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4 Unit Math Homework for Year 12 (Worked Answers)

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Table of contents

1 Harder 3 Unit Topics (Part1) 1

1.1 Geometry of Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Chord Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Angles in a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.3 Circle Quadrilateral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.4 Tangents and Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.5 Direct and indirect Common Tangents . . . . . . . . . . . . . . . . . . . . . . 7

1.1.6 The Alternate Segment Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.7 Intercepts on Intersecting Chords . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.8 Further Circle Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Pass Exam Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

This edition was printed on June 17, 2013.Camera ready copy was prepared with the LATEX2e typesetting system.Copyright © 2000 - 2013 Yimin Math Centre (www.yiminmathcentre.com)

4 Unit Math Homework for Year 12

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Year 12 Topic 1 Worked Answers Page 1 of 19

1 Harder 3 Unit Topics (Part1)

1.1 Geometry of Circle

1.1.1 Chord Properties

Chord Properties of Circle 1:

• A perpendicular drawn to a chord from the centre of a circle bisects the chord, andthe perpendicular bisector of a chord passes through the centre.

• The line from the centre of a circle to the midpoint of the chord meets the chord atright angles.

• When two circles intersect, the line joining their centres bisects their common chordat right angles.

Exercise 1.1.1 Use trigonometry to find x in each diagram (correct to 1 decimals place.).

1. Given that ∠AOM = 50◦, AB = 12 cm.

Solution: AB = 12, cm ⇒ AM = 6 cm, ⇒ x =AM

sin 50◦= 7.8 cm.

2. Given that ∠POQ = 150◦, PQ = 15 cm.

Solution: ∠POQ = 150◦, ⇒ ∠ROQ = 75◦, ⇒ 7.5

sin 75◦= 7.8 cm.

3. Given that ∠MON = 130◦, OM = 8.5 cm.

Solution: ∠MON = 130◦, ⇒ ∠OMN = 25◦, ⇒ x = 2× 8.5× cos 25◦ = 15.4 cm.

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Exercise 1.1.2 For diagram shown below AO = 26 cm, OM = 10 cm. Find the length of AB,giving reasons.

Solution: OA2 = AM2 +OM2 (Phthagoras), ⇒ AM2 = 262 − 102

∴ AM =√576 = 24 cm.

Now AB = 2× AM (as OM is the perpendicular bisector of AB), ⇒ AB = 48 cm.

Exercise 1.1.3

1. Find the radius of a circle in which a chord of length 14 cm subtends and angle of 70◦ at the

centre (correct to 1 decimals place.).

Solution: sin 35◦ =7

r⇒ r =

7

sin 35◦= 12.2 cm.

2. A chord subtends an angle of 110◦ at the centre of a circle of radius 5.6 cm. Find the length of

the chord correct to one decimal place.

Solution:sin 55◦ =

l2

5.6⇒ l = 2× 5.6× sin55◦ = 9.2 cm.

3. A chord of length 12 cm is drawn on a circle of radius 16 cm. How far this chord from the centre

of the circle?

Solution: d =√162 − 62 = 14.8, cm.

4. A chord of length 22 cm has a perpendicular distance of 8 cm from the centre of the circle. What

is the radius of the circle?

Solution: r =√112 + 82 = 13.6 cm


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Exercise 1.1.4 These two circles have as their centres points O and C. PQ is the common chordjoining the points of intersection of the two circles. N is the point where PQ intersects the line OCwhich joins the centres.

1. Prove that the triangles POC and QOC are congruent.

Solution: PC = CQ = r, PO = OQ = R and OC = OC (common)

∴ 4POC ≡ 4QOC (three mattching side are equal).

2. Hence, show that ∠POC = ∠QOC.

Solution: ∵ 4POC ≡ 4QOC

∴ ∠POC = ∠QOC, matching angle of congruent traingles.

3. Now, prove that the triangles PON and QON are congruent.

Solution: PO = QO, (matching sides of congruent traingles, prove above.)

∠POC = ∠QOC, (matching angle of congruent traingles, prove above).

∠OPN = ∠OQN (base angles of an isosceles traingle).

∴ 4PON ≡ 4QON.(ASA)

4. Hence, show that N bisects PQ and that PQ ⊥ OC.

Solution: PN = NQ, (matching sides of congruent traingles, prove above)

and∠PNO = ∠ONQ =180◦

2= 90◦, (matching angles of congruent traingles, prove above)

∴ N bisects PQ and that PQ ⊥ OC.


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Chord Properties of Circles 2:Equal chords of a circle are the same distance from the centre and subtendequal angles at the centre.

∠AOB = ∠COD and OM = ON .

Exercise 1.1.5

1. AB = BC = 25cm, OM = 5.8 cm. Find he length of ON giving reasons.

Solution:ON = 5.8 cm as equal chords of a circle are the same distance from the centre.

2. OM = ON = 3.5 cm, PQ = 12.6 cm. Find the length of RS, giving reasons.

Solution:RS = 12.6 cm as chords that are equidistant from the centre are equal in length.


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1.1.2 Angles in a Circle

Definition: The angle at the centre of a a circle is twice the angle at the circumferencestanding on the same arc.

Definition: The angle in a semicircle is a right angle.

Definition: Angles at the circumference of a circle standing on the same arc are equal.In another words, angles in the same segment of a circle are equal.


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1.1.3 Circle Quadrilateral

A cyclic quadrilateral is a quadrilateral whose vertices lie on the circumference of a circle.

Definition: The opposite angles of a cyclic quadrilateral are supplementary (sum to 180◦).

Exercise 1.1.6 The diagram shows, AOB is the diameter of a circle, centre O. find angle ∠ADC.

Solution: Connect CB to form a cyclic quadrilateral ABCD,

∵ ∠BOC = 60◦ and CO = CB, ⇒ ∴ ∠OBC = ∠OCB = 60◦.

In the quadrilateral ABCD, ∠OBC + ∠ADC = 180◦

∴ ∠ADC = 180◦ − 60◦ = 120◦ (Opposite angles of a cyclic quadrilateral are supplementary).

Exercise 1.1.7 If O is the centre of the circle and ∠CAO = 40◦, find the size of ∠ABC

Solution: Connect AE and CE to form a cyclic quadrilateral ABCE,

∠OAC = ∠OCA, ⇒ ∠AOC = 180◦ − 2× 40◦ = 100◦,

∠AEC = ∠AOC ÷ 2 = 100◦ ÷ 2 = 50◦

∴ ∠ABC = 180◦− 50◦ = 130◦ (The opposite angles of a cyclic quadrilateral are supplementary).


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1.1.4 Tangents and Radii

Definition: The two tangents form a external point have equal lengths.

1.1.5 Direct and indirect Common Tangents

Definition: A common tangent to a pair of circles:

• is called direct, if both circles are on the same side of the tangent,

• is called indirect, if the circles are on the opposite sides of the tangent.

1.1.6 The Alternate Segment Theorem

Definition: The angle between a tangent to a circle and a chord at the point of contactis equal to any angle in the alternate segment.

Exercise 1.1.8 In the figure the tangent TS is parallel to the chord PQ. If ∠TPQ = 40◦, find thesize of ∠PTQ.

Solution: ∠RTP = ∠TPQ = 40◦, and ∠STQ = ∠TQP (Alternate angles, ST ||PQ)

∠STQ = ∠TPQ = 40◦ (alternate segment theorem),

∴ ∠PTQ = 180◦ − 40◦ − 40◦ = 100◦ (angle sum of a straight line).


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1.1.7 Intercepts on Intersecting Chords

Definition: If two chords of a circle intersect, the product of the intercepts on the one chordis equal to the product of the intercepts on the other chord.

AM ×MB = PM ×MQ

Definition: Given a circle and two secants from an external point, the product of the twointervals from the point to the circle on the secant is equal to the product of these twointervals on the other secant.

AM ×MB = PM ×MQ

Exercise 1.1.9 O is the centre of the circle of radius 9 cm. If CD = 6 cm, DP = 4 cm, find PB.

Solution: Since OA = OB = 9 cm, ⇒ AB = 18 cm and CP = CD +DP = 6 + 4 = 10 cm

Let PB = x, AP = AB + x = 18 + x

AP ×BP = CP ×DP, ⇒ (18 + x)× x = 10× 4

Now 18x+ x2 = 40 ⇒ x2 + 18x− 40 = 0 ⇒ (x− 2)(x+ 20) = 0

∴ x = 2 cm, x = −20 (invalid).


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Exercise 1.1.10 O is the centre of the circle. BT is a chord that subtends ∠BAT at the circumfer-ence and ∠TOB at the centre. PT and PB are tangents to the circle.

1. Prove the ∠BOT = 2∠BTP.

Solution: Let ∠BTP be x◦ ⇒∴ ∠BAT = x◦ (angle in the alternate segment)

∠BOT = 2x◦ (angle at centre is twice angle at circum. on the same arc)

∴ ∠BOT = 2∠BTP (Q.E.D)

Q.E.D. quod erat demonstrandum ‘which was to be demonstrated’.

2. Prove that ∠ATQ+ ∠RBA+ ∠PBT = 180◦.

Solution: ∠ABT + ∠RBA+ ∠PBT = 180◦ (∠RBP is a straight angle)

But∠ATQ = ∠ABT (angle in the alternate segment)

∴ ∠ATQ+ ∠RBA+ ∠PBT = 180◦.

3. Prove that ∠BPT = 180◦ − 2∠BAT.

Solution: Let∠BAT be α ∴ ∠BTP = α◦ (angle in the alternate segment)

∠PBT = α◦ (angle in the alternate segment)

∴ ∠BPT + α◦ + α◦ = 180◦ (angle sum of a triangle)

∴ ∠BPT = 180◦ − 2α

= 180◦ − 2∠BAT (Q.E.D)


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1.1.8 Further Circle Properties

• The products of intercepts of intersecting chords or secants are equal.

AX ×BX = CX ×DX and EY × FY = GY ×HY

• The square of the length of a tangent is equal to the product of the intercepts of asecant drawn from an external point.

(PT )2 = AP × PB

Exercise 1.1.11 Find the value of each pronumeral, giving reasons.

Solution: TP 2 = AP ×DP, ⇒ AP = TP 2 ÷DP = 122 ÷ 8 = 18 cm

∴ x = 18− 8− 8 = 2 cm

AC × CD = BC × CT, ⇒ CT = AC × CD ÷ CB

∴ y = 8× 2÷ 2.5 = 6.4 cm.


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Exercise 1.1.12 For the diagram given below find:

1. PT if AB = 9 cm and BP = 3 cm

Solution: AB = 9 andBP = 3 cm ⇒ AP = 3 + 9 = 12 cm

AP ×BP = TP 2, ⇒ PT =√3× 12 = 6 cm.

2. AB if BP = 10 cm and PT = 13 cm.

Solution: Since PT 2 = AP ×BP BP = PT 2 ÷BP

BP = 132 ÷ 10 = 16.9 cm ⇒ AB = 16.9− 10 = 6.9 cm.

3. CD if DP = 5 cm, AB = 8cm and BP = 6 cm.

Solution: Since CP ×DP = AP ×BP, CP = DP × AP ÷BP

CP = 6× (8 + 6)÷ 5 = 16.8 cm ⇒ CD = 16.8− 5 = 11.8 cm

4. EG if GF = 20 cm, CG = 30 cm and DG = 25 cm.

Solution: Since EG×GF = CG×DG ⇒ EG = GF × CG÷DG

∴ EG = 25× 30÷ 20 = 37.5 cm.

5. CD if CG = 16 cm EF = 38 cm and EG = 22 cm.

Solution: Since EG×GF = CG×DG ⇒ DG = GF × CG÷ EG

DG = (38− 22)× 22÷ 16 = 22 cm ⇒ CD = 22 + 16 = 38 cm

6. CD if TP = 18 cm and DP = 9 cm.

Solution: TP 2 = CP ×DP, ⇒ CP = 182 ÷ 9 = 36, ⇒ CD = 36− 9 = 27 cm.


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Exercise 1.1.13

1. AB is the diameter of a circle. AB bisects a chord CD at the point E. Find the length of CE if

AE = 9 cm and BE = 4 cm.

Solution: Because the product of intercepts are equal:

AE ×BE = CE ×DE

Since AB bisects CD ∴ CE = DE

CE2 = AE ×BE

CE =√AE ×BE

CE =√4× 9

CE = 6 cm

2. PT is a direct common tangent of the circles drawn. AB is a common chord that has been produced

to meet the common tangent at C. Use the ‘quare of the tangent’of result to prove that CP = CT .

Solution: By using the square of the tangent: CT 2 = AC ×BC

and PC2 = AC ×BC

Since AB is a common chord ∴ CT 2 = PC2

∴ CP = CT.


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1.2 Pass Exam Questions

Exercise 1.2.1 ADB is a straight line with AD = a and DB = b . A circle is drawn on AB asdiameter. DC is drawn perpendicular to AB to meet this circle at C.

1. Show that4ADC|||4CDB, and hence show that DC =√ab.

Solution: Let O be the center of the circle, and r be its radius

consider the rectangular triangles ACB and CDB.

These triangle have the common angle ∠CAB.

Hence 4ACB are similar 4CDB.

From hereAD

CD=CD

BD⇒ CD2 = AD ×BD;

AD = a, BD = b, ⇒ ∴ CD =√ab.

2. Deduce geometrically that if a > 0 and b > 0, then√ab ≤ a+b

2.

Solution: AB = 2r ⇒ r =a+ b

2;

OC = r and the triangle is rectangular

∴ ⇒ OC > CD ⇒ r >√ab

∴a+ b

2>√ab.


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Exercise 1.2.2 Let A is a point outside a circle, AP is the tangent of the circle, P is the point ofcontact. ABC is a secant. M is a midpoint of BC.

1. Prove that APOM is a cyclic quadrilateral.

Solution: As AP is a tangent ⇒ ∴ OP ⊥ AP.

Since M is the midpoint of the chord BC, ⇒ ∴ OM ⊥ BC

∴ ∠OPA+ ∠OMA = 180◦.

∴ the opposite angles of the quadrilateral are supplementary.

Therefore the quadrilateral APOM is a cyclic quadrilateral.

2. Find the angel sum of ∠OAM and ∠APM .

Solution: As proven above the quadrilateral APOM is a cyclic quadrilateral.

∴ ∠OAM = ∠OPM

Since OP ⊥ AP, ⇒ ∴ ∠OPM + ∠APM = 90◦,

∴ ∠OAM + ∠APM = 90◦.


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Exercise 1.2.3

1. ABC is a triangle. The internal bisectors of ∠ABC and ∠ACB meet at D. DP , DQ and DP

are the perpendiculars from D to BC, CA and AB respectively. Show that DR = DQ and

deduce that the internal bisectors of the three angles of a triangle are concurrent.

Solution: The rectangular triangles BDR and BDP have a common side

and ∠DBR = ∠DBP .

Hence these triangles are congruent.

Therefore DR = DP .

Similarly, the rectangular triangles CDQ and CDP are congruent.

Hence DQ = DP .

We have DR = DP and DQ = DP .

Consequently, DR = DQ.

The rectangular triangles ADP and ADR are congruent,

since they have a common side AD and DQ = DR.

Hence ∠DAQ = ∠DAQ and so AD is the internal bisector of ∠BAC.

2. ABC is a triangle. E and F are the midpoints of AC and AB respectively. BE and CF

intersects at D. Show that the triangle DEF and DBS are similar and hence deduce that the

three medians of a triangle are concurrent.

Solution: Since AE = 12AC, AF = 1

2AB and ∠CAB = ∠EAF ,

the triangles ABC and AEF are similar.

Hence EF = 12CB and these triangles have equal angles.

From here ∠AEF = ∠ACB ⇒ EF ||CB.

Hence ∠FEB = ∠CBE and ∠EFC = ∠BCF .

Therefore4DEF is similar to4DCB ⇒ DEDB

= FDDC

= EFCB

.

But it was shown that EFCB

= 12.

Hence ED = 12

and FD = 12DC ,

(i.e., the point D of intersection of the medians BE and CF divide each

median in accordance with the relation 2 : 1 starting with the top. As it is true for

any pair of the medians, all the medians intersect at a single point).


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Exercise 1.2.4 In the diagram, BD is the diameter of the circle, E is a point on the circle , Point Ais the intersect of CE produced and and BD produced. BC ⊥ AC . and ∠CBE = ∠DBE. Provethat AC is a tangent of the circle.

Solution: Joint the points O,E, ⇒ OE = OB ∴ ∠OEB = ∠OBE.

∵ ∠CBE = ∠DBE ⇒ ∴ ∠CBE = ∠OEB, ⇒ ∴ OE||BCBut BC ⊥ AE, ⇒∴ OE ⊥ AC, ⇒∴ AC is the tangent of the circle.

Exercise 1.2.5 ABC is a triangle inscribed in a circle. P is a point on the minor arc AB. Thepoints L,M, and N are the feet of the perpendiculars from P to CA produced, AB, and BC

respectively. Show that L, M and N are collinear. (The line NL is called the Simpson line.)

Solution: In order to prove that L, M and N are collinear, it is sufficient to show that:

∠LMA = ∠NMB.

For this purpose we hsow , that ∠NMB = ∠BPN = ∠SPA = ∠LMA.

The first step: ∠NMB = ∠BPN .

The4PKM and4BKN are rectangular and

∠PKM = ∠BKN ⇒ 4PKM are similar4BKN ⇒ KBPK

= NKMK

.

But ∠PKB = ∠MKN ⇒ 4PKB|||4MKN ⇒ ∠NMB = ∠BPN.

The second step: ∠BPN = ∠SPA.

The point P lies on the circle⇒ PACB is cyclic quadrilateral

∴ ⇒ ∠PAC + ∠PBC = 180◦.

But ∠PAC + ∠PAL = 180◦.

Hence ∠PBC = ∠PAL.

From here as the4PNB and4OLA are rectangular,

we have4PNB|||4PLA ⇒ ∴ ∠BPN = ∠APL.

The third step: ∠SPA = ∠LMA it is obvious that4ALS|||4PAL,

As these rectangular triangles have the common angle ∠PSM.

Hence PSAS

= MSLS⇒ 4MLS|||4PAS ⇒ ∠SPA = ∠LMA.


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Exercise 1.2.6 The diagram shows that AB is a chord of the circle and CD is a tangent andmeeting the circle at P . AC ⊥ CD, BD ⊥ CD and PQ ⊥ AB. Prove that PQ2 = AC ×BD.

Solution: Joint PA and PB as shown about.

∵ CD meets the circle at P ⇒ ∴ ∠α = ∠β.

∵ AC ⊥ CD, PQ ⊥ AB, ⇒ ∴ ∠ACP = ∠PQB = 90◦,

∴ 4ACP |||4PQB, ⇒ ∴ ACPQ

= APBP

.

Similarly: PQBD

= APBP

.

∴ ACPQ

= PQBD, ⇒ ∴ PQ2 = AC ×BD.

Exercise 1.2.7 ABC is a triangle inscribed in a circle. AD bisects the ∠BAC and produced to E,and E is lies on the circle.

1. Prove that4ABE|||4ADC.

Solution: ∠BAE = ∠CAE = ∠CAD (angles on the same arc CE)

∠AEB = ∠ACB (angles on the same arc CE)

∴ 4ABE|||4ADC.

2. If the area of the triangle ABC is given by A = 12AD × AE, find the size of the ∠BAC.

Solution: As4ABE|||4ADC ⇒ ABAE

= ADAC, ⇒ AB × AC = AD × AE.

But A = 12AB × AC sin∠BAC andA = 1

2AD × AE

∴ AB × AC sin∠BAC = AD × AE.Hence sin∠BAC = 1 ⇒ ∠BAC = 90◦.


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Exercise 1.2.8

1. Given that E lies on the circle and AB is the diameter of the circle. MN is a tangent passes

through E. AM ⊥MN, NB ⊥MNandED ⊥ AB. prove that MD ⊥ ND.

Solution: Jointing AE and BE, ∵ AM ⊥MN, ED ⊥ AB,

∴ ⇒ ∠AME + ∠ADE = 180◦ ⇒ ∠MDA = ∠MEA,

But MN is a tangent ∴ ⇒ ∠MEA = ∠ABE, ∠MDA = ∠ABE.

Similarly ∠BDN = ∠BAE.

∵ AB is the diameter,∴ ⇒ ∠AEB = 90◦,

∴ ∠ABE + ∠BAE = 90◦, ⇒ ∠ADM + ∠BDN = 90◦,

∴ ∠MDN = 90◦, ⇒ ∴ MD ⊥ ND.

2. AB is the diameter of a circle, A tangent passes through D produced and meets the AB produced

at C. If DA = DC, prove that AB = 2BC.

Solution: Jointing OE, ⇒ OB = OE,

∵ ∠AED = ∠OBE

∵ BD is a diameter, ∴ ∠EBD = 90◦, ⇒ ∠OEB + ∠OED = 90◦,

∴ ∠AED + ∠OED = 90◦, ⇒ ∴ OE ⊥ AE,

Therefore AE is the tangent of the circle.


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Cyclic Quadrilaterals:

• Opposite angles of a cyclic quadrilateral are supplementary.

• An exterior angle of a cyclic quadrilateral equals the opposite interior angle.

Exercise 1.2.9 ABC is a triangle inscribed in a circle. AB = AC, point D lies on the circle,produced BD to E.

1. Prove that AD produced bisects the ∠CDE.

Solution: Produced AD to F, ⇒ ∠CDF = ∠ABC (An exterior angle of a cyclic quadrilateral)

and AB = AC, ⇒ ∠ABC = ∠ACB,

∵ ∠ADB = ∠ACB, ∴ ∠ADB = ∠CDF

Since∠EDF = ∠ADB, ⇒ ∴ ∠EDF = ∠CDF

∴ AD produced bisects ∠CDE

2. If ∠BAC = 30◦, It4ABC, if the perpendicular height of base BC is 2 +√3 cm, find the area

of the circle inscribed the4ABC.

Solution: Joint AO and produced to H, ⇒ AH ⊥ BC (height of an isosceles triangle )

Joint OC ⇒ ∠OAC = ∠OCA = 15◦, ⇒ ∠ACB = 75◦

∴ ∠OCH = 60◦

Let r be the radius of the circle, ⇒ r +√32r = 2 +

√3 ⇒ r = 2units.

Therefore the area of the circle is 2πr = 4π units2.


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