
βΉ πΆπ΄2 = 36
βΉ CA = 6m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
16. Find the length of the altitude of an equilateral triangle of side 2a cm.
Sol:
We know that the altitude of an equilateral triangle bisects the side on which it stands and
forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, it will bisects the side BC
β΄ DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
π΄πΆ2 = πΆπ·2 + π·π΄2
βΉ (2π)2 = π2 + π·π΄2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ π·π΄2 = 4π2 β π2
βΉ π·π΄2 = 3π2
βΉ DA = β3π
Hence, the length of the altitude of an equilateral triangle of side 2a cm is β3π ππ
17. βABC~βDEF such that ar(βABC) = 64 cm2 and ar(βDEF) = 169cm2. If BC = 4cm, find EF.
Sol:
We have β ABC ~ β DEF
If two triangles are similar then the ratio of their areas is equal to the ratio of the squares of
their corresponding sides.
β΄ ππππ(βπ΄π΅πΆ)
ππππ(βπ·πΈπΉ)= (
π΅πΆ
πΈπΉ)
2
βΉ 64
169= (
π΅πΆ
πΈπΉ)
2
βΉ (8
13 )
2
= (4
πΈπΉ)
2
βΉ 8
13=
4
πΈπΉ
βΉ EF = 6.5 cm
18. In a trapezium ABCD, it is given that ABβCD and AB = 2CD. Its diagonals AC and BD
intersect at the point O such that ar(βAOB) = 84cm2. Find ar(βCOD).
Sol:
In β AOB and COD
β π΄π΅π = β πΆπ·π (π΄ππ‘ππππ‘π ππππππ ππ π΄π΅ β₯ πΆπ·)
β π΄ππ΅ = β πΆππ· (ππππ‘ππππππ¦ πππππ ππ‘π ππππππ )
By AA similarity criterion, βAOB ~ βCOD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of
their corresponding sides.
β΄ ππππ(βπ΄ππ΅)
ππππ(βπΆππ·)= (
π΄π΅
πΆπ·)
2
βΉ 84
ππππ(βπΆππ·)= (
2 πΆπ·
πΆπ·)
2
βΉ ππππ (β πΆππ· ) = 12 ππ2
19. The corresponding sides of two similar triangles are in the ratio 2: 3. If the area of the smaller
triangle is 48cm2, find the area of the larger triangle.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
If two triangles are similar, then the ratio of their areas is equal to the squares of their
corresponding sides.
β΄ ππππ ππ π ππππππ π‘πππππππ
ππππ ππ πππππ π‘πππππππ= (
ππππ ππ π ππππππ π‘πππππππ
ππππ ππ ππππππ π‘πππππππ)
2
βΉ 48
ππππ ππ ππππππ π‘πππππππ= (
2
3)
2
βΉ ππππ ππ ππππππ π‘πππππππ = 108 ππ2
20. In an equilateral triangle with side a, prove that area = β3
4 π2.
Sol:
We know that the altitude of an equilateral triangle bisects the side on which it stands and
forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB =BC = CA = a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, It will bisects the side BC
β΄ π·πΆ =1
2 π
Now, In right triangle ADC
By using Pythagoras theorem, we have
π΄πΆ2 = πΆπ·2 + π·π΄2
βΉ π2 β (1
2π)
2
+ π·π΄2
βΉ π·π΄2 = π2 β1
4π2
βΉ π·π΄2 =3
4π2
βΉ π·π΄ =β3
2 π
πππ€, ππππ (βπ΄π΅πΆ) =1
2 Γπ΅πΆ Γπ΄π·
= 1
2 Γπ Γ
β3
2 π
= β3
4 π2
21. Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Suppose ABCD is a rhombus.
We know that the diagonals of a rhombus perpendicularly bisect each other.
β΄ β π΄ππ΅ = 900 , π΄π = 12 ππ πππ π΅π = 5 ππ
Now, In right triangle AOB
By using Pythagoras theorem we have
π΄π΅2 = π΄π2 + π΅π2
= 122 + 52
= 144 + 25
= 169
β΄ π΄π΅2 = 169
βΉ AB = 13 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 13 cm
22. Two triangles DEF an GHK are such that 48D and 57H . If DEF GHK then
find the measures of F
Sol:
If two triangles are similar then the corresponding angles of the two triangles are equal.
Here, βDEF ~ βGHK
β΄ β πΈ = β π» = 570
Now, In β DEF
β π· + β πΈ + β πΉ = 1800 (π΄ππππ π π’π πππππππ‘π¦ ππ π‘πππππππ)
βΉ β πΉ = 1800 β 480 β 570 = 750
23. In the given figure MN|| BC and AM: MB= 1: 2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Find
area AMN
area ABC
Sol:
We have
AM : MB = 1 : 2
βΉ ππ΅
π΄π=
2
1
Adding 1 to both sides, we get
βΉ ππ΅
π΄π+ 1 =
2
1+ 1
βΉ ππ΅+π΄π
π΄π=
2+1
1
βΉ π΄π΅
π΄π=
3
1
Now, In βAMN and βABC
β π΄ππ = β π΄π΅πΆ (πΆπππππ πππππππ ππππππ ππ ππ β₯ π΅πΆ)
β π΄ππ = β π΄πΆπ΅ (πΆπππππ πππππππ ππππππ ππ ππ β₯ π΅πΆ)
By AA similarity criterion, βAMN ~ β ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of
their corresponding sides.
β΄ ππππ(βπ΄ππ)
ππππ(βπ΄π΅πΆ)= (
π΄π
π΄π΅)
2
= (1
3)
2
=1
9
24. In triangle BMP and CNR it is given that PB= 5 cm, MP = 6cm BM = 9 cm and NR = 9cm.
If BMP CNR then find the perimeter of CNR
Sol:
When two triangles are similar, then the ratios of the lengths of their corresponding sides
are proportional.
Here, βBMP ~ βCNR
β΄ π΅π
πΆπ=
π΅π
πΆπ =
ππ
ππ . . (1)
πππ€,π΅π
πΆπ=
ππ
ππ [ππ πππ (1)]
βΉ πΆπ =π΅πΓππ
ππ=
9Γ9
6= 13.5 ππ
π΄ππππ,π΅π
πΆπ=
π΅π
πΆπ = [ππ πππ (1)]
βΉ πΆπ =π΅πΓπΆπ
π΅π=
5Γ13.5
9= 7.5 ππ
Perimeter of βCNR = CN + NR + CR = 13.5 + 9 + 7.5 = 30 cm
25. Each of the equal sides of an isosceles triangle is 25 cm. Find the length of its altitude if the
base is 14 cm.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
We know that the altitude drawn from the vertex opposite to the non-equal side bisects the
non-equal side.
Suppose ABC is an isosceles triangle having equal sides AB and BC.
So, the altitude drawn from the vertex will bisect the opposite side.
Now, In right triangle ABD
By using Pythagoras theorem, we have
π΄π΅2 = π΅π·2 + π·π΄2
βΉ 252 = 72 + π·π΄2
βΉ π·π΄2 = 625 β 49
βΉπ·π΄2 = 576
βΉ π·π΄ = 24 ππ
26. A man goes 12m due south and then 35m due west. How far is he from the starting point.
Sol:
In right triangle SOW
By using Pythagoras theorem, we have
ππ2 = ππ2 + ππ2
= 352 + 122
= 1225 + 144
= 1369
β΄ ππ2 = 1369
βΉ ππ = 37 π
Hence, the man is 37 m away from the starting point.
27. If the lengths of the sides BC, CA and AB of a ABC are a, b and c respectively and AD is
the bisector A then find the lengths of BD and DC
Sol:
Let DC = X
β΄ BD = a-X
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
By using angle bisector there in βABC, we have
π΄π΅
π΄πΆ=
π΅π·
π·πΆ
βΉ π
π=
πβπ₯
π₯
βΉ ππ₯ = ππ β ππ₯
βΉ π₯ (π + π) = ππ
βΉ π₯ =ππ
(π+π)
Now, π β π₯ = π βππ
π+π
= ππ+ππβππ
π+π
= ππ
π+π
28. In the given figure, 76AMN MBC . If p, q and r are the lengths of AM, MB and BC
respectively then express the length of MN of terms of P, q and r.
Sol:
In βAMN and βABC
β π΄ππ = β π΄π΅πΆ = 760 (πΊππ£ππ)
β π΄ = β π΄ (πΆπππππ)
By AA similarity criterion, βAMN ~ βABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
β΄ π΄π
π΄π΅=
ππ
π΅πΆ
βΉ π΄π
π΄π+ππ΅=
ππ
π΅πΆ
βΉ π
π+π=
ππ
π
βΉ ππ =ππ
π+π
29. Find the length of each side of a rhombus are 40 cm and 42 cm. find the length of each side
of the rhombus.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Suppose ABCD is a rhombus.
We know that the diagonals of a rhombus perpendicularly bisect each other.
β΄ β π΄ππ΅ = 900, π΄π = 20 ππ πππ π΅π = 21 ππ
Now, In right triangle AOB
By using Pythagoras theorem we have
π΄π΅2 = π΄π2 + ππ΅2
= 202 + 212
= 400 + 441
= 841
β΄ π΄π΅2 = 841
βΉ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm
30. For each of the following statements state whether true(T) or false (F)
(i) Two circles with different radii are similar.
(ii) any two rectangles are similar
(iii) if two triangles are similar then their corresponding angles are equal and their
corresponding sides are equal
(iv) The length of the line segment joining the midpoints of any two sides of a triangles is
equal to half the length of the third side.
(v) In a ABC , AB = 6 cm, 45A and AC = 8 cm and in a DEF , DF = 9 cm 45D
and DE= 12 cm then ABC DEF .
(vi) the polygon formed by joining the midpoints of the sides of a quadrilateral is a rhombus.
(vii) the ratio of the perimeter of two similar triangles is the same as the ratio of the their
corresponding medians.
(ix) if O is any point inside a rectangle ABCD then 2 2 2 2OA OC OB OD
(x) The sum of the squares on the sides of a rhombus is equal to the sum of the squares on
its diagonals.
Sol:
(i)
Two rectangles are similar if their corresponding sides are proportional.
(ii) True
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Two circles of any radii are similar to each other.
(iii)false
If two triangles are similar, their corresponding angles are equal and their corresponding
sides are proportional.
(iv) True
Suppose ABC is a triangle and M, N are
Construction: DE is expanded to F such that EF = DE
To proof = DE = 1
2π΅πΆ
Proof: In βADE and βCEF
AE = EC (E is the mid point of AC)
DE = EF (By construction)
AED = CEF (Vertically Opposite angle)
By SAS criterion, βADE ~= βCEF
CF = AD (CPCT)
βΉ BD = CF
β π΄π·πΈ = β πΈπΉπΆ (πΆππΆπ)
Since, β π΄π·πΈ πππ β πΈπΉπΆ πππ πππ‘πππππ‘π πππππ
Hence, AD β CF and BD β CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
β΄ DF = BC and BD β CF
β΄BDFC is a parallelogram
Hence, DF = BC
βΉ DE + EF = BC
βΉ π·πΈ =1
2 π΅πΆ
(v) False
In βABC, AB = 6 cm, β π΄ = 450 πππ π΄πΆ = 8 ππ πππ ππ βπ·πΈπΉ, π·πΉ = 9 ππ, β π· =
450 πππ π·πΈ = 12 ππ, π‘βππ βπ΄π΅πΆ ~ βπ·πΈπΉ.
In βABC and βDEF
(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a
parallelogram.
(vii) True
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Given: βABC ~ βDEF
To prove = π΄π(βπ΄π΅πΆ)
π΄π(βπ·πΈπΉ)= (
π΄π
π·π)
2
Proof: in βABP and βDEQ
β π΅π΄π = β πΈπ·π (π΄π β π΄ = β π·, π π π‘βπππ π»πππ ππ πππ π πππ’ππ)
β π΅ = β πΈ (β π΄π΅πΆ ~ βπ·πΈπΉ)
By AA criterion, βABP and βDEQ
π΄π΅
π·πΈ=
π΄π
π·π β¦.(1)
Since, βABC ~ βDEF
β΄ π΄π(βπ΄π΅πΆ)
π΄π(βπ·πΈπΉ)= (
π΄π΅
π·πΈ)
2
βΉ π΄π(βπ΄π΅πΆ)
π΄π(βπ·πΈπΉ)= (
π΄π
π·π)
2
[ππ πππ (1)]
(viii)
Given: βABC ~ βDEF
To Prove = πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΄π
π·π
Proof: In βABP and βDEQ
β π΅ = β πΈ (β΄ βπ΄π΅πΆ ~ βπ·πΈπΉ )
β΄ βABC ~ βDEF
β΄ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ
βΉ π΄π΅
π·πΈ=
2 π΅π
2 πΈπ
βΉ π΄π΅
π·πΈ=
π΅π
πΈπ
By SAS criterion, βABP ~ βDEQ
π΄π΅
π·πΈ=
π΄π
π·π β¦ . (1)
Since, βABC ~ βDEF
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
β΄ πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΄π΅
π·πΈ
βΉ πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΄π
π·π [ππ πππ (1)]
(ix) True
Suppose ABCD is a rectangle with O is any point inside it.
Construction: ππ΄2 + ππΆ2 = ππ΅2 + ππ·2
Proof:
ππ΄2 + ππΆ2 = (π΄π2 + ππ2) + (ππ2 + ππΆ2) [Using Pythagoras theorem in right
triangle AOP and COQ]
= (π΅π2 + ππ2) + (ππ2 + π·π2)
= (π΅π2 + ππ2) + (ππ2 + π·π2) [Using Pythagoras theorem in right triangle BOQ
and DOS]
= ππ΅2 + ππ·2
Hence, LHS = RHS
(x) True
Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
π΄π΅2 = (π΄πΆ
2)
2
+ (π΅π·
2)
2
[β΄ Diagonals of a rhombus perpendicularly bisect each other]
βΉ π΄π΅2 =π΄πΆ2
4+
π΅π·2
4
βΉ 4 π΄π΅2 = π΄πΆ2 + π΅π·2
βΉ π΄π΅2 + π΄π΅2 + π΄π΅2 + π΄π΅2 = π΄πΆ2 + π΅π·2
βΉ π΄π΅2 + π΅πΆ2 + πΆπ·2 + π·π΄2 = π΄πΆ2 + π΅π·2 [β΅ π΄ππ π ππππ ππ π πβππππ’π πππ πππ’ππ]
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Exercise β MCQ
1. A man goes 24m due west and then 10m due both. How far is he from the starting point?
(a) 34m (b) 17m (c) 26m (d) 28m
Sol:
(c) 26 m
Suppose, the man starts from point A and goes 24 m due west to point B. From here, he
goes 10 m due north and stops at C.
In right triangle ABC, we have:
AB = 24 m, BC = 10 m
Applying Pythagoras theorem, we get:
π΄πΆ2 = π΄π΅2 + π΅πΆ2 = 242 + 102
π΄πΆ2 = 576 + 100 = 676
π΄πΆ = β676 = 26
2. Two poles of height 13m and 7m respectively stand vertically on a plane ground at a distance
of 8m from each other. The distance between their tops is
(a) 9m (b) 10m (c) 11m (d) 12m
Sol:
(b) 10 m
Let AB and DE be the two poles.
According to the question:
AB = 13 m
DE = 7 m
Distance between their bottoms = BE = 8 m
Draw a perpendicular DC to AB from D, meeting AB at C. We get:
DC = 8m, AC = 6 m
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Applying, Pythagoras theorem in right-angled triangle ACD, we have
π΄π·2 = π·πΆ2 + π΄πΆ2
= 82 + 62 = 64 + 36 = 100
π΄π· = β100 = 10 π
3. A vertical stick 1.8m long casts a shadow 45cm long on the ground. At the same time, what
is the length of the shadow of a pole 6m high?
(a) 2.4m (b) 1.35m (c) 1.5m (d) 13.5m
Sol:
Let AB and AC be the vertical stick and its shadow, respectively.
According to the question:
AB = 1.8 m
AC = 45 cm = 0.45 m
Again, let DE and DF be the pole and its shadow, respectively.
According to the question:
DE = 6 m
DF =?
Now, in right-angled triangles ABC and DEF, we have:
β π΅π΄πΆ = β πΈπ·πΉ = 900
β π΄πΆπ΅ = β π·πΉπΈ (π΄πππ’πππ ππππ£ππ‘πππ ππ π‘βπ ππ’π ππ‘ π‘βπ π πππ π‘πππ)
Therefore, by AA similarity theorem, we get:
β ABC ~ βDFE
βΉ π΄π΅
π΄πΆ=
π·πΈ
π·πΉ
βΉ 1.8
0.45=
6
π·πΉ
βΉ π·πΉ =6Γ0.45
1.8= 1.5 π
4. A vertical pole 6m long casts a shadow of length 3.6m on the ground. What is the height of
a tower which casts a shadow of length 18m at the same time?
(a) 10.8m (b) 28.8m (c) 32.4m (d) 30m
Sol:
(d)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Let AB and AC be the vertical pole and its shadow, respectively.
According to the question:
AB = 6 m
AC = 3.6 m
Again, let DE and DF be the tower and its shadow.
According to the question:
DF = 18 m
DE =?
Now, in right -angled triangles ABC and DEF, we have:
β π΅π΄πΆ = β πΈπ·πΉ = 900
β π΄π΅πΆ = β π·πΉπΈ = (π΄πππ’πππ ππππ£ππ‘πππ ππ π‘βπ π π’π ππ‘ π‘βπ π πππ π‘πππ)
Therefore, by AA similarity theorem, we get:
β ABC - β DEF
βΉ π΄π΅
π΄πΆ=
π·πΈ
π·πΉ
βΉ 6
3.6=
π·πΈ
18
βΉ π·πΈ =6Γ18
3.6= 30 π
5. The shadow of a 5m long stick is 2m long. At the same time the length of the shadow of a
12.5m high tree(in m) is
(a) 3.0 (b) 3.5 (c) 4.5 (d) 5.0
Sol:
Suppose DE is a 5 m long stick and BC is a 12.5 m high tree.
Suppose DA and BA are the shadows of DE and BC respectively.
Now, In βABC and βADE
β π΄π΅πΆ = β π΄π·πΈ = 900
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
β π΄ = β π΄ (πΆπππππ)
By AA- similarity criterion
βABC~ βADE
If two triangles are similar, then the ratio of their corresponding sides are equal.
β΄ π΄π΅
π΄π·=
π΅πΆ
π·πΈ
βΉ π΄π΅
2=
12.5
5
βΉ AB = 5 cm
Hence, the correct answer is option (d).
6. A ladder 25m long just reaches the top of a building 24m high from the ground. What is the
distance of the foot of the ladder from the building?
(a) 7m (b) 14m (c) 21m (d) 24.5m
Sol:
(a) 7 m
Let the ladder BC reaches the building at C.
Let the height of building where the ladder reaches be AC.
According to the question:
BC = 25 m
AC = 24 m
In right-angled triangle CAB, we apply Pythagoras theorem to find the value of AB.
π΅πΆ2 = π΄πΆ2 + π΄π΅2
βΉ π΄π΅2 = π΅πΆ2 β π΄πΆ2 = 252 β 242
βΉ π΄π΅2 = 625 β 576 = 49
βΉ π΄π΅ = β49 = 7 π
7. In the given figure, O is the point inside a MNP such that 90MOP OM = 16 cm and
OP = 12 cm if MN = 21cm and 90NMP then NP= ?
Sol:
Now, In right triangle MOP
By using Pythagoras theorem, we have
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
ππ2 = ππ2 + ππ2
= 122 + 162
= 144 + 256
= 400
β΄ ππ2 = 400
βΉ MO = 20 cm
Now, In right triangle MPN
By using Pythagoras theorem, we have
ππ2 = ππ2 + ππ2
= 212 + 202
= 441 + 400
= 841
β΄ ππ2 = 841
βΉ MP = 29 cm
Hence, the correct answer is option (b).
8. The hypotenuse of a right triangle is 25cm. The other two sides are such that one is 5cm
longer than the other. The lengths of these sides are
(a) 10cm, 15cm (b) 15cm, 20cm
(c) 12cm, 17cm (d) 13cm, 18cm
Sol:
(b) 15 cm, 20 cm
It is given that length of hypotenuse is 25 cm.
Let the other two sides be x cm and (xβ5) cm.
Applying Pythagoras theorem, we get:
252 = π₯2 + (π₯ β 5 )2
βΉ 625 = π₯2 + π₯2 + 25 β 10π₯
βΉ 2π₯2 β 10π₯ β 600 = 0
βΉ π₯2 β 5π₯ β 300 = 0
βΉ π₯2 β 20π₯ + 15π₯ β 300 = 0
βΉ π₯(π₯ β 20) + 15(π₯ β 20) = 0
βΉ (x β 20) (x + 15) =0
βΉ π₯ β 20 = 0 ππ π₯ + 15 = 0
βΉ π₯ = 20 ππ π₯ = β15
Side of a triangle cannot be negative.
Therefore, x = 20 cm
Now,
x β 5 = 20 β 5 = 15 cm
9. The height of an equilateral triangle having each side 12cm, is
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
(a) 6β2 cm (b) 6β3m (c) 3β6m (d) 6β6m
Sol:
(b) 6β3ππ
Let ABC be the equilateral triangle with AD as its altitude from A.
In right-angled triangle ABD, we have
π΄π΅2 = π΄π·2 + π΅π·2
π΄π·2 = π΄π΅2 β π΅π·2
= 122 β 62
= 144 β 36 = 108
AD = β108 = 6β3ππ
10. βABC is an isosceles triangle with AB = AC = 13cm and the length of altitude from A on
BC is 5cm. Then, BC = ?
(a) 12cm (b) 16cm (c) 18cm (d) 24cm
Sol:
(d) 24 cm
In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
π΄π΅2 = π΄π·2 + π΅π·2
βΉ π΅π·2 = π΄π΅2 β π΄π·2
βΉ π΅π·2 = 132 β 52
βΉ π΅π·2 = 169 β 25
βΉ π΅π·2 = 144
βΉ π΅π· = β144 = 12 ππ
Therefore, BC = 2BD = 24 cm
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
11. In a βABC, it is given that AB = 6cm, AC = 8cm and AD is the bisector of β A. Then, BD :
DC = ?
(a) 3 : 4 (b) 9 : 16 (c) 4 : 3 (d) β3 : 2
Sol:
(a) 3 : 4
In β ABD and βACD, we have:
β π΅π΄π· = β πΆπ΄π·
Now,
π΅π·
π·πΆ=
π΄π΅
π΄πΆ=
6
8=
3
4
BD : DC = 3 : 4
12. In βABC, it is given that AD is the internal bisector of β A. If BD = 4cm, DC = 5cm and AB
= 6cm, then AC = ?
(a) 4.5cm (b) 8cm (c) 9cm (d) 7.5cm
Sol:
(d) 7.5 cm
It is given that AD bisects angle A
Therefore, applying angle bisector theorem, we get:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
βΉ 4
5=
6
π₯
βΉ π₯ =5 Γ6
4= 7.5
Hence, AC = 7.5 cm
13. In a βABC, it is given that AD is the internal bisector of β A. If AB = 10cm, AC = 14cm and
BC = 6cm, then CD = ?
(a) 4.8cm (b) 3.5cm (c) 7cm (d) 10.5cm
Sol:
By using angle bisector in βABC, we have
π΄π΅
π΄πΆ=
π΅π·
π·πΆ
βΉ 10
14=
6βπ₯
π₯
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ 10x = 84 β 14x
βΉ 24x = 84
βΉ x = 3.5
Hence, the correct answer is option (b).
14. In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is
(a) right-angled (b) isosceles
(c) scalene (d) obtuse-angled
Sol:
(b) Isosceles
In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.
15. In an equilateral triangle ABC, if AD β₯ BC, then which of the following is true?
(a) 2AB2 = 3AD2 (b) 4AB2 = 3AD2
(c) 3AB2 = 4AD2 (d) 3AB2 = 2AD2
Sol:
(c) 3π΄π΅2 = 4π΄π·2
Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:
π΄π΅2 = π΄π·2 + π΅π·2
βΉ π΄π΅2 = (1
2 π΄π΅)
2
+ π΄π·2 (β΅ βπ΄π΅πΆ ππ πππ’ππππ‘ππππ πππ π΄π· =1
2 π΄π΅)
βΉ π΄π΅2 =1
4 π΄π΅2 + π΄π·2
βΉ π΄π΅2 β1
4 π΄π΅2 = π΄π·2
βΉ 3
4 π΄π΅2 = π΄π·2
βΉ 3π΄π΅2 = 4π΄π·2
16. In a rhombus of side 10cm, one of the diagonals is 12cm long. The length of the second
diagonal is
(a) 20cm (b) 18cm (c) 16cm (d) 22cm
Sol:
(c) 16 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
Also, diagonals of a rhombus bisect each other at right angles.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
If AC = 12 cm, AO = 6 cm
Applying Pythagoras theorem in right-angled triangle AOB. We get:
π΄π΅2 = π΄π2 + π΅π2
βΉ π΅π2 = π΄π΅2 β π΄π2
βΉ π΅π2 = 102 β 62 = 100 β 36 = 64
βΉ π΅π = β64 = 8
βΉ π΅π· = 2 Γπ΅π = 2 Γ8 = 16 ππ
Hence, the length of the second diagonal BD is 16 cm.
17. The lengths of the diagonals of a rhombus are 24cm and 10cm. The length of each side of
the rhombus is
(a) 12cm (b) 13cm (c) 14cm (d) 17cm
Sol:
(b) 13 cm
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We have:
AC = 24 cm and BD = 10 cm
We know that diagonals of a rhombus bisect each other at right angles.
Therefore applying Pythagoras theorem in right-angled triangle AOB, we get:
π΄π΅2 = π΄π2 + π΅π2 = 122 + 52
= 144 + 25 = 169
π΄π΅ = β169 = 13
Hence, the length of each side of the rhombus is 13 cm.
18. If the diagonals of a quadrilateral divide each other proportionally, then it is a
(a) parallelogram (b) trapezium
(c) rectangle (d) square
Sol:
(b) trapezium
Diagonals of a trapezium divide each other proportionally.
19. The line segments joining the midpoints of the adjacent sides of a quadrilateral form
(a) parallelogram (b) trapezium
(c) rectangle (d) square
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
(a) A parallelogram
The line segments joining the midpoints of the adjacent sides of a quadrilateral form a
parallelogram.
20. If the bisector of an angle of a triangle bisects the opposite side, then the triangle is
(a) scalene (b) equilateral
(c) isosceles (d) right-angled
Sol:
(c) isosceles
Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
π΄π΅
π΄πΆ=
π΅π·
π·πΆ
It is given that AD bisects BC.
Therefore, BD = DC
βΉ π΄π΅
π΄πΆ= 1
βΉ AB = AC
Therefore, the triangle is isosceles.
21. In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such
that OA = (3x β 1) cm, OB = (2x + 1)cm, OC = (5x β 3)cm and OD = (6x β 5)cm. Then,
x = ?
(a) 2 (b) 3 (c) 2.5 (d) 4
Sol:
(a) 2
We know that the diagonals of a trapezium are proportional.
Therefore ππ΄
ππΆ=
ππ΅
ππ·
βΉ 3π₯β1
5π₯β3=
2π₯+1
6π₯β5
βΉ (3X β 1) (6X β 5) = (2X + 1) (5X -3)
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ 18π2 β 15π β 6π + 5 = 10π2 β 6π + 5π β 3
βΉ 18π2 β 21π + 5 = 10π2 β π β 3
βΉ 18π2 β 21π + 5 β 10π2 + π + 3 = 0
βΉ 8π2 β 20π + 8 = 0
βΉ 4 (2π2 β 5π + 2) = 0
βΉ 2π2 β 5π + 2 = 0
βΉ 2π2 β 4π β π + 2 = 0
βΉ 2π(π β 2) β 1 (π β 2) = 0
βΉ (π β 2)(2π β 1) = 0
βΉ πΈππ‘βππ π₯ β 2 = 0 ππ 2π₯ β 1 = 0
βΉ πΈππ‘βππ π₯ = 2 ππ π₯ =1
2
πβππ π₯ =1
2, 6π₯ β 5 = β2 < 0, π€βππβ ππ πππ‘ πππ π ππππ.
Therefore, x = 2
22. In βABC, it is given that π΄π΅
π΄πΆ=
π΅π·
π·πΆ. If β B = 700 and β C = 500, then β BAD = ?
(a) 300 (b) 400 (c) 450 (d) 500
Sol:
(a) 300
We have:
π΄π΅
π΄πΆ=
π΅π·
π·πΆ
Applying angle bisector theorem, we can conclude that AD bisects β π΄.
In βABC,
β π΄ + β π΅ + β πΆ = 1800
βΉ β π΄ = 180 β β π΅ β β πΆ
βΉ β π΄ = 180 β 70 β 50 = 600
β΅ β π΅π΄π· = β πΆπ΄π· =1
2 β π΅π΄πΆ
β΄ β π΅π΄π· =1
2 Γ60 = 300
23. In βABC, DEβBC so that AD = 2.4cm, AE = 3.2cm and EC = 4.8cm. Then, AB = ?
(a) 3.6cm (b) 6cm (c) 6.4cm (d) 7.2cm
Sol:
(b) 6 cm
It is given that DE || BC.
Applying basic proportionality theorem, we have:
π΄π·
π΅π·=
π΄πΈ
πΈπΆ
βΉ 2.4
π΅π·=
3.2
4.8
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ π΅π· =2.4Γ4.8
3.2= 3.6 ππ
Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm
24. In a βABC, if DE is drawn parallel to BC, cutting AB and AC at D and E respectively such
that AB = 7.2cm, AC = 6.4cm and AD = 4.5cm. Then, AE = ?
(a) 5.4cm (b) 4cm (c) 3.6cm (d) 3.2cm
Sol:
(b) 4cm
It is given that DE || BC.
Applying basic proportionality theorem, we get:
π΄π·
π΄π΅=
π΄πΈ
π΄πΆ
βΉ 4.5
7.2=
π΄πΈ
6.4
βΉ π΄πΈ =4.5Γ6.4
7.2= 4 ππ
25. In βABC, DE β BC so that AD = (7x β 4)cm, AE = (5x β 2)cm, DB = (3x + 4)cm and EC =
3x cm. Then, we have:
(a) x = 3 (b) x = 5 (c) x = 4 (d) x = 2.5
Sol:
(c) x = 4
It is given DE || BC.
Applying Thalesβ theorem. We get:
π΄π·
π΅π·=
π΄πΈ
πΈπΆ
βΉ 7π₯β4
3π₯+4=
5π₯β2
3π₯
βΉ 3π₯(7π₯ β 4) = (5π₯ β 2)(3π₯ + 4)
βΉ 21π₯2 β 12π₯ = 15π₯2 + 20π₯ β 6π₯ β 8
βΉ 21π₯2 β 12π₯ = 15π₯2 + 14π₯ β 8
βΉ 6π₯2 β 26π₯ + 8 = 0
βΉ 2 (3π₯2 β 13π₯ + 4) = 0
βΉ 3π₯2 β 13π₯ + 4 = 0
βΉ 3π₯2 β 12π₯ β π₯ + 4 = 0
βΉ 3π₯ (π₯ β 4) β 1 (π₯ β 4) = 0
βΉ (π₯ β 4)(3π₯ β 1) = 0
βΉ π₯ β 4 = 0 ππ 3π₯ β 1 = 0
βΉ π₯ β 4 ππ π₯ =1
3
If π₯ =1
3 , 7π₯ β 4 = β
5
3 < 0 ; ππ‘ ππ πππ‘ πππ π ππππ.
Therefore, x = 4
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
26. In βABC, DE β BC such that π΄π·
π·π΅=
3
5. If AC = 5.6cm, then AE = ?
(a) 4.2cm (b) 3.1cm (c) 2.8cm (d) 2.1cm
Sol:
(d) 2.1 cm
It is given that DE || BC.
Applying Thalesβ theorem, we get:
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
Let AE be x cm.
Therefore, EC = (5.6 β x) cm
βΉ 3
5=
π₯
5.6βπ₯
βΉ 3(5.6 β π₯) = 5π₯
βΉ 16.8 β 3π₯ = 5π₯
βΉ 8π₯ = 16.8
βΉ x = 2.1 cm
27. βABC~βDEF and the perimeters of βABC and βDEF are 30cm and 18cm respectively. If
BC = 9cm, then EF = ?
(a) 6.3cm (b) 5.4cm (c) 7.2cm (d) 4.5cm
Sol:
(b) 5.4 cm
βABC ~ βDEF
Therefore,
πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΅πΆ
πΈπΉ
βΉ 30
18=
9
πΈπΉ
βΉ πΈπΉ =9Γ18
30= 5.4 ππ
28. βABC~βDEF such that AB = 9.1cm and DE = 6.5cm. If the perimeter of βDEF is 25cm,
what is the perimeter of βABC?
(a) 35cm (b) 28cm (c) 42cm (d) 40cm
Sol:
(a) 35 cm
β΅βABC ~ βDEF
β΄ πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΄π΅
π·πΈ
βΉ πππππππ‘ππ(βπ΄π΅πΆ)
25=
9.1
6.5
βΉ πππππππ‘ππ (βπ΄π΅πΆ) =9.1Γ25
6.5= 35 ππ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
29. In βABC, it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, βDEF is given such
that EF = 8cm and βDEF~βABC. Then, perimeter of βDEF is
(a) 22.5cm (b) 25cm (c) 27cm (d) 30cm
Sol:
(d) 30 cm
Perimeter of βABC = AB + BC + CA = 9 + 6 + 7.5 = 22.5 cm
β΅ βDEF ~ βABC
β΄ πππππππ‘ππβ(π΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΅πΆ
πΈπΉ
βΉ 22.5
πππππππ‘ππ(βπ·πΈπΉ)=
6
8
Perimeter(βDEF) = 22.5Γ8
6= 30 ππ
30. ABC and BDE are two equilateral triangles such that D is the midpoint of BC. Ratio of these
area of triangles ABC and BDE is
(a) 2 : 1 (b) 1 : 4 (c) 1 : 2 (d) 4 : 1
Sol:
Give: ABC and BDE are two equilateral triangles
Since, D is the midpoint of BC and BDE is also an equilateral triangle.
Hence, E is also the midpoint of AB.
Now, D and E are the midpoint of BC and AB.
In a triangle, the line segment that joins midpoint of the two sides of a triangle is parallel to
the third side and is half of it.
π·πΈ Η πΆπ΄ πππ π·πΈ =1
2 πΆπ΄
Now, in βABC and βEBD
β π΅πΈπ· = β π΅π΄πΆ (πΆπππππ πππππππ ππππππ )
β π΅ = β π΅ (πΆπππππ)
By AA-similarity criterion
βABC ~ βEBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of
their corresponding sides.
β΄ ππππ(βπ΄π΅πΆ)
ππππ(βπ·π΅πΈ)= (
π΄πΆ
πΈπ·)
2
= (2πΈπ·
πΈπ·)
2
=4
1
Hence, the correct answer is option (d).
31. It is given that βABC~βDEF. If β A = 300, β C = 500, AB = 5cm, AC = 8cm and DF = 7.5cm,
then which of the following is true?
(a) DE = 12cm, β F = 500 (b) DE = 12cm, β F = 1000
(c) DE = 12cm, β D = 1000 (d) EF = 12cm, β D = 300
Sol:
(b) DE = 12 cm, β πΉ = 1000
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Disclaimer: In the question, it should be βABC ~ βDFE instead of βABC ~ βDEF.
In triangle ABC,
β π΄ + β π΅ + β πΆ = 1800
β΄ β π΅ = 180 β 30 β 50 = 1000
β΅ βABC ~ βDFE
β΄ β π· = β π΄ = 300
β πΉ = β π΅ = 1000
And β πΈ = β πΆ = 500
Also,
π΄π΅
π·πΉ=
π΄πΆ
π·πΈ βΉ
5
7.5=
8
π·πΈ
βΉ π·πΈ =8Γ7.5
5= 12 ππ
32. In the given figure, β BAC = 900 and ADβ₯BC. Then,
(a) BC.CD = BC2
(b) AB.AC = BC2
(c) BD.CD = AD2
(d) AB.AC = AD2
Sol:
(c) BD . CD = π΄π·2
In β BDA and βADC, we have:
β π΅π·π΄ = β π΄π·πΆ = 900
β π΄π΅π· = 900 β β π·π΄π΅
= 900 β (900 β β π·π΄πΆ)
= 900 β 900 + β π·π΄πΆ
= β π·π΄πΆ
Applying AA similarity, we conclude that βπ΅π·π΄ β βπ΄π·πΆ.
βΉ π΅π·
π΄π·=
π΄π·
πΆπ·
βΉ π΄π·2 = π΅π·. πΆπ·
33. In ABC , 6 3AB , AC = 12 cm and BC = 6cm. Then B is
Sol:
π΄π΅ = 6β3ππ
βΉ π΄π΅2 = 108 ππ2
AC = 12 cm
βΉ π΄πΆ2 = 144 ππ2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
BC = 6 cm
βΉ π΅πΆ2 = 36 ππ
β΄ π΄πΆ2 = π΄π΅2 + π΅πΆ2
Since, the square of the longest side is equal to the sum of two sides, so βABC is a right
angled triangle.
β΄ The angle opposite to β 90Β°
Hence, the correct answer is option (c)
34. In βABC and βDEF, it is given that π΄π΅
π·πΈ=
π΅πΆ
πΉπ·, then
(a) β B = β E (b) β A = β D (c) β B = β D (d) β A = β F
Sol:
(c)β π΅ = β π·
Disclaimer: In the question, the ratio should be π΄π΅
π·πΈ=
π΅πΆ
πΉπ·=
π΄πΆ
πΈπΉ .
We can write it as:
π΄π΅
πΈπ·=
π΅πΆ
π·πΉ=
π΄πΆ
πΉπΈ
Therefore, β ABC β EDF
Hence, the corresponding angles, i.e.,β π΅ πππ β π·, π€πππ ππ πππ’ππ.
π. π. , β π΅ = β π·
35. In βDEF and βPQR, it is given that β D = β Q and β R = β E, then which of the following is
not true?
(a) πΈπΉ
ππ =
π·πΉ
ππ (b)
π·πΈ
ππ=
πΈπΉ
π π (c)
π·πΈ
ππ =
π·πΉ
ππ (d)
πΈπΉ
π π=
π·πΈ
ππ
Sol:
(b) π·πΈ
ππ=
πΈπΉ
π π
In βDEF and βPQR, we have:
β π· = β π πππ β π = β πΈ
Applying AA similarity theorem, we conclude that βDEF ~ βQRP.
π»ππππ,π·πΈ
ππ =
π·πΉ
ππ=
πΈπΉ
ππ
36. If βABC~βEDF and βABC is not similar to βDEF, then which of the following is not true?
(a) BC.EF = AC.FD
(b) AB.EF = AC.DE
(c) BC.DE = AB.EF
(d) BC.DE = AB.FD
Sol:
(c) BC. DE = AB. EF
βABC ~ βEDF
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Therefore,
π΄π΅
π·πΈ=
π΄πΆ
πΈπΉ=
π΅πΆ
π·πΉ
βΉ BC. DE β AB. EF
37. In βABC and βDEF, it is given that β B = β E, β F = β C and AB = 3DE, then the two triangles
are
(a) congruent but not similar (b) similar but not congruent
(c) neither congruent nor similar (d) similar as well as congruent
Sol:
(b) similar but not congruent
In βABC and βDEF, we have:
β π΅ = β πΈ πππ β πΉ = β πΆ
Applying AA similarity theorem, we conclude that βABC - βDEF.
Also,
AB = 3DE
βΉ AB β DE
Therefore, βABC and βDEF are not congruent.
38. If in βABC and βPQR, we have: π΄π΅
ππ =
π΅πΆ
ππ =
πΆπ΄
ππ, then
(a) βPQR ~ βCAB (b) βPQR ~ βABC
(c) βCBA ~ βPQR (d) βBCA ~ βPQR
Sol:
(a) βPQR ~ βCAB
In βABC and βPQR, we have:
π΄π΅
ππ =
π΅πΆ
ππ =
πΆπ΄
ππ
βΉ βABC ~ βQRP
We can also write it as βPQR ~ βCAB.
39. In the given figure, two line segment AC and BD intersect each other at the point P such that
PA = 6cm, PB = 3cm, PC = 2.5cm, PD = 5cm, β APB = 500 and β CDP = 300, then β PBA =
?
(a) 500 (b) 300
(c) 600 (b) 1000
Sol:
(d)1000
In β APB and β DPC, we have:
β π΄ππ΅ = β π·ππΆ = 500
π΄π
π΅π=
6
3= 2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π·π
πΆπ=
5
2.5= 2
Hence, π΄π
π΅π=
π·π
πΆπ
Applying SAS theorem, we conclude that β APB- β DPC.
β΄ β ππ΅π΄ = β ππΆπ·
In β DPC, we have:
β πΆπ·π + β πΆππ· + β ππΆπ· = 1800
βΉβ ππΆπ· = 1800 β β πΆπ·π β β πΆππ·
βΉ β ππΆπ· = 1800 β 300 β 500
βΉβ ππΆπ· = 1000
Therefore, β ππ΅π΄ = 1000
40. Corresponding sides of two similar triangles are in the ratio 4:9 Areas of these triangles are
in the ration
(a) 2:3 (b) 4:9 (c) 9:4 (d) 16:81
Sol:
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of
their corresponding sides.
β΄ ππππ ππ ππππ π‘ π‘πππππππ
ππππ ππ π πππππ π‘πππππππ= (
ππππ ππ ππππ π‘ π‘πππππππ
ππππ ππ π πππππ π‘πππππππ)
2
= (4
9)
2
=16
81
Hence, the correct answer is option (d).
41. It is given that βABC~βPQR and π΅πΆ
ππ =
2
3, then
ππ(βπππ )
ππ(βπ΄π΅πΆ)= ?
(a) 2
3 (b)
3
2 (c)
4
9 (d)
9
4
Sol:
(d) 9:4
It is given that βABC ~ βPQR and π΅πΆ
ππ =
2
3
Therefore,
ππ(βπππ )
ππ(βπ΄π΅πΆ)=
ππ 2
π΅πΆ2 = (3
2)
2
=9
4
42. In an equilateral βABC, D is the midpoint of AB and E is the midpoint of AC. Then,
ar(βABC) : ar(βADE) = ?
(a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 4
Sol:
(b) 4:1
In βABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem,
Also, by Basic Proportionality Theorem,
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
Also,
AB = AC = BC (β΅βABC is an equilateral triangle)
So, π΄π·
π·π΅=
π΄πΈ
πΈπΆ= 1
In βABC and βADE, we have:
β π΄ = β π΄
π΄π·
π΄π΅=
π΄πΈ
π΄πΆ=
1
2
β΄ β ABC - βADE (SAS criterion)
β΄ ππ (βπ΄π΅πΆ) βΆ ππ (βπ΄π·πΈ) = (π΄π΅)2 βΆ (π΄π·)2
βΉ ππ (βπ΄π΅πΆ) βΆ ππ (βπ΄π·πΈ) = 22 βΆ 12
βΉ ππ(βπ΄π΅πΆ): ππ (βπ΄π·πΈ) = 4: 1
43. In βABC and βDEF, we have: π΄π΅
π·πΈ =
π΅πΆ
πΈπΉ =
π΄πΆ
π·πΉ =
5
7, then ar(βABC) : β(DEF) = ?
(a) 5 : 7 (b) 25 : 49 (c) 49 : 25 (d) 125 : 343
Sol:
(b) 25 :49
In βABC and βDEF, we have :
π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ=
π΄πΆ
π·πΈ=
5
7
Therefore, by SSS criterion, we conclude that βABC ~ βDEF.
βΉ ππ(βπ΄π΅πΆ)
ππ(βπ·πΈπΉ)=
π΄π΅2
π·πΈ2 = (5
7)
2
=25
49= 25 βΆ 49
44. βABC~βDEF such that ar(βABC) = 36cm2 and ar(βDEF) = 49cm2. Then, the ratio of their
corresponding sides is
(a) 36 : 49 (b) 6 : 7 (c) 7 : 6 (d) β6 : β7
Sol:
(b) 6:7
β΅ βABC ~ βDEF
β΄ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ=
π΄πΆ
π·πΉ β¦ (π)
Also,
ππ(βπ΄π΅πΆ)
ππ(βπ·πΈπΉ)=
π΄π΅2
π·πΈ2
βΉ 36
49=
π΄π΅2
π·πΈ2
βΉ 6
7=
π΄π΅
π·πΈ
βΉ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ=
π΄πΆ
π·πΉ=
6
7 (ππππ (π))
Thus, the ratio of corresponding sides is 6 : 7.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
45. Two isosceles triangles have their corresponding angles equal and their areas are in the ratio
25: 36. The ratio of their corresponding heights is
(a) 25 : 36 (b) 36 : 25 (c) 5 : 6 (d) 6: 5
Sol:
(c) 5:6
Let x and y be the corresponding heights of the two triangles.
It is given that the corresponding angles of the triangles are equal.
Therefore, the triangles are similar. (By AA criterion)
Hence,
ππ(β1)
ππ(β2)=
25
36=
π₯2
π¦2
βπ₯2
π¦2 =25
36
βΉ π₯2
π¦2 = β25
36=
5
6= 5 βΆ 6
46. The line segments joining the midpoints of the sides of a triangle form four triangles, each
of which is
(a) congruent to the original triangle
(b) similar to the original triangle
(c) an isosceles triangle
(d) an equilateral triangle
Sol:
(b) similar to the original triangle
The line segments joining the midpoint of the sides of a triangle form four triangles, each
of which is similar to the original triangle.
47. If βABC~βQRP, ππ(βπ΄π΅πΆ)
ππ(βππ π)=
9
4, AB = 18cm and BC = 15cm, then PR = ?
(a) 18cm (b) 10cm
(c) 12 cm (d) 20
3 cm
Sol:
(b) 10 cm
β΅ βABC ~ βQRP
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
β΄ π΄π΅
ππ =
π΅πΆ
ππ
Now,
ππ(βπ΄π΅πΆ)
ππ(βππ π)=
9
4
βΉ (π΄π΅
ππ )
2
=9
4
βΉ π΄π΅
ππ =
π΅πΆ
ππ =
3
2
Hence, 3PR = 2BC = 2 Γ 15 = 30
PR = 10 cm
48. In the given figure, O is the point of intersection of two chords AB and CD such that
OB = OD and β AOC = 450. Then, βOAC and βODB are
(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar
Sol:
(c) isosceles and similar
In βAOC and βODB, we have:
β π΄ππΆ = β π·ππ΅ (ππππ‘ππππππ¦ πππππ ππ‘π ππππππ )
πππ β ππ΄πΆ = β ππ·π΅ (π΄πππππ ππ π‘βπ π πππ π ππππππ‘ )
πβπππππππ, ππ¦ π΄π΄ π ππππππππ‘π¦ π‘βπππππ, π€π ππππππ’ππ π‘βππ‘ βπ΄ππΆ β βπ·ππ΅.
βΉ ππΆ
ππ΅=
ππ΄
ππ· =
π΄πΆ
π΅π·
Now, OB = OD
βΉ ππΆ
ππ΄=
ππ΅
ππ·= 1
βΉ OC = OA
Hence, βOAC and βODB are isosceles and similar.
49. In an isosceles βABC, if AC = BC and π΄π΅2 = 2π΄πΆ2, then β C = ?
(a) 300 (b) 450
(c) 600 (d) 900
Sol:
(d) 900
Given:
AC = BC
π΄π΅2 = 2π΄πΆ2 = π΄πΆ2 + π΄πΆ2 = π΄πΆ2 + π΅πΆ2
Applying Pythagoras theorem, we conclude that βABC is right angled at C.
Or, β πΆ = 900
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
50. In βABC, if AB = 16cm, BC = 12cm and AC = 20cm, then βABC is
(a) acute-angled (b) right-angled (c) obtuse-angled
Sol:
(b) right-angled
We have:
π΄π΅2 + π΅πΆ2 = 162 + 122 = 256 + 144 = 400
πππ, π΄πΆ2 = 202 = 400
β΄ π΄π΅2 + π΅πΆ2 = π΄πΆ2
Hence, βABC is a right-angled triangle.
51. Which of the following is a true statement?
(a) Two similar triangles are always congruent
(b) Two figures are similar if they have the same shape and size.
(c)Two triangles are similar if their corresponding sides are proportional.
(d) Two polygons are similar if their corresponding sides are proportional.
Sol:
(c)Two triangles are similar if their corresponding sides are proportional.
According to the statement:
βABC~ βDEF
ππ π΄π΅
π·πΈ=
π΄πΆ
π·πΉ=
π΅πΆ
πΈπΉ
52. Which of the following is a false statement?
(a) If the areas of two similar triangles are equal, then the triangles are congruent.
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding
sides.
(c) The ratio of the areas of two similar triangles is equal to the ratio of squares of their
corresponding.
(d) The ratio of the areas of two similar triangles is equal to the ratio of squares of their
corresponding altitudes.
Sol:
(b) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding
sides.
Because the ratio of the areas of two similar triangles is equal to the ratio of squares of
their corresponding sides.
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
53. Match the following columns:
Column I Column II
(a) In a given βABC, DE βBC and π΄π·
π·π΅ =
3
5. If AC = 5.6cm, then AE =
β¦β¦..cm.
(b) If βABC~βDEF such that 2AB =
3DE and BC = 6cm, then EF =
β¦β¦.cm.
(c) If βABC~βPQR such that
ar(βABC) : ar(βPQR) = 9 : 16 and
BC = 4.5cm, then QR = β¦β¦..cm.
(d) In the given figure, ABβCD and
OA = (2x + 4)cm, OB = (9x β
21)cm, OC = (2x β 1)cm and OD =
3cm. Then x = ?
(p) 6
(q) 4
(r) 3
(s) 2.1
The correct answer is:
(a) - β¦β¦, (b)-β¦β¦, (c)-β¦β¦, (d)-β¦..,
Sol:
(a) β(s)
Let AE be X.
Therefore, EC = 5.6 β X
It is given that DE || BC.
Therefore, by B.P.T.,we get:
π΄π·
π·π΅=
π΄πΈ
πΈπΆ
βΉ 3
5=
π₯
5.6βπ₯
βΉ 3(5.6 β π₯) = 5π₯
βΉ 16.8 β 3π₯ = 5π₯
βΉ 8π₯ = 16.8
βΉ π₯ = 2.1 ππ
(b) β(q)
β΅ βABC-βDEF
β΄ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ
βΉ 3
2=
6
πΈπΉ
πΈπΉ =6Γ2
3= 4 ππ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
(c) β(p)
β΅ βABC ~ βPQR
β΄ ππ(βπ΄π΅πΆ)
ππ(βπππ )=
π΅πΆ2
ππ 2
βΉ 9
16=
4.52
ππ 2 βΉ ππ = β4.5 Γ4.5 Γ16
9=
4.5Γ4
3= 6 ππ
(d) β(r)
β΅ AB || CD
β΄ ππ΄
ππ΅=
ππΆ
ππ· (πβππππ β²π‘βπππππ)
βΉ 2π₯+4
9π₯β21=
2π₯β1
3
3 2 4 2 1 9 21x x x
βΉ 6π₯ + 12 = 18π₯2 β 42π₯ β 9π₯ + 21
βΉ 18π₯2 β 57π₯ + 9 = 0
βΉ 6π₯2 β 19π₯ + 3 = 0
βΉ 6π₯2 β 18π₯ β π₯ + 3 = 0
βΉ (6x β 1) (x -3) = 0
βΉ π₯ = 3 ππ π₯ = β1
6
π΅π’π‘ π₯ = β1
6πππππ (2π₯ β 1) < 0, π€βππβ ππ πππ‘ πππ π ππππ.
Therefore, x= 3
54. Match the following columns:
Column I Column II
(a) A man goes 10m due east and then
20m due north. His distance from the
starting point is β¦β¦m.
(b) In an equilateral triangle with each side
10cm, the altitude is β¦..cm.
(c) The area of an equilateral triangle
having each side 10cm is β¦..cm2.
(d) The length of a diagonal of a rectangle
having length 8m and breadth 6m is β¦.m.
(p) 25β3
(q) 5β3
(r) 10β5
(s) 10
The correct answer is:
(a) - β¦β¦, (b)-β¦β¦, (c)-β¦β¦, (d)-β¦..,
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
(a) β(r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
π΄π΅2 + π΅πΆ2 = π΄πΆ2
βΉ π΄πΆ = β102 + 202 = β100 + 200 = 10β3
Hence, the man is 10 β3π ππ€ππ¦ ππππ π‘βπ π π‘πππππ πππππ‘.
(b) β(q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABC, we have:
π΄π΅2 = π΄π·2 + π΅π·2
βΉ π΄π·2 = 102 β 52 (β΅ π΅π· =1
2 π΅πΆ)
βΉ π΄π· = β100 β 25 = β75 = 5β3 ππ
(c) β (p)
Area of an equilateral triangle with side a = β3
4 π2 =
β3
4 Γ102 = β3 Γ5Γ5
= 25β3 ππ2
(d) β (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
π΄πΆ2 = π΄π΅2 + π΅πΆ2 = 82 + 62 = 64 + 36
βΉ π΄πΆ = β100 = 10 π
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Exercise β Formative Assessment
1. βABC ~ βDEF and the perimeters of βABC and ~ βDEF are 32cm and 24cm respectively. If
AB = 10cm, then DE = ?
(a) 8cm (b) 7.5cm (c) 15cm (d) 5β3cm
Sol:
(b) 7.5 cm
β΅ βABC ~ βDEF
β΄ πππππππ‘ππ(βπ΄π΅πΆ)
πππππππ‘ππ(βπ·πΈπΉ)=
π΄π΅
π·πΈ
βΉ 32
24=
10
π·πΈ
βΉ π·πΈ =10Γ24
32= 7.5 ππ
2. In βABC, DEβBC. If DE = 5cm, BC = 8cm and AD = 3.5cm, then AB = ?
(a) 5.6cm (b) 4.8cm (c) 5.2cm (d) 6.4cm
Sol:
(a) 5.6 cm
β΅ DE β BC
β΄ π΄π·
π΄π΅=
π΄πΈ
π΄πΆ=
π·πΈ
π΅πΆ (πβππππ β²π‘βπππππ)
βΉ 3.5
π΄π΅=
5
8
βΉ π΄π΅ =3.5Γ8
5= 5.6 ππ
3. Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between
their feet is 12m, find the distance between their tops.
(a) 12m (b) 13m (c) 14m (d) 15m
Sol:
(b)13 m
Let the poles be and CD
It is given that:
AB = 6 m and CD = 11 m
Let AC be 12 m
Draw a perpendicular farm Bon CD, meeting CD at E
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
Then,
BE = 12 m
We have to find BD.
Applying Pythagoras theorem in right-angled triangle BED, we have:
π΅π·2 = π΅πΈ2 + πΈπ·2
= 122 + 52 (β΅ πΈπ· = πΆπ· β πΆπΈ = 11 β 6)
= 144 + 25 = 169
BD = 13 m
4. The areas of two similar triangles are 25cm2 and 36cm2 respectively. If the altitude of the
first triangle is 3.5cm, then the corresponding altitude of the other triangle.
(a) 5.6cm (b) 6.3cm (c) 4.2cm (d) 7cm
Sol:
(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their
corresponding altitudes.
Let Ι¦ be the altitude of the other triangle.
Therefore,
25
36=
(3.5)2
β2
βΉ β2 =(3.5)2Γ36
25
βΉ β2 = 17.64
βΉ β = 4.2 ππ
5. If βABC~βDEF such that 2AB = DE and BC = 6cm, find EF.
Sol:
β΅ βABC ~ β DEF
β΄ π΄π΅
π·πΈ=
π΅πΆ
πΈπΉ
βΉ 1
2=
6
πΈπΉ
βΉ πΈπΉ = 12 ππ
6. In the given figure, DEβBC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and
EC = (3x + 19) cm. Find the value of x.
Sol:
β΅ DE || BC
β΄ π΄π·
π·π΅=
π΄πΈ
πΈπΆ (π΅ππ ππ πππππππ‘ππππππ‘π¦ π‘βπππππ)
π₯
3π₯+4=
π₯+3
3π₯+19
βΉ π₯ (3π₯ + 19) = (π₯ + 3)(3π₯ + 4)
βΉ 3π₯2 + 19π₯ = 3π₯2 + 4π₯ + 9π₯ + 12
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ 19π₯ β 13π₯ = 12
βΉ 6x = 12
βΉ π₯ = 2
7. A ladder 10m long reaches the window of a house 8m above the ground. Find the distance
of the foot of the ladder from the base of the wall.
Sol:
Let the ladder be AB and BC be the height of the window from the ground.
We have:
AB 10 m and BC = 8 m
Applying theorem in right-angled triangle ACB, we have:
π΄π΅2 = π΄πΆ2 + π΅πΆ2
βΉ π΄πΆ2 = π΄π΅2 β π΅πΆ2 = 102 β 82 = 100 β 64 = 36
βΉ π΄πΆ = 6 π
Hence, the ladder is 6 m away from the base of the wall.
8. Find the length of the altitude of an equilateral triangle of side 2a cm.
Sol:
Let the triangle be ABC with AD as its altitude. Then, D is the midpoint of BC.
In right-angled triangle ABD, we have:
π΄π΅2 = π΄π·2 + π·π΅2
βΉ π΄π·2 = π΄π΅2 β π·π΅2 = 4π2 β π2 (β΅ π΅π· =1
2 π΅πΆ)
= 3π2
π΄π· = β3π
Hence, the length of the altitude of an equilateral triangle of side 2a cm is β3π ππ.
9. βABC~βDEF such that ar(βABC) = 64cm2 and ar(βDEF) = 169cm2. If BC = 4cm, find EF.
Sol:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
β΅ βABC ~ βDEF
β΄ ππ(βπ΄π΅πΆ)
ππ(βπ·πΈπΉ)=
π΅πΆ2
πΈπΉ2
βΉ 64
169=
42
πΈπΉ2
βΉ πΈπΉ2 =16Γ169
64
βΉ πΈπΉ =4Γ13
8= 6.5 ππ
10. In a trapezium ABCD, it is given that ABβCD and AB = 2CD. Its diagonals AC and BD
intersect at the point O such that ar(βAOB) = 84cm2. Find ar(βCOD).
Sol:
In βAOB and βCOD, we have:
β π΄ππ΅ = β πΆππ· (ππππ‘ππππππ¦ πππππ ππ‘π ππππππ )
β ππ΄π΅ = β ππΆπ· (π΄ππ‘πππππ‘π ππππππ ππ π΄π΅ Η πΆπ· )
Applying AA similarity criterion, we get :
β AOB - βCOD
β΄ ππ(βπ΄ππ΅)
ππ(βπΆππ·)=
π΄π΅2
πΆπ·2
βΉ 84
ππ(βπΆππ·)= (
π΄π΅
πΆπ·)
2
βΉ 84
ππ(βπΆππ·)= (
2πΆπ·
πΆπ·)
2
βΉ ππ(βπΆππ·) =84
4= 21 ππ2
11. The corresponding sides of two similar triangles are in the ratio 2: 3. If the area of the smaller
triangle is 48cm2, find the area of the larger triangle.
Sol:
It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their
corresponding sides.
β΄ 48
π΄πππ ππ ππππππ π‘πππππππ=
22
32
βΉ 48
π΄πππ ππ ππππππ π‘πππππππ=
4
9
βΉ π΄πππ ππ ππππππ π‘πππππππ =48Γ9
4= 108 ππ2
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
12. In the given figure, LMβCB and LNβCD. Prove that π΄π
π΄π΅=
π΄π
π΄π·.
Sol:
LM || CB and LN || CD
Therefore, applying Thalesβ theorem, we have:
π΄π΅
π΄π=
π΄πΆ
π΄πΏ πππ
π΄π·
π΄π=
π΄πΆ
π΄πΏ
βΉ π΄π΅
π΄π=
π΄π·
π΄π
β΄ π΄π
π΄π΅=
π΄π
π΄π·
This completes the proof.
13. Prove that the internal bisector of an angle of a triangle divides the opposite side internally
in the ratio of the sides containing the angle.
Sol:
Let the triangle be ABC with AD as the bisector of β π΄ which meets BC at D.
We have to prove:
π΅π·
π·πΆ=
π΄π΅
π΄πΆ
Draw CE || DA, meeting BA produced at E.
CE || DA
Therefore,
β 2 = β 3 (π΄ππ‘πππππ‘π ππππππ )
πππ β 1 = β 4 (πΆπππππ πππππππ ππππππ )
But,
β 1 = β 2
Therefore,
β 3 = β 4
βΉ AE = AC
In βBCE, DA || CE.
Applying Thalesβ theorem, we gave:
π΅π·
π·πΆ=
π΄π΅
π΄πΈ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ π΅π·
π·πΆ=
π΄π΅
π΄πΆ
This completes the proof.
14. In an equilateral triangle with side a, prove that area = β3
4 π2.
Sol:
Let ABC be the equilateral triangle with each side equal to a.
Let AD be the altitude from A, meeting BC at D.
Therefore, D is the midpoint of BC.
Let AD be h.
Applying Pythagoras theorem in right-angled ABD, we have:
π΄π΅2 = π΄π·2 + π΅π·2
βΉ π2 = β2 + (π
2)
2
βΉ β2 = π2 βπ2
4=
3
4π2
βΉ β =β3
2 π
Therefore,
π΄πππ ππ π‘πππππππ π΄π΅πΆ =1
2 Γπππ π Γβπππβπ‘ =
1
2 Γπ Γ
β3
2 π
This completes the proof.
15. Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.
Sol:
Let ABCD be the rhombus with diagonals AC and BD intersecting each other at O.
We know that the diagonals of a rhombus bisect each other at right angles.
β΄ If AC β 24 cm and BD = 10 cm, AO = 12 cm and BO = 5 cm
Applying Pythagoras theorem in right-angled triangle AOB, we get:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π΄π΅2 = π΄π2 + π΅π2 = 122 + 52 = 144 + 25 = 169
AB = 13 cm
Hence, the length of each side of the given rhombus is 13 cm.
16. Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their
corresponding sides.
Sol:
Let the two triangles be ABC and PQR.
We have:
βABC ~ βPQR,
Here,
BC = a, AC = b and AB = c
PQ = r, PR = q and QR = p
We have to prove:
π
π=
π
π=
π
π=
π+π+π
π+π+π
βABC ~ βPQR; therefore, their corresponding sides will be proportional.
βΉ π
π=
π
π=
π
π= π (π ππ¦) β¦ (π)
βΉ π = ππ, π = ππ πππ π = ππ
β΄ πππππππ‘ππ ππ βπ΄π΅πΆ
πππππππ‘ππ ππ βπππ =
π+π+π
π+π+π=
ππ+ππ+ππ
π+π+π= π β¦ (ππ)
πΉπππ (π)πππ (ππ), π€π πππ‘:
π
π=
π
π=
π
π=
π+π+π
π+π+π=
πππππππ‘ππ ππ βπ΄π΅πΆ
πππππππ‘ππ ππ βπππ
This completes the proof.
17. In the given figure, βABC and βDBC have the same base BC. If AD and BC intersect at O,
prove that ππ(βπ΄π΅πΆ)
ππ(βπ·π΅πΆ)=
π΄π
π·π.
Sol:
πΆπππ π‘ππ’ππ‘πππ βΆ π·πππ€ π΄π β₯ πΆπ πππ π·π β₯ π΅π.
As,
ππ(βπ΄π΅πΆ)
ππ(βπ·π΅πΆ)=
12
Γπ΄π Γπ΅πΆ
12
Γπ·π Γπ΅πΆ
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
βΉ ππ(βπ΄π΅πΆ)
ππ(βπ·π΅πΆ) =
π΄π
π·πβ¦ (π)
πΌπ β π΄π΅πΆ πππ βπ·π΅πΆ, β π΄ππ = β π·ππ = 900 (π΅π ππππ π‘ππ’ππ‘ππ)β π΄ππ =
β π·ππ (ππππ‘ππππππ¦ πππππ ππ‘π πππππ ) β΄ βπ΄ππ~βπ·ππ(π΅π π΄π΄ ππππ‘πππππ) β΄π΄π
π·π
= π΄π
π·π (πβππππ β²π π‘βπππππ) β¦ (ππ)πΉπππ (π)πππ (ππ), π€π βππ£π βΆ
ππ(βπ΄π΅πΆ)
ππ(βπ·π΅πΆ)
= π΄π
π·π=
π΄π
π·π ππ,
ππ(βπ΄π΅πΆ)
ππ(βπ·π΅πΆ)=
π΄π
π·π
This completes the proof.
18. In the given figure, XYβAC and XY divides βABC into two regions, equal in area. Show
that π΄π
π΄π΅=
(2β β2)
2.
Sol:
In β ABC and βBXY, we have:
β π΅ = β π΅
β π΅ππ = β π΅π΄πΆ (πΆπππππ πππππππ ππππππ )
πβπ’π , βπ΄π΅πΆ β βπ΅ππ (π΄π΄ ππππ‘πππππ )
β΄ ππ(βπ΄π΅πΆ)
ππ(βπ΅ππ)=
π΄π΅2
π΅π2 =π΄π΅2
(π΄π΅βπ΄π)2 β¦ (π)
π΄ππ π,ππ(βπ΄π΅πΆ)
ππ(βπ΅ππ)=
2
1 {β΄ ππ (βπ΅ππ) = ππ(π‘πππππ§ππ’π π΄πππ)} β¦ (ππ)
πΉπππ (π)πππ (ππ), π€π βππ£π:
π΄π΅2
(π΄π΅βπ΄π)2 =2
1
βΉ π΄π΅
(π΄π΅βπ΄π)= β2
βΉ (π΄π΅βπ΄π)
π΄π΅=
1
β2
βΉ 1 βπ΄π
π΄π΅=
1
β2
βΉ π΄π
π΄π΅= 1 β
1
β2=
β2β1
β2=
(2β β2)
2
19. In the given figure, βABC is an obtuse triangle, obtuse-angled at B. If ADβ₯CB, prove that
π΄πΆ2 = π΄π΅2 + π΅πΆ2 + 2π΅πΆ. π΅π·.
Sol:
Applying Pythagoras theorem in right-angled triangle ADC, we get:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
π΄πΆ2 = π΄π·2 + π·πΆ2
βΉ π΄πΆ2 β π·πΆ2 = π΄π·2
βΉ π΄π·2 = π΄πΆ2 β π·π΅2 β¦(1)
Applying Pythagoras theorem in right-angled triangle ADB, we get:
π΄π΅2 = π΄π·2 + π·π΅2
βΉ π΄π΅2 β π·π΅2 = π΄π·2
βΉ π΄π·2 = π΄π΅2 β π·π΅2 β¦(2)
πΉπππ πππ’ππ‘πππ (1)πππ (2), π€π βππ£π:
π΄πΆ2 β π·πΆ2 = π΄π΅2 β π·π΅2
βΉ π΄πΆ2 = π΄π΅2 + π·πΆ2 β π·π΅2
βΉ π΄πΆ2 = π΄π΅2 + (π·π΅ + π΅πΆ)2 β π·π΅2 (β΅ π·π΅ + π΅πΆ = π·πΆ)
βΉ π΄πΆ2 = π΄π΅2 + π·π΅2 + π΅πΆ2 + 2π·π΅. π΅πΆ β π·π΅2
βΉ π΄πΆ2 = π΄π΅2 + π΅πΆ2 + 2π΅πΆ. π΅π·
This completes the proof.
20. In the given figure, each one of PA, QB and RC is perpendicular to AC. If AP = x, QB = z,
RC = y, AB = a and BC = b, show that 1
π₯+
1
π¦=
1
π§.
Sol:
In β PAC and βQBC, we have:
β π΄ = β π΅ (π΅ππ‘β ππππππ πππ 900)
β π = β π (πΆπππππ πππππππ ππππππ )
And
β πΆ = β πΆ (ππππππ ππππππ )
πβπππππππ, βππ΄πΆ~ βππ΅πΆ
π΄π
π΅π=
π΄πΆ
π΅πΆ
βΉ π₯
2=
π+π
π
βΉ π + π =ππ¦
π§ β¦ (1)
In β RCA and βQBA, we have:
β πΆ = β π΅ (π΅ππ‘β ππππππ πππ 900)
β π = β π (πΆπππππ πππππππ ππππππ )
And
β π΄ = β π΄ (ππππππ ππππππ )
πβπππππππ, βπ πΆπ΄ ~ βππ΅π΄ π πΆ
π΅π=
π΄πΆ
π΄π΅
βΉ π¦
π§=
π+π
π
βΉ π + π =ππ¦
π§ β¦ (2)
From equation (1) and (2), we have:
Class X Chapter 4 β Triangles Maths
______________________________________________________________________________
ππ₯
π§=
ππ¦
π§
βΉbx = ay
βΉ π
π=
π₯
π¦ β¦ (3)
Also,
π₯
π§=
π+π
π
βΉ π₯
π§=
π
π+ 1
ππ πππ π‘βπ π£πππ’π πππ
π ππππ πππ’ππ‘πππ (3), π€π βππ£π:
βΉ π₯
π§=
π₯
π¦+ 1
π·ππ£πππππ πππ‘β π ππππ ππ¦ π₯, π€π πππ‘:
1
π§=
1
π¦+
1
π₯
β΄ 1
π₯+
1
π¦=
1
π§
πβππ πππππππ‘ππ π‘βπ πππππ.