This exercise deals with problems based on all the concepts of the chapter. These exercise problems can be effectively solved with guidance of the Selina Solutions for Class 10 Maths. It is also a wonderful resource for students to clarify their doubts and boost the exam preparations. The solutions of the Concise Selina Solutions for Class 10 Maths Chapter 18 Tangents and Intersecting Chords Exercise 18(C) are available in a PDF, which is provided in the link given below.
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Access Selina Solutions Concise Maths Class 10 Chapter 18 Tangents and Intersecting Chords Exercise 18(C)
1. Prove that, of any two chords of a circle, the greater chord is nearer to the center.
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
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.
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.
(i) Given, radius = 10 cm
In rhombus OABC,
OC = 10 cm
OE = ½ x OB = ½ x 10 = 5 cm
Now, in right ∆OCE
OC2 = OE2 + EC2
102 = 52 + EC2
EC2 = 100 – 25 = 75
EC = √75 = 5√3
Hence, AC = 2 x EC = 2 x 5√3 = 10√3
We know that,
Area of rhombus = ½ x OB x AC
= ½ x 10 x 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 x (√3/4) r2
Where r is the side of the equilateral triangle OAB.
2 x (√3/4) r2 = 32√3
√3/2 r2 = 32√3
r2 = 64
r = 8
Therefore, 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.
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
PQ = 2 CP
= 2 x 2√6 cm
= 4√6 cm
Therefore, 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.
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.
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
∠APB + ∠APC = 180
2∠APB = 180
[∠APB = ∠APC]
∠APB = 90
Now, BP = CP and ∠APB = 90
Therefore, 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?
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.
Therefore, 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
Therefore, 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.
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
∠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.
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
∠BDC = 40o
8. In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5 ∠B. Find the measure of each angle of the quadrilateral.
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
∠A = 3 ∠C = 3 x 45o
∠A = 135o
∠B + ∠D = 180o [As ∠D = 5 ∠B]
∠B + 5∠B = 180o
6∠B = 180o
∠B = 30o
∠D = 5∠B = 5 x 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.
Let’s join AD.
And. AB is the diameter.
We have ∠ADB = 90º [Angle in a semi-circle]
∠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
Therefore, 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
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.
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.
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
AM = AN and CL = CN = R[Tangents from the point outside the circle]
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.
First join AP and BP.
As TPS is a tangent and PA is the chord of the circle.
∠BPT = ∠PAB [Angles in alternate segments]
∠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.
From P, AP is the tangent and PMN is the secant for first circle.
AP2 = PM x PN …. (1)
Again from P, PB is the tangent and PMN is the secant for second circle.
PB2 = PM x 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:
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
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 x AD = AB2
ii) BD2 = AD x DC.
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 x 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.
i) CP = EP
ii) BP = DP
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
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
∠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 x 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
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.
(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.
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.
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.