ICSE Solutions for Selina Concise Chapter 18 Tangents and Intersecting Chords Class 10 Maths
Exercise 18(A)
1. The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.
Solution
Given, a circle with centre O and radius 8 cm.
An external point P from where a tangent is drawn to meet the circle at T.
OP = 10 cm; radius OT = 8 cm
As OT ⊥ PT
In right ∆OTP, we have
OP2 = OT2 + PT2 [By Pythagoras Theorem]
⇒ 102 = 82 + PT2
⇒ PT2 = 100 – 64 = 36
So, PT = 6
∴ length of tangent = 6 cm.
2. In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Solution
Given,
AB = 15 cm, AC = 7.5 cm
Let’s assume the radius of the circle to be ‘r’.
So, AO = AC + OC = 7.5 + r
In right ∆AOB, we have
AO2 = AB2 + OB2 [By Pythagoras Theorem]
⇒ (7.5 + r)2 = 152 + r2
⇒ 56.25 + r2 + 15r = 225 + r2
⇒ 15r = 225 – 56.25
⇒ r = 168.75/ 15
Thus, r = 11.25 cm
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.
Solution
Let Q be the point from which, QA and QP are two tangents to the circle with centre O
So, QA = QP ….(i)
Similarly, from point Q, QB and QP are two tangents to the circle with centre O’
So, QB = QP …(ii)
From (i) and (ii), we have
QA = QB
∴ tangents QA and QB are equal.
Hence, Proved
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.
Solution
Let Q be the point on the common tangent from which, two tangents QA and QP are drawn to the circle with centre O.
So, QA = QP …(1)
Similarly, from point Q, QB and QP are two tangents to the circle with centre O’
So, QB = QP …(2)
From (1) and (2), we have
QA = QB
∴ tangents QA and QB are equal.
Hence, Proved
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.
Solution
Given,
OS = 5 cm and OT = 3 cm
In right triangle OST, we have
ST2 = OS2 – OT2
= 25 – 9
= 16
So, ST = 4 cm
As we know, OT is perpendicular to SP and OT bisects chord SP
Hence, SP = 2×ST = 8 cm
6. Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution
Let ABC be the triangle formed when centres of 3 circles are joined.
Given,
AB = 6 cm, AC = 8 cm and BC = 9 cm
And let the radii of the circles having centres A, B and C be r1, r2 and r3 respectively.
So, we have
r1 + r3 = 8
r3 + r2 = 9
r2 + r1 = 6
Adding all the above equations, we get
r1 + r3 + r3 + r2 + r2 + r1 = 8 + 9 + 6
2(r1 + r2 + r3) = 23
So,
r1 + r2 + r3 = 11.5 cm
Now,
r1 + 9 = 11.5 (As r2 + r3 = 9)
r1 = 2.5 cm
And,
r2 + 6 = 11.5 (As r1 + r3 = 6)
r2 = 5.5 cm
Lastly, r3 + 8 = 11.5 (As r2 + r1 = 8)
r3 = 3.5 cm
∴ the radii of the circles are r1 = 2.5 cm, r2 = 5.5 cm and r3 = 3.5 cm.
7. If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Solution
Let a circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.
As, AP and AS are tangents to the circle from an external point A, we have
AP = AS …(1)
Similarly, we also get
BP = BQ …(2)
CR = CQ …(3)
DR = DS …(4)
Adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
∴ AB + CD = AD + BC
Hence, Proved
8. If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Solution
Let a circle touch the sides AB, BC, CD and DA of parallelogram ABCD at P, Q, R and S respectively.
Now, from point A, AP and AS are tangents to the circle.
So, AP = AS …(1)
Similarly, we also have
BP = BQ ...(2)
CR = CQ …(3)
DR = DS ....(4)
Adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
∴ AB + CD = AD + BC
But AB = CD and BC = AD …(5) [Opposite sides of a parallelogram]
Hence, AB + AB = BC + BC
⇒ 2AB = 2 BC
AB = BC …(6)
From (5) and (6), we conclude that
AB = BC = CD = DA
Thus, ABCD is a rhombus.
9. From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.
Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.
Solution
As from point B, BQ and BP are the tangents to the circle
We have, BQ = BP …(1)
Similarly, we also get
AP = AR …(2)
And, CR = CQ …(3)
Adding (1), (2) and (3) we get,
AP + BQ + CR = BP + CQ + AR …(4)
Now, adding AP + BQ + CR to both sides in (4), we get
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR
⇒ 2(AP + BQ + CR) = AB + BC + CA
∴ we get
AP + BQ + CR = ½ ×(AB + BC + CA)
i.e. AP + BQ + CR = ½ ×perimeter of triangle ABC
10. In the figure, if AB = AC then prove that BQ = CQ.
Solution
As, from point A
AP and AR are the tangents to the circle
So, we have AP = AR
Similarly, we also have
BP = BQ and CR = CQ [From points B and C]
Now adding the above equations, we get
AP + BP + CQ = AR + BQ + CR
⇒ (AP + BP) + CQ = (AR + CR) + BQ
⇒ AB + CQ = AC + BQ …(i)
But, as AB = AC [Given]
∴ from (i)
CQ = BQ or BQ = CQ
11. Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
(i) they touch each other externally.
(ii) they touch each other internally.
Solution
Given,
Radius of bigger circle = 6.3 cm and of smaller circle = 3.6 cm
(i)
When the two circles touch each other at P externally. O and O’ are the centers of the circles. Join OP and O’P.
So, OP = 6.3 cm, O’P = 3.6 cm
Hence, the distance between their centres (OO’) is given by
OO’ = OP + O’P = 6.3 + 3.6 = 9.9 cm
(ii)
When the two circles touch each other at P internally. O and O’ are the centers of the circles. Join OP and O’P
So, OP = 6.3 cm, O’P = 3.6 cm
Hence, the distance between their centres (OO’) is given by
OO’ = OP – O’P = 6.3 – 3.6 = 2.7 cm
12. From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP
(ii) OP is the ⊥ bisector of chord AB.
Solution
(i) In ∆AOP and ∆BOP, we have
AP = BP [Tangents from P to the circle]
OP = OP [Common]
OA = OB [Radii of the same circle]
Hence, by SAS criterion of congruence
ΔAOP ≅ ΔBOP
So, by C.P.C.T we have
∠AOP = ∠BOP
(ii) In ∆OAM and ∆OBM, we have
OA = OB [Radii of the same circle]
∠AOM = ∠BOM [Proved ∠AOP = ∠BOP]
OM = OM [Common]
Hence, by SAS criterion of congruence
ΔOAM ≅ ΔOBM
So, by C.P.C.T we have
AM = MB
And ∠OMA = ∠OMB
But,
∠OMA + ∠OMB = 180o
Thus, ∠OMA = ∠OMB = 90o
∴ OM or OP is the perpendicular bisector of chord AB.
Hence, Proved
Exercise 18(B)
1. (i) In the given figure, 3×CP = PD = 9 cm and AP = 4.5 cm. Find BP.
(ii) In the given figure, 5 x PA = 3×AB = 30 cm and PC = 4cm. Find CD.
(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Solution
(i) As the two chords AB and CD intersect each other at P, we have
AP ×PB = CP×PD
⇒ 4.5×PB = 3×9 [3CP = 9 cm so, CP = 3 cm]
PB = (3×9)/4.5 = 6 cm
(ii) As the two chords AB and CD intersect each other at P, we have
AP×PB = CP×PD
But, 5×PA = 3×AB = 30 cm
So, PA = 30/5 = 6 cm and AB = 30/3 = 10 cm
And, BP = PA + AB = 6 + 10 = 16 cm
Now, as
AP×PB = CP ×PD
⇒ 6×16 = 4×PD
⇒ PD = (6×16)/4 = 24 cm
⇒ CD = PD – PC = 24 – 4 = 20 cm
(iii) As PAB is the secant and PT is the tangent, we have
PT2 = PA×PB
⇒ 12.52 = 10×PB
⇒ PB = (12.5×12.5)/ 10 = 15.625 cm
⇒ AB = PB – PA = 15.625 – 10 = 5.625 cm
2. In the given figure, 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.
Solution
(i) PA = AB + BP = (AB + 4) cm
PC = PD + CD = 5 + 7.8 = 12.8 cm
As PA×PB = PC×PD
⇒ (AB + 4)×4 = 12.8×5
⇒ AB + 4 = (12.8 ×5)/4
⇒ AB + 4 = 16
Hence, AB = 12 cm
(ii) As we know,
PT2 = PC ×PD
⇒ PT2 = 12.8 ×5 = 64
Thus, PT = 8 cm
3. In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30o and ∠CBD = 60o; calculate:
(i) ∠QAB
(ii) ∠PAD
(iii) ∠CDB
Solution
(i) Given, PAQ is a tangent and AB is the chord
∠QAB = ∠ADB = 30o [Angles in the alternate segment]
(ii) OA = OD [radii of the same circle]
So, ∠OAD = ∠ODA = 30o
But, as OA ⊥ PQ
∠PAD = ∠OAP – ∠OAD = 90o – 30o = 60o
(iii) As BD is the diameter, we have
∠BCD = 90o [Angle in a semi-circle]
Now in ∆BCD,
∠CDB + ∠CBD + ∠BCD = 180o
⇒ ∠CDB + 60o + 90o = 180o
Thus, ∠CDB = 180o – 150o = 30o
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 ∠TRQ = 30o
Solution
(i) As PQ is the tangent and OR is the radius.
So, OR ⊥ PQ
∠ORT = 90o
⇒ ∠TRQ = 90o – 30o = 60o
But in ∆OTR, we have
OT = OR [Radii of same circle]
⇒ ∠OTR = 60o or ∠STR = 60o
But, ∠PRS = ∠STR = 60o [Angles in the alternate segment]
(ii) In ∆OTR,
∠ORT = 60o
∠OTR = 60o
Thus, ∠ROT = 180o – (60o + 60o) = 180o – 120o = 60o
5. AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Solution
Join OC.
∠BCD = ∠BAC = 30o [Angles in the alternate segment]
It’s seen that, arc BC subtends ∠DOC at the center of the circle and ∠BAC at the remaining part of the circle.
So, ∠BOC = 2∠BAC = 2 ×30o = 60o
Now, in ∆OCD
∠BOC or ∠DOC = 60o
∠OCD = 90o [OC ⊥ CD]
⇒ ∠DOC + ∠ODC = 90o
⇒ ∠ODC = 90o – 60o = 30o
Now, in ∆BCD
As ∠ODC or ∠BDC = ∠BCD = 30o
∴ BC = BD
6. Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Solution
Let DE be the tangent to the circle at P.
And, DE || QR [Given]
∠EPR = ∠PRQ [Alternate angles are equal]
∠DPQ = ∠PQR [Alternate angles are equal] ….(i)
Let ∠DPQ = x and ∠EPR = y
As the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have
∠DPQ = ∠PRQ …(ii) [DE is tangent and PQ is chord]
So, from (i) and (ii),
∠PQR = ∠PRQ
PQ = PR
∴ triangle PQR is an isosceles triangle.
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.
Solution
Join OA, OB, O’A, O’B and O’O.
CD is the tangent and AO is the chord.
∠OAC = ∠OBA …(i) [Angles in alternate segment]
In ∆OAB,
OA = OB [Radii of the same circle]
∠OAB = ∠OBA …(ii)
From (i) and (ii), we have
∠OAC = ∠OAB
Thus, OA is the bisector of ∠BAC.
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
Solution
Let’s draw a tangent TS at P to the circles given.
As TPS is the tangent and PD is the chord, we have
∠PAB = ∠BPS …(i) [Angles in alternate segment]
Similarly,
∠PCD = ∠DPS …(ii)
Now, subtracting (i) from (ii) we have
∠PCD – ∠PAB = ∠DPS – ∠BPS
But in ∆PAC,
Ext. ∠PCD = ∠PAB + ∠CPA
⇒ ∠PAB + ∠CPA – ∠PAB = ∠DPS – ∠BPS
Thus, ∠CPA = ∠DPB
Exercise 18(C)
1. Prove that, of any two chords of a circle, the greater chord is nearer to the center.
Solution
Given: A circle with center O and radius r. AB and CD are two chords such that AB > CD. Also, OM ⊥ AB and ON ⊥ CD.
Required to prove: OM < ON
Proof:
Join OA and OC.
Then in right ∆AOM, we have
AO2 = AM2 + OM2
⇒ r2 = (½AB)2 + OM2
⇒ r2 = ¼ AB2 + OM2 …(i)
Again, in right ∆ONC, we have
OC2 = NC2 + ON2
⇒ r2 = (½CD)2 + ON2
⇒ r2 = ¼ CD2 + ON2 …(ii)
On equating (i) and (ii), we get
¼ AB2 + OM2 = ¼ CD2 + ON2
But, AB > CD [Given]
So, ON will be greater than OM to be equal on both sides.
Thus,
OM < ON
Hence, AB is nearer to the centre than CD.
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.
Solution
(i) Given, radius = 10 cm
In rhombus OABC,
OC = 10 cm
So,
OE = ½ ×OB = ½ ×10 = 5 cm
Now, in right ∆OCE
OC2 = OE2 + EC2
⇒ 102 = 52 + EC2
⇒ EC2 = 100 – 25 = 75
⇒ EC = √75 = 5√3
Hence, AC = 2×EC = 2×5√3 = 10√3
We know that,
Area of rhombus = ½ ×OB ×AC
= ½ ×10×10√3
= 50√3 cm2 ≈ 86.6 cm2
(ii) We have the area of rhombus = 32√3 cm2
But area of rhombus OABC = 2 x area of ∆OAB
Area of rhombus OABC = 2×(√3/4) r2
Where r is the side of the equilateral triangle OAB.
2×(√3/4) r2 = 32√3
⇒ √3/2 r2 = 32√3
⇒ r2 = 64
⇒ r = 8
∴ the radius of the circle is 8 cm.
3. Two circles with centers 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.
Solution
We know that,
If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 – 3) cm = 2 cm.
Also, the common chord PQ is the perpendicular bisector of AB.
Thus, AC = CB = ½ AB = 1 cm
In right ∆ACP, we have
AP2 = AC2 + CP2 [Pythagoras Theorem]
⇒ 52 = 12 + CP2
⇒ CP2 = 25 – 1 = 24
⇒ CP = √24 cm = 2√6 cm
Now,
PQ = 2 CP
= 2 × 2√6 cm
= 4√6 cm
∴ the length of PQ is 4√6 cm.
4. Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.
Solution
Given: AB and AC are two equal chords of C (O, r).
Required to prove: Centre, O lies on the bisector of ∠BAC.
Construction: Join BC. Let the bisector of ∠BAC intersects BC in P.
Proof:
In ∆APB and ∆APC,
AB = AC [Given]
∠BAP = ∠CAP [Given]
AP = AP [Common]
Hence, ∆APB ≅ ∆APC by SAA congruence criterion
So, by CPCT we have
BP = CP and ∠APB = ∠APC
And,
∠APB + ∠APC = 180° [Linear pair]
⇒ 2∠APB = 180° [∠APB = ∠APC]
⇒ ∠APB = 90°
Now, BP = CP and ∠APB = 90°
∴ AP is the perpendicular bisector of chord BC.
Hence, AP passes through the centre, O of the circle.
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 center of the circle?
Solution
We have, AB as the diameter and AC as the chord.
Now, draw OL ⊥ AC
Since OL ⊥ AC and hence it bisects AC, O is the centre of the circle.
∴ OA = 10 cm and AL = 6 cm
Now, in right ∆OLA
AO2 = AL2 + OL2 [By Pythagoras Theorem]
⇒ 102 = 62 + OL2
⇒ OL2 = 100 – 36 = 64
⇒ OL = 8 cm
∴ the chord is at a distance of 8 cm from the centre of the circle.
6. ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110o and angle BAC = 50o. Find angle DAC and angle DCA.
Solution
Given, ABCD is a cyclic quadrilateral in which AD || BC
And, ∠ADC = 110o, ∠BAC = 50o
We know that,
∠B + ∠D = 180o [Sum of opposite angles of a quadrilateral]
⇒ ∠B + 110o = 180o
So, ∠B = 70o
Now in ∆ADC, we have
∠BAC + ∠ABC + ∠ACB = 180o
⇒ 50o + 70o + ∠ACB = 180o
⇒ ∠ACB = 180o – 120o = 60o
And, as AD || BC we have
∠DAC = ∠ACB = 60o [Alternate angles]
Now in ∆ADC,
∠DAC + ∠ADC + ∠DCA = 180o
⇒ 60o + 110o + ∠DCA = 180o
Thus, ∠DCA = 180o – 170o = 10o
7. In the given figure, C and D are points on the semi-circle described on AB as diameter. Given angle BAD = 70o and angle DBC = 30o, calculate angle BDC.
Solution
As ABCD is a cyclic quadrilateral, we have
∠BCD + ∠BAD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
⇒ ∠BCD + 70o = 180°
⇒ ∠BCD = 180o – 70o = 110o
And, by angle sum property of ∆BCD we have
∠CBD + ∠BCD + ∠BDC = 180o
⇒ 30o + 110o + ∠BDC = 180o
⇒ ∠BDC = 180o – 140o
Thus, ∠BDC = 40o
8. In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5 ∠B. Find the measure of each angle of the quadrilateral.
Solution
Given, cyclic quadrilateral ABCD
So, ∠A + ∠C = 180o [Opposite angles in a cyclic quadrilateral is supplementary]
⇒ 3∠C + ∠C = 180o [As ∠A = 3 ∠C]
⇒ ∠C = 45o
Now,
∠A = 3 ∠C = 3×45o
⇒ ∠A = 135o
Similarly,
∠B + ∠D = 180o [As ∠D = 5 ∠B]
⇒ ∠B + 5∠B = 180o
⇒ 6∠B = 180o
⇒ ∠B = 30o
Now, ∠D = 5∠B = 5×30o
⇒ ∠D = 150o
∴ ∠A = 135o, ∠B = 30o, ∠C = 45o, ∠D = 150o
9. Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution
Let’s join AD.
And. AB is the diameter.
We have ∠ADB = 90º [Angle in a semi-circle]
But,
∠ADB + ∠ADC = 180º [Linear pair]
So, ∠ADC = 90º
Now, in ∆ABD and ∆ACD we have
∠ADB = ∠ADC [each 90º]
AB = AC [Given]
AD = AD [Common]
Hence, ∆ABD ≅ ∆ACD by RHS congruence criterion
So, by C.P.C.T
BD = DC
∴ the circle bisects base BC at D.
10. Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90o – ½ ∠A
Solution
Join ED, EF and DF. Also join BF, FA, AE and EC.
∠EBF = ∠ECF = ∠EDF …(i) [Angle in the same segment]
In cyclic quadrilateral AFBE,
∠EBF + ∠EAF = 180o ...(ii)
[Sum of opposite angles in a cyclic quadrilateral is supplementary]
Similarly in cyclic quadrilateral CEAF,
∠EAF + ∠ECF = 180o …(iii)
Adding (ii) and (iii) we get,
∠EBF + ∠ECF + 2∠EAF = 360o
⇒ ∠EDF + ∠EDF + 2∠EAF = 360o [From (i)]
⇒ ∠EDF + ∠EAF = 180o
⇒ ∠EDF + ∠1 + ∠BAC + ∠2 = 180o
But, ∠1 = ∠3 and ∠2 and ∠4 [Angles in the same segment]
∠EDF + ∠3 + ∠BAC + ∠4 = 180o
But, ∠4 = ½ ∠C, ∠3 = ½ ∠B
Thus, ∠EDF + ½ ∠B + ∠BAC + ½ ∠C = 180o
⇒ ∠EDF + ½ ∠B + 2 x ½ ∠A + ½ ∠C = 180o
⇒ ∠EDF + ½ (∠A + ∠B + ∠C) + ½ ∠A = 180o
⇒ ∠EDF + ½ (180o) + ½ ∠A = 180o
⇒ ∠EDF + 90o + ½ ∠A = 180o
⇒ ∠EDF = 180o – (90o + ½ ∠A)
⇒ ∠EDF = 90o – ½ ∠A
11. In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20o, find angle AOD.
Solution
Join OB.
In ∆OBC, we have
BC = OD = OB [Radii of the same circle]
∠BOC = ∠BCO = 20o
And ext. ∠ABO = ∠BCO + ∠BOC
Ext. ∠ABO = 20o + 20o = 40o ….(1)
Now in ∆OAB,
OA = OB [Radii of the same circle]
⇒ ∠OAB = ∠OBA = 40o [from (1)]
⇒ ∠AOB = 180o – 40o – 40o = 100o
As DOC is a straight line,
∠AOD + ∠AOB + ∠BOC = 180o
⇒ ∠AOD + 100o + 20o = 180o
⇒ ∠AOD = 180o – 120o
Thus, ∠AOD = 60o
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.
Solution
Let’s join OL, OM and ON.
And, let D and d be the diameter of the circumcircle and incircle.
Also, let R and r be the radius of the circumcircle and incircle.
Now, in circumcircle of ∆ABC,
∠B = 90o
Thus, AC is the diameter of the circumcircle i.e. AC = D
Let the radius of the incircle be ‘r’
OL = OM = ON = r
Now, from B, BL and BM are the tangents to the incircle.
So, BL = BM = r
Similarly,
AM = AN and CL = CN = R
[Tangents from the point outside the circle]
Now,
AB + BC + CA = AM + BM + BL + CL + CA
= AN + r + r + CN + CA
= AN + CN + 2r + CA
= AC + AC + 2r
= 2AC + 2r
= 2D + d
Hence, Proved
13. P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution
First join AP and BP.
As TPS is a tangent and PA is the chord of the circle.
∠BPT = ∠PAB [Angles in alternate segments]
But,
∠PBA = ∠PAB [Since PA = PB]
Thus, ∠BPT = ∠PBA
But these are alternate angles,
Hence, TPS || AB
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.
Solution
From P, AP is the tangent and PMN is the secant for first circle.
AP2 = PM×PN …(1)
Again from P, PB is the tangent and PMN is the secant for second circle.
PB2 = PM×PN …(2)
From (i) and (ii), we have
AP2 = PB2
⇒ AP = PB
Thus, P is the midpoint of AB.
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 = 40o and ∠ABD = 60o, find:
(i) ∠DBC
(ii) ∠BCP
(iii) ∠ADB
Solution
PQ is a tangent and CD is a chord.
∠DCQ = ∠DBC [Angles in the alternate segment]
∠DBC = 40o [As ∠DCQ = 40o]
(ii) ∠DCQ + ∠DCB + ∠BCP = 180o
⇒ 40o + 90o + ∠BCP = 180o [As ∠DCB = 90o]
⇒ ∠BCP = 180o – 130o = 50o
(iii) In ∆ABD,
∠BAD = 90o [Angle in a semi-circle]
∠ABD = 60o [Given]
⇒ ∠ADB = 180o – (90o + 60o)
⇒ ∠ADB = 180o – 150o = 30o
16. The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90o
Solution
Let’s join OC.
BCD is the tangent and OC is the radius.
As, OC ⊥ BD
∠OCD = 90o
⇒ ∠OCD + ∠ACD = 90o …(i)
But, in ∆OCA
OA = OC [Radii of the same circle]
Thus, ∠OCA = ∠OAC
Substituting in (i), we get
∠OAC + ∠ACD = 90o
Hence, ∠BAC + ∠ACD = 90o
17. ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D. Prove that:
(i) AC×AD = AB2
(ii) BD2 = AD×DC.
Solution
(i) In ∆ABC, we have
∠B = 90o and BC is the diameter of the circle.
Hence, AB is the tangent to the circle at B.
Now, as AB is tangent and ADC is the secant we have
AB2 = AD×AC
(ii) In ∆ADB,
∠D = 90o
So, ∠A + ∠ABD = 90o ...(i)
But in ∆ABC, ∠B = 90o
∠A + ∠C = 90o …(ii)
From (i) and (ii),
∠C = ∠ABD
Now in ∆ABD and ∆CBD, we have
∠BDA = ∠BDC = 90o
∠ABD = ∠BCD
Hence, ∆ABD ~ ∆CBD by AA postulate
So, we have
BD/DC = AD/BD
∴ BD2 = AD x DC
18. In the given figure, AC = AE.
Show that:
(i) CP = EP
(ii) BP = DP
Solution
In ∆ADC and ∆ABE,
∠ACD = ∠AEB [Angles in the same segment]
AC = AE [Given]
∠A = ∠A [Common]
Hence, ∆ADC ≅ ∆ABE by ASA postulate
So, by C.P.C.T we have
AB = AD
But, AC = AE [Given]
So, AC – AB = AE – AD
BC = DE
In ∆BPC and ∆DPE,
∠C = ∠E [Angles in the same segment]
BC = DE
∠CBP = ∠CDE [Angles in the same segment]
Hence, ∆BPC ≅ ∆DPE by ASA postulate
So, by C.P.C.T we have
BP = DP and CP = PE
19. ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120o
Calculate:
(i) ∠BEC
(ii) ∠BED
Solution
(i) Join OC and OB.
AB = BC = CD and ∠ABC = 120o [Given]
So, ∠BCD = ∠ABC = 120o
OB and OC are the bisectors of ∠ABC and ∠BCD and respectively.
So, ∠OBC = ∠BCO = 60o
In ∆BOC,
∠BOC = 180o – (∠OBC + ∠BOC)
⇒ ∠BOC = 180o – (60o + 60o) = 180o – 120o
⇒ ∠BOC = 60o
Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.
∠BEC = ½ ∠BOC = ½ ×60o = 30o
(ii) In cyclic quadrilateral BCDE, we have
∠BED + ∠BCD = 180o
⇒ ∠BED + 120o = 180o
Thus, ∠BED = 60o
20. In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30o, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB
Solution
In the given fig, O is the centre of the circle and, CA and CB are the tangents to the circle from C. Also, ∠ACO = 30o
P is any point on the circle. P and B are joined.
To find:
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Proof:
(i) In ∆OAC and ∆OBC, we have
OC = OC [Common]
OA = OB [Radii of the same circle]
CA = CB [Tangents to the circle]
Hence, ∆OAC ≅ ∆OBC by SSS congruence criterion
Thus, ∠ACO = ∠BCO = 30o
(ii) As ∠ACB = 30o + 30o = 60o
And, ∠AOB + ∠ACB = 180o
⇒ ∠AOB + 60o = 180o
⇒ ∠AOB = 180o – 60o
⇒ ∠AOB = 120o
(iii) Arc AB subtends ∠AOB at the center and ∠APB is the remaining part of the circle.
∠APB = ½ ∠AOB = ½ x 120o = 60o
21. ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centers. Find the radii of the three circles.
Solution
Given: ABC is a triangle with AB = 10 cm, BC= 8 cm, AC = 6 cm. Three circles are drawn with centre A, B and C touch each other at P, Q and R respectively.
So, we need to find the radii of the three circles.
Let,
PA = AQ = x
QC = CR = y
RB = BP = z
So, we have
x + z = 10 …(i)
z + y = 8 …(ii)
y + x = 6 ...(iii)
Adding all the three equations, we have
2(x + y + z) = 24
⇒ x + y + z = 24/2 = 12 …(iv)
Subtracting (i), (ii) and (iii) from (iv) we get
y = 12 – 10 = 2
x = 12 – 8 = 4
z = 12 – 6 = 6
Thus, radii of the three circles are 2 cm, 4 cm and 6 cm.