# Tangents and Intersecting Chords Chapter-18 Concise Maths Solutions

## Tangent Properties of Circles Concise Mathematics Solutions

Tangents and Intersecting Chords Chapter-18 Concise Maths Solutions. Tangents and Chords properties of Circles for Concise Maths Solutions. Solutions of Exercise – 18 (A), Exercise – 18 (B), Exercise – 18 (C) for Concise Selina Maths of ICSE Board Class 10th. Concise Maths Solutions Tangents and Intersecting Chords Chapter-18 for ICSE Maths Class 10 is available here.

## Tangents and Intersecting Chords Chapter-18 Concise Maths Solutions

All Solutions of Concise Selina Maths of Tangents and Intersecting Chords Chapter-18 has been solved according instruction given by council. This is the  Solutions of Tangents and Intersecting Chords Chapter-18 for ICSE Class 10th. ICSE Maths text book of Concise is In series of famous ICSE writer in maths publications. Concise is most famous among students

### Concise Maths Solutions Chapter-18 Tangents and Intersecting Chords

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Exercise – 18 (A), Exercise – 18 (B), Exercise – 18 (C)

Concise Solutions Chapter-18 Tangents and Intersecting Chords

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### EXERCISE -18 (A), Tangents and Intersecting Chords Concise Solutions for ICSE Maths Class 10

#### Question 1

The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre.

OP = 10 cm,
∵ OT ⊥ PT
∴ In right ΔOTP,
OP= OT2+PT2
⇒  (10)2 =(8)2+PT2
and ⇒  100 = 64+PT2
⇒ PT2 = 100-64 = 36
∴ PT = $\sqrt{36}$ = 6 cm

#### Question 2.

In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)
and ⇒  OB2 = OA2 – AB2 ⇒ r2 = (r + 7.5)2 – 152
so ⇒ r2 = r2 + 56.25 + 15r – 225 168.75
hence ⇒   15r= 168.75
⇒  r = $\frac { 168.75 }{ 75 }$ ⇒ r=11.25
Hence, radius of the circle = 11.25 cm

#### Question 3.

Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Given: Two circles with centre O and O’ touches at P externally. Q is a point on the common tangent through P.
QA and QB are tangents from Q to the circles respectively.
To Prove: QA=QB.
Proof: From Q, QA and QP are the tangents to the circle with centre O
∴  QA=QP ….(i)
Similarly, QP and QB are the tangents to the circle with centre O’
∴ QP=QB   ….(ii)
From (i) and (ii)
QA=QB                           Q.ED.

#### Question 4.

Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent, are equal in length.

Given: Two circles with centre O and O’ touch each other internally at P. Q is a point on the common tangent through P. QP an QB are tangents from Q to the circles respectively.

#### Question 5.

Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Given: Two concentric circles with radius 5 cm and 3 cm with centre O. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.
OL=3 cm, OP = 5 cm

#### Question 6.

Three circles touch each other externally. A triangle is formed when the centres of these circles are j oined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.

Three circles touches each other externally
Δ ABC is formed by joining the centres A, B and C of the circles.
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centres A, B and C be r1, r2, r3 respectively

#### Question 7.

If the sides of a quadrilateral ABCD touch a circle, prove that: AB + CD = BC + AD.

Given: A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q,R and S respectively.

To Prove: AB + CD = BC + AD
Proof: Since AP and AS are the tangents to the circle from external point A.
∴ AP = AS     ….(i)
Similarly, we can prove that,
BP=BQ               ….(ii)
CR=CQ             …(iii)
DR=DS          ….(iv)
AP + BP + CR + DR = AS + BQ + CQ + DS
and AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence AB + CD = BC + AD.     Q.E.D.

#### Question 8.

If the sides of a parallelogram touch a circle (refer figure of Q/7) prove that the parallelogram is a rhombus.

Given : The sides AB, BC, CD and DA of ||gm ABCD touches the circle at P, Q, R and S respectively.

To Prove : ABCD is a rhombus.
Proof : From A, AP and AS are the tangents to the circle.
∴ AP = AS    ….(i)

#### Question QUESTION 9.

From the given figure, prove that:
AP + BQ + CR = BP + CQ + AR.

Given: In the figure, sides of Δ ABC touch a circle at P, Q, R.
To Prove:
(i) AP + BQ + CR = BP + CQ + AR
(ii) AP + BQ + CR = $\frac { 1 }{ 2 }$ Perimeter of Δ ABC.
Proof :
∵  From B, BQ and BP are the tangents to the circle.
∴ BQ = BP   ….(i)
Similarly we can prove that
AP= AR    ……… (ii)
and CR = CQ  …..(iii)
AP + BQ + CR = BP + CQ + AR ….(iv)
Adding AP + BQ + CR both sides,
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR+CR.
⇒  2 (AP + BQ + CR) = AB + BC + CA
∴ AP + BQ + CR = $\frac { 1 }{ 2 }$ (AB + BC + CA) $\frac { 1 }{ 2 }$ Perimeter of Δ ABC.            Q.E.D.

#### Question 10.

In the figure of Q.9 if AB = AC then prove that BQ = CQ.

Given:  A circle touches the sides AB, BC, CA of Δ ABC at P, Q and R respectively, and AB = AC

#### Question 11.

Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centres if
(i) they touch each other externally,
(ii) they touch each other internally.

Radius of bigger circle = 6.3 cm

and raduis of smaller circle = 3.6 cm.
(i) Two circles touch each other at P externally. 0 and O’ are the centres of the circles.
Join OP and OP’
OP = 3.6 cm, O’P = 6.3 cm.
OO’ = OP + O’P = 3.6 + 6.3 = 9.9 cm
(ii) If the circles touch each other internally at P.
OP = 3.6 cm and O’P = 6.3 cm.

∴ OO’ = O’P – OP
= 6.3 – 3.6 = 2.7 cm

#### Question 12.

From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP,
(ii) OP is the ⊥ bisector of chord AB.

Given:  A circle with centre 0. A point P out side the circle. From P, PA and PB are the tangents to the circle, OP and AB are joined.

To prove:
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of chord AB.
Proof : In ∆ AOP and ∆ BOP,
AP = BP (Tangents from P to the circle.)
OP = OP (Common)
OA = OB (Radii of the same circle)
∴ ∆ AOP s ∆ BOP (SSS postulate)
∴∠AOP = ∠BOP (C.P.C.T.)
Now in ∆ OAM and ∆ OBM,
OA = OB (Radii of the same circle)
OM = OM (Common)
∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)
∴ ∆ OAM = ∆ OBM (S.A.S. Postulate)
∴ AM = MB (C.P.C.T.)
and ∠OMA = ∠OMB (C.P.C.T.)
But ∠OMA + ∠OMB = 180° (Linear pair)
∴ ∠OMA = ∠OMB = 90°
Hence OM or OP is the perpendicular bisector of AB. Q.E.D.

#### Question 13.

In the given figure, two circles touch each other externally at point P, AB is the direct common tangent of these circles. Prove that:
(i) tangent at point P bisects AB.
(ii) Angle APB = 90°

Given : Two circles with centre O and O’ touch each other at P externally. AB is the direct common tangent touching the circles at A and B respectively.

AP, BP are joined. TPT’ is the common tangent to the circles.
To Prove : (i) TPT’ bisects AB (ii) ∠APB = 90°
Proof :
∵ TA and TP arc the tangents to the circle
∴ TA = TP …(i)
Similarly TP. = TB ….(ii)
From (i) and (ii)
TA = TB
∴ TPT’ is the bisector of AB.
Now in ∆ ATP
TA = TP
∴ ∠TAP = ∠TPA
Similarly in A BTP.
∠TBP = ∠TPB
∠TAP + ∠TBP = ∠APB
But ∠TAP + ∠TBP + ∠APB = 180°
∴ ∠APB = ∠TAP + ∠TBP = 90°. Q.E.D.

#### Question 14.

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ

Given: A circle with centre O. two tangents PA and QA are drawn from a point A out side the circle OP, OQ. OA and PQ are joined.

#### Question 15.

Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.

Given: A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined

#### Question 16.

ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.

Calculate the value of x, the radius of the inscribed circle.

In ∆ ABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and
ON ⊥ AC.

#### Question 17.

In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
(i) ∠ QOR
(ii) ∠ QPR given that ∠ A = 60°.

#### Question 18.

In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60°, calculate:
(i) ∠ QOR
(ii) ∠ OQR
(iii) ∠ QSR.

#### Question 19.

In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.

#### Question 20.

In quadrilateral ABCD; angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle. |1990]

BQ and BR arc the tangenls from B to the circle.

∴ BR = BQ = 27 cm.
∴ RC = 38-27 = 11 cm.
Since CR and CS are the tangents from C to the circle
∴ CS = CR= 11 cm.
∴ DS = 25 – 11 = 14 cm.
DS and DP are the tangents to the circle
∴ DS = DP
∴ ∠ PDS = 90° (given)
and OP ⊥ AD, OS ⊥ DC
∴ Radius = DS = 14 cm

#### Question 21.

In the given, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given ∠ SPR = x° and ∠ QRP = y°;
Prove that
(i) ∠ ORS = y°
(ii) Write an expression connecting x and y. 

∠ QRP = ∠ OSR = y (Angles in the alternate segment)

But OS = OR (radii of the same circle)
and ∴ ∠ ORS = ∠ OSR = y°
so ∴ OQ = OR (radii of the same circle)
hence ∴ ∠ OQR = ∠ ORQ = 90° – y° ….(i) (OR ⊥ PT)
But in ∆ PQR,
Ext. ∠ OQR = x° + y° ….(ii)
from (i) and (ii)
x° + y° = 90° – y°
⇒ x° + 2y° = 90°

#### Question 22.

PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50°; calculate :

#### Question 23.

In the given figure. O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC 80°; find the angle BAC. 

Join OC
∴ PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
∴ ∠ APC + ∠ AOC = 180°
⇒ 80° + ∠ AOC = 180°
∴ ∠ AOC = 180° – 80° = 100°
∠ BOC = 360° – (∠ AOB + ∠ AOC)
= 360° – (140° + 100°)
= 360° – 240° = 120°
Now arc BC subtends ∠ BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠ BAC = $\frac { 1 }{ 2 }$∠BOC= $\frac { 1 }{ 2 }$ x 120° = 60°

#### Question 24.

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°

Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle

### EXERCISE – 18 (B), Tangents and Intersecting Chords Concise Solutions ICSE Maths

#### Question 1.

(i) In the given figure 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

(ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4 cm. Find CD.

(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

#### Question 2.

In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. (2014)

PT is tangent and PDC is secant out to the circle
∴ PT² = PC x PD
PT² = (5 + 7.8) x 5 = 12.8 x 5
PT² = 64 ⇒ PT = 8 cm
In ΔOTP
PT² + OT² = OP²
8²+x² = (x + 4)²
⇒ 64 +x² = x² + 16 + 8x
64- 16 = 8x
⇒ 48 = 8x
x = $\frac { 48 }{ 8 }$ = 6 cm
AB = 2 x 6 = 12 cm

#### Question 3.

In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60°, calculate
(i) ∠ QAB
(iii) ∠ CDB.

(i) PAQ is a tangent and AB is the chord, ∠ QAB = ∠ ADB (Angles in the alternate segment)
= 30°
(ii) OA = OD (Radii of the same circle)
∴ ∠ OAD = ∠ ODA = 30°
But OA ⊥ PQ
∴ ∠ PAD = ∠ OAP – ∠OAD = 90° – 30° = 60°
(iii) BD is diameter
∴ ∠ BCD = 90° (Angle in semi circle)
Now in ∆ BCD,
∠ CDB + ∠ CBD + ∠ BCD = 180°
⇒ ∠ CDB + 60° + 90° = 180°
⇒ ∠ CDB = 180°- (60° + 90°) = 180° – 150° = 30°

#### Question 4.

If PQ is a tangent to the circle at R; calculate:
(i) ∠ PRS
(ii) ∠ ROT.

Given O is the centre of the circle and angle TRQ = 30°.

PQ is tangent and OR is the radius
∴ OR ⊥ PQ
∴ ∠ORT = 90° =
∠TRQ = 90° – 30° = 60°
But in ∆ OTR, OT = OR (Radii of the circle)
∴ ∠ OTR = 60° or ∠STR = 60°
But ∠PRS = ∠STR (Angles in the alternate segment) = 60°
In ∆ ORT, ∠ OTR = 60°, ∠ TOR = 60°
∴ ∠ROT= 180°-(60°+ 60°)= 180°-120° = 60°

#### Question 5.

AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersect AB produced in D. Show that BC = BD.

Given: In a circle, O is the centre,
AB is the diameter, a chord AC such that ∠ BAC = 30°
and a tangent from C, meets AB in D on producing. BC is joined.
To Prove: BC = BD
Construction: Join OC
Proof : ∠ BCD = ∠ BAC =30°(Angle in alternate segment)
Arc BC subtends ∠ DOC at the centre of the circle and ∠ BAC at the remaining part of the circle.
∴ ∠ BOC = 2 ∠ BAC = 2 x 30° = 60°
Now in ∆ OCD,
∠ BOC or ∠ DOC = 60° (Proved)
∠ OCD = 90° (∵ OC ⊥ CD)
∴ ∠ DOC + ∠ ODC = 90°
⇒ 60° + ∠ ODC = 90°
∴ ∠ ODC = 90°- 60° = 30°
Now in ∆BCD,
∵∠ ODC or ∠ BDC = ∠ BCD (Each = 30°)
∴ BC = BD Q.E.D.

#### Question 6.

Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that ∆PQR is isosceles.

In a circumcircle of ∆PQR, a tangent TPS is drawn through P which is parallel to QR
To prove : ∆PQR is an isosceles triangle.
Proof:
∵ TS $\parallel$ QR
∠TPQ = ∠PQR (Alternate angles) ….(i)
∵ TS is tangent and PQ is the chord of the circle
∴ ∠TPQ = ∠RP (Angles in the alternate segment) ….(ii)
From (i) and (ii),
∠PQR = ∠QRP
∴ PQ = PR (Opposite sides of equal angles)
∴ ∆PQR is an isosceles triangle
Hence proved

#### Question 7.

Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.

Given: Two circles with centre O and O’ intersect each other at A and B, O lies on the circumference, of the other circle. CD is a tangent at A to the second circle. AB, OA are joined.
To Prove: OA bisects ∠ BAC.
Construction: Join OB, O’A, O’B and OO’
Proof: CD is the tangent and AO is the chord
∠ OAC = ∠ OBA …(i)
(Angles in alt. segment)
In ∆ OAB, OA = OB (Radii of the same circle)
∴ ∠ OAB = ∠ OBA ….(ii)
From (i) and (ii),
∠ OAC = ∠ OAB
∴ OA is the bisector of ∠ BAC Q.E.D.

#### Question 8.

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠ CPA = ∠DPB.

Given:Two circles touch each other internally at P. A chord AB of the bigger circle intersects the smaller circle at C and D. AP, BP, CP and DP are joined.
To Prove: ∠ CPA = ∠ DPB
Construction: Draw a tangent TS at P to the circles given.
Proof:
∵ TPS is the tangent, PD is the chord.
∴ ∠ PAB = ∠ BPS …(i) ( Angles in alt. segment)
Similarly we can prove that
∠ PCD = ∠ DPS …(ii)
Subtracting (i) from (ii), we gel
∠ PCD – ∠ PAB = ∠ DPS – ∠ BPS
But in ∆ PAC,
Ext. ∠ PCD = ∠ PAB + ∠ CPA
∴ ∠ PAB + ∠ CPA – ∠ PAB = ∠ DPS – ∠ BPS
∠ CPA = ∠ DPB           Q.E.D.

#### Question 9.

In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.

Given: ABCD is a cyclic quadrilateral and diagonal AC bisects ∠ BCD. AT A. a tangent TAS is drawn. BD is joined.
To Prove: TS || BD.
Proof: ∠ ADB = ∠ ACB …….(i) (Angles in the same segment)
Similarly ∠ ABD = ∠ ACD ……..(ii)
But ∠ ACB = ∠ ACD (AC is the bisector of ∠ BCD)
∴ ∠ ADB = ∠ ABD |From (i) and (ii)]
TAS is a tangent and AB is chord
∴ ∠ BAS = ∠ ADB (Angles in all segment)
But ∠ ADB = ∠ ABD (Proved)
∴ ∠ BAS = ∠ ABD
But these are alternate angles.
∴ TS || BD.      Q.E.D.

#### Question 10.

In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠ BCG = 108° and O is the centre of the circle,
Find:
(i) angle BCT
(ii) angle DOC

Join OC, OD and AC.
(i) ∠ BCG + ∠ BCD = 180° (Linear pair)
⇒ 108° + ∠ BCD = 180°
(∵∠ BCG = 108° given)
∴∠ BCD = 180° – 108° = 72°
BC = CD (given)
∴ ∠ DCP = ∠ BCT
But ∠ BCT + ∠ BCD + ∠ DCP = 180°
∴∠ BCT + ∠ BCT + 72° = 180°
(∵∠ DCP = ∠ BCT)
2 ∠ BCT = 180° – 72° = 108°
∴∠ BCT = $\frac { { 108 }^{ \circ } }{ 2 }$ = 54°
(ii) PCT is the tangent and CA is chord
∴ ∠ CAD = ∠ BCT = 54°
But arc DC subtends ∠ DOC at the centre and
∠ CAD at the remaining part of the circle
∴ ∠ DOC = 2 ∠ CAD = 2 x 54° = 108°.

#### Question 11.

Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T arc concyclic.

Given: Two circles intersect each other at point A and B. PAQ is a line which intersects circles at P, A and Q. At P and Q, tangents are drawn to the circles which meet at T.
To Prove: P, B, Q, T are concyclic.

Construction: Join AB, BP and BQ.
Proof: TP is the tangent and PA. a chord
∴ ∠ TPA = ∠ ABP
(angles in alt. segment)
Similarly we can prove that
∠ TQA = ∠ ABQ …,(ii)
Adding (i) and (ii), we get
∠ TPA + ∠ TQA = ∠ ABP + ∠ ABQ
But in ∆ PTQ,
∠ TPA + ∠ TQA + ∠ PTQ = 180°
and ⇒ ∠ TPA + ∠ TQA = 180° – ∠ PTQ
so ⇒ ∠ PBQ = 180°- ∠ PTQ
hence ⇒ ∠ PBQ + ∠PTQ = 180°
But there are the opposite angles of the quadrilateral
∴ Quad. PBQT is a cyclic
Hence P, B. Q and T are concyclic     Q.E.D.

#### Question 12.

In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC.
Show that triangle PAD is an isosceles triangle. Also shaw that:

Given: In a circle PA is the tangent, PBC is the secant and AD is the bisector of ∠BAC which meets the secant at D.
To Prove:
(i) ∆ PAD is an isosceles triangle.
(ii) ∠CAD = $\frac { 1 }{ 2 }$ [(∠PBA – ∠PAB)]
Proof:
(i) PA is the tangent and AB is chord.
∠PAB = ∠C ….(i)
(Angles in the alt. segment)
AD is the bisector is ∠BAC
∴ ∠1 = ∠2 ….(ii)
Ext. ∠ADP = ∠C + ∠1
= ∠PAB + ∠2 = ∠PAD
∴ ∆ PAD is an isosceles triangle.
(ii) In A ABC,
Ext. ∠PBA = ∠C + ∠BAC
∴∠BAC = ∠PBA – ∠C
and ⇒ ∠1 + ∠2 = ∠PBA – ∠PAB [from (i)]
so ⇒ 2 ∠1 = ∠PBA – ∠PAB
therefore ⇒ ∠1 = – [∠PBA – ∠PAB]
⇒ ∠CAD = – [∠PBA – ∠PAB] Q.E.D.

#### Question 13.

Two circles intersect each other at points A and B. Their common tangent touches the circles v at points P and Q as shown in the figure . Show that the angles PAQ and PBQ arc supplementary. 

Given: Two circles intersect each other at A and
B. A common tangent touches the circles at P and
Q. PA. PB, QA and QB are joined.
To Prove: ∠ PAQ + ∠ PBQ = 180°
or ∠ PAQ and ∠ PBQ are supplementary.
Construction: Join AB.
Proof: PQ is the tangent and AB is the chord
∴ ∠ QPA = ∠ PBA (alternate segment) ,…(i)
Similarly we can prove that
∠ PQA = ∠ QBA ,…(ii)
Adding (i) and (ii), we get
∠ QPA + ∠ PQA = ∠ PBA + ∠ QBA
But ∠ QPA + ∠ PQA = 180° – ∠ PAQ ,…(iii) (In ∆ PAQ)
and ∠ PBA + ∠ QBA = ∠ PBQ ,…(iv)
from (iii) and (iv)
∠ PBQ = 180° – ∠ PAQ
⇒ ∠ PBQ + ∠ PAQ = 180°
= ∠ PAQ + ∠ PBQ = 180°
Hence ∠ PAQ and ∠ PBQ arc supplementary Q.E.D.

#### Question 14.

In the figure, chords AE and BC intersect each other at point D.
(i) If ∠ CDE = 90°.
AB = 5 cm, BD = 4 cm and CD 9 cm;
Find DE.
(ii) If AD = BD, show that AE = BC.

#### Question 15.

Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles arc congruent; show that CE = BD.

Given: Two circles with centre P and Q intersect each other at A and B. CBD is a line segment and EBM is tangent to the circle with centre Q, at B. Radii of the cirlces are equal.
To Prove: CE = BD
Proof: EBM is the tangent and BD is the chord
∴ ∠ DBM = ∠ BAD (Anglesi in alt. segment)
But ∠ DBM = ∠ CBE (Vertically opposite angles)
∴ ∠ BAD = ∠ CBE
∵ In the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
∴ CE = BD Q.E.D.

#### Question 16.

In the following figure O is the centre of ti.e circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO

Given: ∠BDC = 65° and AB is tangent to circle with centre O.
⇒ OB ⊥ AB
In ∆BDC
∠DBC + ∠BDC + ∠B CD = 180°
90° + 65° + ∠BCD =180°
⇒ ∠BCD = 25°
∵ OE = OC = radius
⇒ ∠OEC = ∠OCE
⇒ ∠OEC = 25°
Also, ∠BOE = ∠OEC + ∠OCE
[Exterior angle = sum of opposite interior angles in a ∆]
⇒ ∠BOE = 25°+ 25°
⇒∠BOE = 50°
⇒ ∠BOA = 50°
In ∆AOB
∠AOB + ∠BAO + ∠OBA = 180°
50° + ∠BAO + 90° = 180°
⇒ ∠BAO = 40°

### EXERCISE – 18 (C), Concise Maths Solutions Tangents and Intersecting Chords

#### Question 1.

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Given: In circle with centre O and radius r.
OM ⊥ AB and ON ⊥ CD and AB > CD
To Prove: OM < ON
Construction: Join OA, OC

#### Question 2.

OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(i) If the radius of the circle is 10 cm, find the area of the rhombus.
(ii) If the area of the rhombus is 32√3 cm² find the radius of the circle.

#### Question 3.

Two circles with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

Two circles with centres A and B touch each other at C internally.
PQ is the perpendicular bisector of AB meeting the bigger circles at P and Q. Join AP.
and radius BC = 3 cm.
AB = AC – BC = 5-3 = 2 cm.

#### Question 4.

Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of angle BAC.

Given: A circle in which two chords AC and AB are equal in length. AL is the bisector of ∠ BAC.
To Prove: O lies on the bisector of ∠ BAC

AB = AC (Given)
∴ BD = DC (C.P.C.T.)
∴ AD is the perpendicular bisector of chord BC.
∵ The perpendicular bisector of a chord passes through the centre of the circle.
∴ AD is the bisector of ∠ BAC passes through the centre O of the circle. Q.E.D.

#### Question 5.

The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle ?

AB is the diameter and AC is the chord
∴ AB = 20 cm and AC = 12 cm
Draw OL ⊥ AC
∵ OL ⊥ AC and hence it bisects AC, O is the center of the circle.

#### Question 6.

ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.

∠ ADC = 110°, ∠ BAC = 50°
∠B + ∠D= 180° (Sum of opposite angles of a cyclic quad.)
⇒∠B + 110°= 180°
∴ ∠ B or ∠ ABC = 180° – 110° – 70″
Now. in ∆ ABC,
∠ BAC + ∠ ABC – ∠ ACB = 180°
⇒ 50° + 70° + ∠ ACB – 180°
⇒ 120° – ∠ ACB = 180°
∴ ∠ ACB = 180° – 120° = 60″
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
∴ ∠ DAC = ∠ ACB (Alternate angles)
= 60°
∠ DAC + ∠ ADC + ∠ DCA -= 180°
60°+ 110° + ∠ DCA = 180°
170° + ∠ DCA – 180°
∴ ∠ DCA = 180″ – 170° = 10°

#### Question 7.

In the given figure, C and D arc points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°. calculate angle BDC.

∴ ABCD is a cyclic quad.
∠ BAD ∠ BCD 180 (Sum of opposite angles)
⇒ 70° + ∠ BCD = 180°
⇒ ∠ BCD = 180°- 70° = 110°
Now in ∆ BCD,
∠ BCD + ∠ DBC + ∠ BDC = 180°
⇒ 30°+ 110° + ∠ BDC = 180°
⇒ 140°+ ∠ BDC = 180°
∴ ∠ BDC = 180°- 140° = 40°

#### Question 8.

In cyclic quadrilateral ABCD, ∠ A = 3 ∠ C and ∠ D = 5 ∠ B. Find the measure of each angle of the quadrilateral.

#### Question 9.

Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Given: ∆ABC in which AB=AC and AB as diameter, a circle is drawn which intersects BC at D.
To Prove: BD = DC
Proof: AB is the diameter
∴ ∠ ADB = 90° (Angle in a semi-circle)
Now in right angled ∆s ABD and ACD,
Hypotenuse AB = AC (given)
∴ ∆ ABD ≅ ∆ ACD (RHS postulate)
BD = DC (C.P.C.T.)
Hence the circle bisects base BC at D. Q.ED.

#### Question 10.

Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that.
angle EDF = 90° – $\frac { 1 }{ 2 }$∠A.

Given: A ∆ ABC whose bisectors of angles A, B and C intersect the circumcircle at D, E and F respectively. ED, EF and DF are joined.
∠ EDF = 90° – $\frac { 1 }{ 2 }$∠A
Construction: Join BF, FA, AE and EC.
Proof: ∠ EBF = ∠ ECF = ∠ EDF ……..(i)
(Angles in the same segment)

Question 11.

In the flgure; AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠ C = 20°, find angle AOD.

Join OB
In ∆ OBC, BC = OD = OB (radii of the circle)
∴ ∠ BOC = ∠ BCO = 20°
and Ext. ∠ ABO = ∠ BCO + ∠ BOC
= 20°+ 20° = 40° ….(i)
In ∆ OAB, OA = OB (radii of the circle)
∴ ∠ OAB = ∠ OBA = 40° [from (i)]
∠ AOB = 180°- ∠ OAB – ∠ OBA
= 180°-40°-40°= 100°
∵ DOC is a line
∴ ∠ AOD + ∠ AOB + ∠ BOC = 180°
⇒ ∠ AOD + 100° + 20° = 180°
⇒ ∠ AOD + 120° = 180°
∴ ∠ AOD = 180° – 120° = 60°

#### Question 12.

Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Given: In ∆ ABC, ∠ B = 90° which is inscribed in a circle and O is the incentre of the incircle of ∆ABC.
D and d are the diameters of circumcircle and incircle of ∆ ABC.
To Prove: AB +BC + CA = 2D + d.
Construction: Join OL, OM and ON.
Proof: In circumcircle of ∆ ABC,
∠ B = 90° (given)
∴ AB is the diameter of circumcircle i.e. AB = D.
Let radius of incircle = r
∴ OL = OM = ON = r
Now from B, BL, BM are the tangents to the incircle
∴ BL = OM = r
Similarly we can prove that:
AM = AN and CL = CN = R (radius)
(Tagents from the point outside the circle)
Now AB + BC + CA = AM + BM + BL + CL + CA
and = AN + r + r + CN + CA
so = AN + CN + 2r + CA
hence = AC + AC + 2r = 2 AC + 2r = 2D + d Q.E.D.

#### Question 13.

P is the mid point of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

Given: A circle with centre O, AB is an arc whose mid point is P and AB is chord. TPS is the tangent at P.
To Prove: TPS ||AB.
Construction: Join AP and BP.
Prove: TPS is tangent and PA is chord of the circle

∠ APT = ∠ PBA (angles in the alternate segment)
But ∠ PBA = ∠ PAB (∵ PA = BP)
∴ ∠ APT = ∠ PAB
But these are alternate angles
∴ TPS || AB Q.E.D

#### Question 14.

In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent Prove that the line NM produced bisects AB at P.

Prove that the line NM produced bisects AB at P.

Given: Two circles intersect each other at M and N. AB is their common tangent, chord MN intersect the tangent at P.
To Prove: P is mid point of AB.
Proof: From P, AP is the tangent and PMN is the secant of first circle.
∴ AP2 = PM x PN ….(i)
Again from P, PB is the tangent and PMN is the secant of the second circle.
PB2 = PM x PN ….(ii)
from (i) and (ii)
AP2 = PB2 ⇒ AP = PB
∴ P is the mid point of AB. Q.E.D.

#### Question 15.

In the given figure, ABCD is a cyclic- quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠ DCQ = 40° and ∠ ABD = 60°, find :
(i) ∠ DBC
(ii) ∠ BCP

(i) PQ is the tangent and CD is the chord
∴ ∠ DCQ = ∠ DBC (angles in the alternate segment)
∴ ∠ DBC = 40° (∵ ∠ DCQ = 40°)
(ii) ∠ DCQ + ∠ DCB + ∠ BCP = 180°
⇒ 40° + 900 + ∠ BCP = 180°(∵ ∠ DCB = 90°)
⇒ 130°+ ∠ BCP = 180°
∵ ∠ BCP =180° -130° = 50°
(iii) In Δ ABD, ∠ BAD = 90° (Angle in a semi circle) and ∠ ABD = 60°
∴ ∠ ADB = 180°- (60° + 90°)
⇒ 1800- 150° = 30°

#### Question 16.

The given figure shows a circle with centre O and BCD is tangent to it at C. Show that:
∠ ACD + ∠ BAC = 90°.

Given: A circle with centie O and BCD is a tangent at C.

To Prove: ∠ ACD + ∠ BAC = 90°
Construction: Join OC.
Proof: BCD is the tangent and OC is the radius
∴ OC ⊥ BD
⇒ ∠ OCD = 90°
⇒ ∠ OCA + ∠ ACD = 90° ….(i)
But in ∆ OCA
OA = OC (radii of the same circle)
∴ ∠ OCA = ∠ OAC [from (i)]
∠OAC + ∠ACD = 90°
⇒ ∠ BAC + ∠ ACD = 90° Q.E.D.

#### Question 17.

ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that
(i) AC x AD = AB2
(ii) BD2 = AD x DC.

Given: A circle with BC as diameter meets the hypotenuse of right ∆ ABC with ∠ B = 90° meets at D. BD is joined.
To Prove:
(i) AC x AD = AB2
(ii) BD2 = AD x DC
Proof:
(i) In ∆ABC, ∠ B = 90° and BC is the diameter of the circle.

#### Question 18.

In the given figure. AC = AE.
Show that :
(i) CP = EP
(ii) BP = DP.

Given: In the figure, AC = AE
To Prove: (i) CP = EP (ii) BP = DP
Proof: In ∆ ADC and ∆ ABE,
AC = AE (given)
∠ ACD = ∠ AEB (Angles in the same segment)
∠ A = ∠ A (Common)
∆ ADC ≅ ∆ ABE (ASA postulate)
But AC = AE (given)
∴ AC – AB = AE – AD
⇒ BC = DE
Now in ∆ BPC and ∆ DPE,
BC = DE (proved)
∠ C = ∠ E (Angles in the same segment)
∠ CBP = ∠ CDE (Angles in the same segment)
∴ ∆ BPC ≅ ∆ DPE (S.A.S. postulate)
∴ BP=DP (C.P.C.T.)
CP = PE (C.P.C.T.) Q.E.D.

#### Question 19.

ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.
Calculate :
(i) ∠ BEC
(ii) ∠ BED

(i) In cyclic pentagon, O is the centre of circle.
Join OB, OC
AB = BC = CD (given)
and ∠ ABC = 120°.
∴ ∠ BCD = ∠ ABC = 120°
OB and OC are the bisectors of ∠ ABC and ∠ BCD respectively.
∴ ∠ OBC = ∠ BCO = 60°
In ∆ BOC,

#### Question 20.

In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO=30°, find
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB

Given: In the fig. O is the centre of the circle CA and CB are the tangents to the circle from C. ∠ACO = 30°
P is any point on the circle. PA and PB are joined.
To find:
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Proof:
(i) In ∆ OAC and ∆OBC,
OC=OC (common)
OA = OB (radius of the circle)
CA = CB (tangents to the circle)
∴ ∆OAC ≅ ∆OBC (SSS axion)
∴ ∠ACO = ∠BCO = 30°
(ii) ∴ ∠ACB = 30° + 30° = 60°
∴ ∠AOB + ∠ACB = 180°
⇒ ∠AOB+ 60° =180°
∴ ∠AOB = 180° – 60° = 120°
(iii) Arc AB, subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle
∴ ∠APB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ x 120° = 60°

#### Question 21.

ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. (2011)

Given: ABC is a triangle with AB = 10 cm, BC = 8 cm, AC = 6 cm. Three circle are drawn with centre A, B and C touch each other at P, Q and R respectively

#### Question 22.

In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L.
Show that :
(i) ∠ ONL + ∠ OML = 180°
(ii) ∠ BAM = ∠ BMA
(iii) ALOB is a cyclic quadrilateral

#### Question 23.

The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Join PB
∠ BPQ + ∠ BCQ = 180°
⇒ ∠ BPQ + 140° = 180°
∴ ∠ BPQ = 180° – 140° = 40°
Now, in ∆ PBQ,
∠ PBQ + ∠ BPQ + ∠ BQP = 180°
⇒ 90° + 40° + ∠ BQP = 180°
(∠ PBQ = 90° angle in a semicircle)
⇒ 130° + ∠ BQP = 180°
∴ ∠ BQP = 180° – 130°= 50°
∠ PQB + ∠ PAB = 180°
⇒ 50° + ∠ PAB = 180°
∴ ∠ PAB = 180° – 50° = 130°
(ii) Now, in ∆ PAB,
∠ PAB + ∠ APB + ∠ ABP = 180°
⇒ 130° + ∠ APB + ∠ ABP = 180°
⇒ ∠ APB + ∠ ABP = 180° – 130° = 50°
But ∠ APB = ∠ ABP = 25°
∴ (PA = AB)
∠ BAQ = ∠ BPQ = 40°
(Angles in the same segment)
Now, in ∆ ABQ,
∠ AQB + ∠ QAB + ∠ ABQ = 180°
and ⇒ ∠ AQB + 40° + 115° = 180°
so ⇒ ∠ AQB + 155° = 180°
therefore ⇒ ∠ AQB = 180° – 155° = 25°
(iii) Arc AQ subtends ∠ AOQ at the centre and ∠APQ at the remaining part of the circle,
∠ AOQ = 2 ∠ APQ = 2 x 65° = 130°
Now, in ∆ AOQ,
∠ OAQ = ∠ OQA
(∵ OA = OQ radii of the same circle)
But ∠ OAQ + ∠ OQA + ∠ AOQ = 180°
⇒ ∠ OAQ + ∠ OAQ + 130° = 180°
2 ∠ OAQ = 180° – 130° = 50°
∴ ∠ OAQ = 25°
∵ ∠ OAQ = ∠ AQB (each = 25°)
But these are alternate angles.
∴ AO || BQ.
Q.E.D.

#### Question 24.

The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate
(i) angle QTR
(ii) angle QRP
(iii) angle QRS
(iv) angle STR

#### Question 25.

In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segemcnt, show that :
(i) ∠ BAP = ∠ ADQ
(ii) ∠ AOB = 2 ∠ ADQ

Given: PAT is the tangent to the circle with centre O, at A. Chord BC || PAT is drawn.

CDQ is a line segment which intersects the circle at C and D and meets the tangent PAT at Q.
To Prove:
(i) ∠ BAP = ∠ ADQ (ii)∠ AOB = 2 ∠ ADQ
Proof:
(i) ∵ PAT || BC
∴ ∠ PAB = ∠ ABC (Alternate angles) ….(i)
∴ ∠ PAB = ∠ ADQ (from (i) and (ii))
(ii) Arc AB subtends ∠ AOB at the centre and ∠ADB at the remaining part of the circle.
∴ ∠ AOB = 2 ∠ ADB
= 2 ∠ PAB (In the alt. segment)
= 2 ∠ ADQ [proved in (i)]
(iii) ∵ ∠ BAP = ∠ ADB
(Angles in the alt. segment)
Bui ∠ BAP = ∠ ADQ (Proved in (i))

#### Question 26.

AB is a line segment and M is its mid-point. Three semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semi-circles. Show that: AB = 6 x r

:

Given: A line segment AB whose mid-point is M. Three circles are drawn on AB, AM and MB as diameler A circle with radius r is drawn which touches (he three circles externally at L, R and N respectively. M, P, Q are the centres of the three circles.
To Prove: AB= 6r
Construction: Join OP and OQ.
Proof: OM = ON = r

#### Question 27.

TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.

Given: A circle with centre C. From a point T outside th circle, TA and TB are two tangent to the circle OT intersects the circle at P, AP and AB are joined.
To Prove: AP is the bisector of ∠ TAB
Construction: Join PB.

#### Question 28.

Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.

Given: Two circles intersect each other at P and Q. From P, a secant intersects the circles at A and B respectively. From A and B tangents are drawn which intersect each other at T.
To Prove: A. Q, B and T on a circle.
Construction: Join PQ.
Proof: AT is the tangent and AP is chord
∴ ∠ TAP = ∠ AQP (Angles in all. segment) …(i)
Similarly ∠ TBP = ∠ BQP ….(ii)
∠ TAP + ∠ TBP = ∠ AQP + ∠ BQP = ∠ AQB …..(iii)
Now, in ∆ TAB,
∠ ATB + ∠ TAP + ∠ TBP = 180°
⇒ ∠ ATB + ∠ AQB = 180° (from (iii)
∴ AQBT is a cyclic quadrilateral.
Hence A, Q, B and T lie on the same circle. Q.E.D.

#### Question 29.

Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle).

ABCDE is a regular pentagon.

To Prove: Any four vertices lie on the same circle.
Construction: Join AC.
Proof: Each angle of a regular pentagon

#### Question 30.

Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD. 

Let CD = x
∴ chords AB and CD intersect each other at outside the circle.
∴ AX.XB = CX.XD
and ⇒(4+6)x6 = (x + 5)x5
so ⇒ 10 x 6 = 5x + 25
hence ⇒ 60 = 5x + 25
⇒ 5x = 60 – 25 = 35
∴ x = $\frac { 35 }{ 5 }$ = 7
CD = 7 cm

#### Question 31.

in the given figure. find TP if AT = 16 cm AB = 12 cm.

In the figure.
PT is the tangent and TBA is the secant of the circle.
∴ TP2 = TA x TB = 16 x (16 – 12) = 16 x 4 = 64 = (8)2
Hence, TP = 8 cm.

#### Question 32.

In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm. QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle. (1990)

A circle with centre is inscribed (see the fig.)
in a quadrilateral ABCD. BC = 38 cm, QB = 27cm,
DC = 25 cm and AD ⊥ BC.
Join OP and QS.
and ∵ OP and OS are the radii of the circle
so ∴ OP ⊥ AD and OS ⊥ CD
hence∴ OPDS is a square
∴ OP = OS – DP = DS.
Let length of radius of the circle = r
then DP = DS = r
and ∴ CS = 25 – r
so ∵ EQ = BR = 27 cm (tangents to the circle from B)
therefore ∴ CR = BC – BR = 38 – 27 = 11 cm
Similarly CR = CS
∴ 25 – r = 11 ⇒ r = 25 – 11 = 14
∴ Radius of the circle = 14 cm

#### Question 33.

In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.

If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.

In the above figure,
XY is a diameter of the circle PQ is tangent to the circle at Y.
∠AXB = 50° and ∠ABX = 70°
(i) In ∆AXB,
∠XAB + ∠ABX + ∠AXB = 180° (Angles of a triangle)
and ⇒ ∠XAB + 70° + 50° = 180°
so ⇒ ∠XAB + 120° =180°
threfore ⇒ ∠XAB = 180° – 120° = 60°
But, ∠XAY = 90° (Angle in a semicircle)
∴ ∠BAY = ∠XAY-∠XAB=90°- 60° = 30°
(ii) Similarly ∠XBY=90°(Angle in a semicircle) and ∠CXB = 70°
∴ ∠PBY = ∠XBY-∠XBA =90° – 70° = 20°
∵ ∠BYA = 180° – ∠AXB ( ∵ ∠BYA + ∠AYB = 180°) = 180°- 50° = 130°
∠PYA =∠ABY (Angles in the alternate segment) = ∠PBY = 20°
and ∠PYB = ∠PYA + ∠AYB
= 20° + 130° = 150°
∴ ∠APY = 180°-(∠PYA + ∠ABY)
= 180° -(150° +20°) =180° – 170° = 10°

#### Question 34.

In the given figi QAP is the tangent point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°, find:
(i) ∠ BAP
(ii) ∠ABD
(iv) ∠ BCD

In the given figure, QAP is the tangent to the circle at A and PBD is a B straight line.

#### Question 35.

In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If ∠ CAB = 34°,
find :
(i) ∠CBA
(ii) ∠CQB

In ∆ ABC,
we have ∠ ACB = 90°
[Angle in a semicircle is 90°]
(i) Also ∠ CBA + ∠ CAB + ∠ ACB = 180° [Angle sum property of a ∆ ]
⇒ ∠ CBA =180°- ∠ CAB – ∠ ACB = 180°-34°-90° = 180°-124° = 56°
(ii) CQ is a tangent at C and CB is a chord of the circle.
⇒ ∴ ∠ QCB = ∠ BAC = 34° [Angles in the alternate segments]
∠ CBQ =180°- ∠ ABC [Linear pair]
⇒ ∠ CBQ = 180°- 56° = 124° [From (i)]
In ∆ BCQ, we have
⇒ ∠ CQB = 180° – (∠ QCB + ∠ CBQ) [Angle sum property of a ∆ ]
= 180° -(34° + 124°) = 180°- 158° = 22°
Hence, ∠CQB = 22°

#### Question 36.

In the given figure, A O is the centre of the circle. The tangents at B and D intersect each other at point P.

If AB is parallel to CD and ∠ ABC = 55
find :
(i) ∠ BOD
(ii) ∠ BPD

In the given figure,
O is the centre of the circle AB || CD,
∠ ABC = 55° tangents at B and D are drawn which meet at P.
∵ AB || CD
∴ ∠ ABC = ∠ BCD (Alternate angles)
∴ ∠ ABC = 55° (Given)
(i) Now arc BD subtands ∠ BOD at the centre and ∠ BCD at the remaining part of the circle.
∴ ∠BOD = 2∠BCD = 2 x 55° = 110°
∠OBP = ∠ODP = 90° (∵ BP and DP are tangents)
∴ ∠BOD + ∠BPD = 180°
⇒ 110° +∠BPD =180°
⇒ ∠BPD =180°-110°= 70°
Hence, ∠BOD = 110° and ∠BPD = 70°

#### Question 37.

In the figure given below PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
(i) ∠QOP
(ii) ∠QCP

#### Question 38.

In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.

Given: Two concentric circles with centre O AB and CD are two cords of outer circle which touch the inner circle at P and Q respectively
To prove: AB = CD
Construction : Join OA, OC, OP and OQ
Proof: ∵ OP and OQ are the radii of the inner circle and AB and CD are tangents
∴ OP ⊥ AB and OQ ⊥ CD
and P and Q are the midpoints of AB and CD Now in right AOAP and OCQ,
Side OP = OQ (radii of the inner circle)
Hyp. OA = OC (radii of the outer circle)
∴ ∆OAP = ∆OCQ (R.H.S. axiom)
∴ AP = CQ (c.p.c.t.)
But AP = $\frac { 1 }{ 2 }$ AB and CQ = $\frac { 1 }{ 2 }$ CD
∴ AB = CD Hence proved.

#### Question 39.

In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.

Given that AB = 5 cm , Calculate PQ.

In the figure, two circles with centres P and Q and radii 6 cm and 3 cm respectively
ABC is the common transverse tangent to the two circles. AB = 8 cm
Join AP and CQ
∵ AC is the tangents to the two circles and PA and QC are the radii

#### Question 40.

In the figure, given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

In the figure, a circle with centre O, is the circumcircle of ∆XYZ.
∠XOZ =140° (given)
Tangents from X and Y to the circle meet at T such that ∠XTY = 80°
∵ ∠XTY = 80°
∴ ∠XOY= 180°-80°= 100°
But ∠XOY + ∠YOZ + ∠XOZ = 360° (Angles at a point)
and ⇒ 100°+∠YOZ+ 140o = 360o
so ⇒ 240o+∠YOZ =360°
hence ⇒ ∠YOZ =360°- 240°
⇒∠YOZ =120°
Now arc YZ subtends ∠YOZ at the centre and ∠YXZ at the remaining part of the circle
∴ ∠YOZ = 2 ∠YXZ
⇒ ∠YXZ= $\frac { 1 }{ 2 }$ ∠YOZ ⇒ ∠YXZ = $\frac { 1 }{ 2 }$ x 120° = 60°

#### Question 41.

In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, findAE.

In the given circle,
Chords AE and BC interesect each other at D at right angle i.e., ∠CDE = 90°, AB is joined AB = 5cm, BD = 4 cm, CD = 9 cm
Now we have to find AE.
Let DE=xm
Now in right ∆ABD,
AB= AD2 + BD2 (Pythagoras Theorem)
⇒ 25 = AD2 + 16
⇒ AD2 = 25-16 = 9 = (3)2
∵ Chords AE and BC intersect eachothcr at D inside the circle
∴ AD x DE = BD x DC
⇒ 3 x x = 4 x 9
⇒ x= $\frac { 4x9 }{ 3 }$ = 12cm;
∴ AE=AD + DE = 3 + 12 = 15 cm

#### Question 42.

In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

∴ ∠ABC + ∠ADC = 180°
∠ADC = 180°- 100° = 80°
∠ACD + ∠CDA + ∠D AC =180°
40° + 80° +∠D AC = 180°
∠D AC = 180° – 80° – 40° = 60°
Now ∠DAC = 60°
⇒ ∠DCT = 60° [angle in alt. segment]

#### Question 43.

In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x,y and z. (2015)

In the given figure,
O is the centre of the circle.
SP is tangent ∠SRT =65°.
To find the values of x, y and z
(i) In ∆STR,
∠S = 90° (∵ OS is radius and ST is tangent)
∴ ∠T + ∠R = 90°
⇒ x + 65° = 90°
⇒ x = 90° – 65° = 25°
(ii) Arc CQ subtends ∠SOQ at the centre and
∠STQ at the remaining part of the circle.
∠y = ∠QOS = 2∠T = 2∠x = 2 x 25° = 50°
(iii) In ∆OSP,
∠S = 90°
∴ ∠SOQ or ∠SOP + ∠P = 90°
and ⇒ y+z=90o
so ⇒ 50° + z = 90°
hence ⇒ z = 90°-50° = 40°
x = 25°, y = 50° and z = 40°

### Chapter Wise Concise Maths Solutions for ICSE Class 10th Selina Publications

1. GST ( Goods and Services Tax)
2. Banking
3. Shares and Dividends
4. Linear In equations in One Variable
6. Solving Simple Problems (Based on Quadratic Equations)
7. Ratio and Proportion ( Including Properties and Uses )
8. Reminder and Factor Theorems
9. Matrices
10. Arithmetic Progression
11. Geometric Progression
12. Reflection
13. Section And Mid Point Formula
14. Equation of a Line
15. Similarity (With Applications to Maps and Models )
16. Loci (Locus and its Constructions)
17. Circles
18. Tangents and Intersecting Chords (Currently Open )
19. Constructions ( Circles)
20. Cylinder, Cone and Spheres (Surface Area and Volume )
21. Trigonometrical Identities
22. Heights and Distances
23. Graphical Representation (Histograms and Ogives)
24. Measures of Central Tendency (Mean , Median, Quartiles and Mode)
25. Probability

### Other Subject Chapter-Wise Solutions for ICSE Class 10

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