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Revision Question Bank Triangles

1. In the figure, DEAC and DC AP. Prove that BE BC

EC CP.

Solution:

In BPA, we have

DCAP [given]

Therefore, by basic proportionality theorem, we have

BC BD

CP DA ..(i)

In BCA , we have

DE AC [given]


BE BD

EC DA ..(ii)

From Eqs. (i) and (ii)

BC BE BE BCor

CP EC EC CP Hence proved.

2. In figure, if EDC~ ESA BEC = 115° and EDC=70°. Find DEC, DCE, EAB,

AEB and EBA.

Solution:

Since, BD is a line and EC is a ray on it.

DEC + BEC = 180° [linear pair axiom]

DEC +1150=1800 [ BEC =1150,given]

DEC = 180° – 115° = 65°

But AEB = DEC [vertically opposite angles]

AEB=65°

In CDE, we have CDE + DEC+ DCE = 180°

[by property of sum of the angles of a triangle is 1800 ]



70° + 65° + DCE =180°

DCE = 180° – 135° = 45°

It is given that, EDC EBA

EBA = EDC, EAB = ECD

EBA = 70° and EAB = 45°

[ ECD = DCE =45°]

Hence, DEC = 65°, DCE = 45°, EAB = 45°,

AEB = 65° and EBA = 70°

3. In the figure, ABCD is a quadrilateral with BA parallel to CD.

AC and BD meet at X, where CX=8 cm and XA = 10 cm.

(i) Given that, BD=27cm,find the length of BX.

(ii) Find the ratio

Area of a BXC: Area of AXD

Solution: AXB CXD

BX AX

XD XC

BX 10 5

XD 8 4

Since, BD=27cm

5

BX9

× 27 =15 cm

(ii) Consider BXA and AXD, sharing common height

ar BXA 15 5

ar AXD 12 4

[ area of triangle =1

2 × b × h]

Consider BXC and AXD,BXC 8 4

AXD 10 5

Ratio of area of BXC : Area of AXD

= 4 : 5

4. In the given figure, ABC is a right angled triangle, right angled at C and DE AB. Prove

that ABC ADE and hence find the lengths of AE and DE.



Solution:

In ABC and ADE, we have

A= A [common angle]

C = E [each 90°]

ABC ADE

[by AA similarity criterion]

In ABC, we have

AB2 =BC2 +AC2 [Pythagoras theorem]

AB2 =122 +52 =144+25=169

[ BC = 12 cm and AC = 5 cm given]

AB = 13 cm

Since, ABC ADE

AB BC AC

AD DE AE

[by basic proportionality theorem]

13 12 5

3 DE AE

36

DE cm13

and 15

AE13

cm

5. In the figure, AB = 8 cm, BC=3 cm and BE is parallel to CD.

(i) Find the values of

(a) BE

CD (b)

Area of ABE

Area of ACD (c)

Area of ABE

Area of Quadrilateral BCDE

(ii) What is the special name given to the quadrilateral BCDE?

Solution:

(i)(a) ABE and ACD ,

ABE = ACD

[corresponding angles, BE CD]

AEB = ADC [corresponding angles, BE CD]



and BAE = CAD [common angle]

ABE and ACD is similar.

BE AB 8 8

CD AC 8 3 11

(b)

2Area of ABE Length of side AB

Area of ACD Length of side AC

= 2

8 64

11 121

(c) Area of ABE 64 64

Area of Quadrilateral CDE 121 64 57

(ii) It is a trapezium, which has 2 parallel sides BE and CD.

6. In the figure, ABC= AED.

(i) Explain why ABC and AED are similar.

(ii) Given, also that AD = 3cm, AE =5cm and EC=2cm. Calculate

(a) BD

(b) Area of AED

Area of ABC Solution:

(i) In ABC and AED ABC= AED [given]

BAC= EAD [common angle]

ACB ADE [by AAA similarity criterion]

(ii) (a) Since, ABC AED

AB BC AC

AE ED AD


AB 7

5 3 [ AE = 5 cm, AC = 7cm and AD = 3cm given)

3AB = 35

35 2

AB 113 3

cm

BD = AB – AD = 2

113

– 3 = 82

3cm



(b) Since, AD 3

AC 7

2

Area of AED 3 9

Area of ABC 7 49

7. Two poles of heights p and q m are standing on a level ground, a metre apart. Prove that

the height of the point of intersection of the line segments joining the top of each pole to

the foot of the opposite pole is given by pq

p qm.

Solution:

In CQPand CAB,

C = C [common angle]

CQP = CAB [each90°]

CQP CAB [by AA similarity criterion]

CQ PQ x h

CA AB a q ...(i)

In AQP and ACD,

A = A [common angle]

AQP = ACD [each 90°]

AQP ACD [by AA similarity criterion]

AQ PQ a x h

AC CD a p

[by basic proportionality theorem

x h

1a p

..(ii)

On adding Eqs. (i) and (ii), we get

h h ph qh1 1

q p pq

h p q pq

1 hpq p q

m Hence proved.



8. In the figure, PQ BC and PR CD. Prove that

(i) AR AQ

AD AB (ii)

QB DR

AQ AR Solution:

In ABC, we have

PQBC [given]


AQ AP

AB AC ..(i)

In ACD,

PR CD


AP AR

AC AD ...(ii)

From Eqs.(i) and (ii),

AQ AR AR ADor

AB AD AD AB

From Eqs. (i) and (ii),

AQ AR AB AD

AB AD AQ AR

[reciprocal the above equation]

AQ QB AR RD QB

1AQ AR AQ

= RD QB DR

1AR AQ AR

Hence proved.

9. In the figure. A, B and C are points on OP, OQ and Off respectively, such that ABPQ and

BC QR. Show that AC PR.

Solution:

In OPQ, we have

ABPQ



OA OB

AP BQ ...(i)


In OQR, we have

BC QR

OB OC

BQ CR ...(ii)


From Eqs. (i) and (ii),

OA OC

AP CR

Thus, A and C are points on sides OP and OR, respectively of PQR, such that

OA OCAC PR

AP CR

[by converse of basic proportionality theorem]

10. D, E and F are the points on sides BC, CA and AB respectively, such that AD bisects A,

BE bisects B and F bisects C. If 48 = 5 cm, SC = 8 cm and A=4 cm, then determine

AF,CE and BD.

Solution:

Given in ABC, CF bisects C.

We know that, the internal bisector of an angle of a triangle divides the opposite side

internally in the ratio of the sides containing the angle.

AF AC AF 4

FB BC 5 AF 8

[ FB = AB – AF = 5 – AF]

AF 1

5 AF 2 2AF = 5 – AF

2AF + 5 3AF = 5



AF = 5

3 cm

Again, in ABC, BE bisects B.

AE AB

EC BC [by above theorem]

4 CE 5

CE 8

[ AE =AC– CE =4 – CE, AB = 5cm and BC = 8 cm, given]

8(4 – CE) = 5CE

32 – 8CF =5CE 32=13CE

CE = 32

13 cm

Similarly, in ABC, AD bisects A.

BD AB BD 5

DC AC 8 BD 4

[ DC = BC – BD = 8 – BD, AB = 5 cm, AC = 4 cm]

4BD = 5(8 – BD)

4BD = 40 – 5BD 9BD = 40

BD = 40

9 cm

Hence, AF = 5

3 cm, CE =

32

13 cm and BD =

40

9 cm



Chapter Test {Triangles}

M: Marks: 40 M: Time: 40 Min.

1. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. [4] Solution: Given A ABC in which AB = BC = CA and AD BC.

To prove 3AB2 =4AD2 Proof In ADB and ADC, AB = AC [given]

B = C [each 60°] and ADB = ADC [each 90°]

ADB ADC [by SAA congruence rule]

1BD DC BC

2 ..(i)

In right ADB, by Pythagoras theorem, AB2 =AD2+ BD2

AB2 = AD2 + 2

1BC

2 [ from Eq. (i)]

2 2 21AB AD BC

4

4AB2 = 4AD2 + BC2 4AB2 – ABD2 = 4AD2 [given, BC=AB]

3AB2 = 4AD2 Hence proved.

2. If the area of two similar triangles are equal, then prove that they are congruent. [4] Solution: Given ABC DEF such that ar( ABC) = ar( DEF)



To prove ABC DEF Proof Since, ABC - DEF and ar( ABC) = ar( DEF)

B E

and C F ..(i)

Now, 2 2

2 2

ar ABC ar ABCBC BC

ar DEF EF ar ABC EF

[ ar ABC ar DEF , given]

2

2 2

2

BC1 EF BC

EF

EF = BC [taking positive square root] Now, in ABC and DEF, we get B = E [from Eq. (i)] C = F [from Eq. (i)] and BC = EF [proved above]

ABC DEF

[by ASA congruence rule] Hence proved.

3. If the sides AB and AC and median AD of a ABC are proportional to sides PQ and PR and median PM of another PQR. Then, prove that ABC PQR. [4] Solution: Sol : Given Two ABC and PQR in which AD and PM are medians such that AB AC AD

PQ PR PM

To prove ABC PQR Construction Produce AD to E and PM to N such that AD = DE and PM = MN. Join CE and NR.

Proof In ADB and EDC, AD = ED [by construction] BD = CD [since, D is mid-point of BC] and ADB = CDE [vertically opposite angles]



ADB EDC [by SAS congruence rule] AB = EC [by CPCT] ...(i) Similarly, in PMQ and NMR, PQ = NR . ..(ii)

Now, AB AC AD

PQ PR PM [given]

EC AC 2AD

NR PR 2PM

[ from Eqs. (i) and (ii)]

EC AC AE

NR PR PN

[2AD = AE and 2Pm = PN] ACE PRN [by SSS similarity criterion] 1 = 2 ...(iii)

[corresponding angles are equal] Similarly, 3 = 4 ...(iv) On adding Eqs. (iii) and (iv), we get

1 + 3 = 2 + 4 or A = P Now, in ABC and PQR,

A = P and AB AC

PQ PR [given]

ABC PQR [by SAS criterion rule] Hence proved. Or Given In the given figure, ABDC and AC and PQ intersect each other at the point O.

To prove OA . CQ = OC . AP Proof Since, AB DC and AC is a transversal.

OCQ = OAB ..(i) [alternate angles] Again, PQ is also a transversal.

OQC = APO ... (ii) [alternate angles] Now, in OAP and OCQ,

OAP = OCO, [from Eq. (i)] APO = OQC [from Eq. (ii)]

and AOP= QOC [vertically opposite angles] OAP OCQ [by AAA similarity criterion]

OA OP AP

OC OQ CQ



OA AP

OC CQ

OA = CQ = QC. AP Hence proved.

4. In the given figure, in ABC, XY AC and XY divides the ABC into two regions such

that ar BXY = 2 ar (ACY X). DetermineAX

AB. [4]

Solution: We have, ar( BXY) = 2ar( CYX)

ar( BXY)=2[ar( BAC) – ar( BXY)] 3ar( BAY)=2ar( BAC)

ar BXY 2

ar BAC 3

In BXY and BAC, B = B [common angle] BXY = BAC [corresponding angles]

and BVX = BCA [corresponding angles] BXY BAC [by AAA similarity criterion]

2

2

ar BXY BX

ar BAC BA

2

2

2 BX

3 BA from Eq. (i)]

BX 2

BA 3 [taking positive square root]

BX 2

BA 3 [multiplying both sides by –1]

BX 2

1 1BA 3

[adding l on both sides]

BA BX 3 2

BA 3

AX 3 2

BA 3



5. In the given figure, DEFG is a square and BAC = 900 show that DE2 = BD × EC. [4]

Solution: Given In the given figure, DEFG is a square and BAC = 90°. To prove DE2 = BD × EC Proof In DBG and ECF

3 + 1 = 90° = 3 + 4 3 + 1 = 3 + 4

1 = 4 Also D = E [each 90°]

DBG DEF [by AA similarity criterion]

BD EF

DG EC [by basic proportionality theorem]

BD × EC = EF × DG But DG = EF = DE [side of a square]

BD × EC = DE × DE DE2 = BD × EC

Hence proved. 6. P is the mid-point of side BC of ABC. Q is the mid-point of AP, BQ when produced

meets AC at L. Prove that 1

AL AC3

. [4]

Solution: Given In ABC, P is the mid-point of BC and Q is the mid-point of AP. BQ when produced meets AC at L.

To prove AL=1

3AC

Construction Draw PM BL meeting AC at M.

Proof In ABLC, PM SL

BP LM

PC MC [by basic proportionality theorem]

But BP= PC [ P is mid-point of BC]



LM

1 LM MCMC

...(i)

Now, in APM, Q is the mid-point of AP and QLPM

AQ AL

QP LM [by basic proportionality theorem]

But AQ=QP [Q is mid-point of AP]

AL

1 AL LMLM

..(ii)

From Eqs. (i) and (ii),we get

AL=LM=MC AL = 1

3 AC

Hence proved. 7. Gopal is walking away from the base of a lamp-post at a speed of 2 m/s and his height

(MN) is 0.9 m. Suppose the lamp-post 4.5 m above from the ground as shown in adjoining figure.

(i) Find the distance of Gopal from the base of lamp-post in 5s. (ii) Is the PQR and MNR similar? (iii) Find the length of his shadow after 5 s. [4] Solution: (i) Let N be the position of Gopal after 5 s, then distance of Gopal From the base of lamp-post, QN = 2 × 5=10m (ii) In PQR and MNR, LQ = LN [since, each angle is 90° because lamp-post as well as Gopal is standing vertically]

R = R [common angle] PQR MNR [by AA similarity criterion]

(iii) Let length of his shadow after 5 s in NR = x Since, PQR – MNR

Therefore, QR PQ QN NR PQ

NR MN NR MN

10 x 4.5

x 0.9

[QN = 10 m, PQ = 4.5 m, MN = 0.9 m] 9 + 0.9x = 4.5X 9 = 4.5x - 0.9x 9 = 3.6x



9 90 10

x 2.5m36 36 4

Hence, the shadow of Gopal after 5 s is 2.5 m. 8. In the given figure, FEC GBD and 1 = 2. Prove that ADE ABC. [4]

Solution: Given FEC GBD and 1 = 2

To prove ADE ABC

Proof Since, FEC GBD

DG=EF [by CPCT] EFG = DGB

i.e., 5= 6 [angles opposite to equal sides are equal] Also, BD=EC [by CPCT]

DBC= ECB i.e., 3 = 4 and BC=FC [by CPCT]

BDC= CEF But also 1= 2 [given]

AE= AE [sides opposite to equal angles are equal] In ADE and ABC, AD=AE Also, BD=EC

AD AE

BD EC

ADE ABC

[by converse of basic proportionality theorem] 9. State and prove the basic proportionality theorem. [4]

Solution. Given A triangle ABC in which DE BC, and intersects AB and D and AC in E.

To prove AD AE

DB EC



Construction Join BE, CD and draw EF BA and DG CA. Proof Since EF is perpendicular to AB. Therefore, EF is the height of triangles ADE and DBE.

Now, Area 1

ADE2

(base × height) = 1

2(AD. EF)

and, Area DBE = 1

2 (base × height) =

1

2 (DB. EF)

1AD.EFArea ADE AD2

1Area ABE DBDB.EF2

..(i)

Similarly, we have

1AE. DGArea ADE AE2

1Area DEC ECEC. DG2

..(ii)

But, DAC are on the same base DE and between the same parallels DE and BC.

Area DBE = Area DEC

1 1

Area DBE Area DEC [Taking reciprocals of both sides]

Area ADE Area ADE

Area DBE Area DEC [Multiplying both sides by Area ( ADE)]

AD AE

DB EC [Using (i) and (ii)]

10. In the given figure, ABCD is a quadrilateral P, Q, R and S are the points of trisection of the sides AB, BC, CD and DA, respectively. Prove that PQRS is a parallelogram. [4]



Solution. Given A quadrilateral ABCD in which P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C. To prove PQRS is a parallelogram i.e., PQ SR and QR PS.

Construction Join AC. Proof Since, P, Q, R and S are the points of trisection of AB, BC, CD and DA respectively.

BP = 2PA, BQ = 2 QC, DR = 2 RC and, DS = 2 SA In ADC, we have DS 2SA DR 2RC

2 and , 2SA SA RC RC

DS DR

SA RC

S and R divide the sides DA and DC respectively in the same ratio. SR AC ..(i)

[By the converse of Thale’s Theorem]

In ABC, we have BP 2PA BQ 2QC

2 and 2PA PA QC QC

BP BQ

PA QC

P and Q divide the sides BA and BC respectively in the same ratio. PQ AC ..(ii) [By the converse of Thale’s Theorem]

From equations (i) and (ii), we have



SR AC and PQ AC

SR PQ

Similarly, by joining BD, we can prove that QR PS

Hence, PQRS is a parallelogram.

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