mathamatics xam idea

13. It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P? [NCERT Exemplar]

14. A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 10.5
cm PA = 4.5 cm, BR = 8 cm and PB = 6 cm. Is AB || QR ?

15. If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the
other triangle, can you say that the two triangles will be similar?

16. L and M are respectively the points on the sides DE and DF of a triangle DEF, such that DL = 4,
4
LE = 3 , DM = 6 and DF = 8. Is LM ||EF ? Give reason.

L and M are points on the sides DE and DF respectively of a D DEF. For the following cases

(Q. 17 and Q. 18), state whether LM || EF

17. DL = 3.9 cm, LE = 3 cm, DM = 3.6 cm and MF = 2.4 cm.

18. DE = 8 cm, DF = 15 cm, LE = 3.2 cm and MF = 6 cm.

19. The ratio of the corresponding altitudes of two similar triangles is 2 . Is it correct to say that ratio
5
2
of their areas is also 5 ? Why?

20. If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what

is the length of QR?

QQ Short Answer Questions-I: [2 marks each]

21. X is a point on the side BC of DABC. XM and XN are drawn parallel to AB and AC respectively
meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX2 = TB × TC.

[CBSE 2018 (C) (30/1)]
22. The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.

23. In Fig. 7.72, OA = OD . Prove that ∠A = ∠C and ∠B = ∠D.
OC OB

Fig. 7.72

24. Sides of triangles are given below. Determine which of them are right triangles. In case of a right
triangle, write the length of its hypotenuse.

(i) 13 cm, 12 cm, 5 cm (ii) 20 cm, 25 cm, 30 cm.

25. ∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. Find the ratio
of the areas of triangles ABC and BDE.

26. AD is the bisector of ∠BAC in ∆ABC. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find BD.

27. The perimeters of two similar triangles ABC and PQR are 60 cm and 36 cm respectively.
If PQ = 9 cm, then find the length of AB.

28. Two poles of height 9 m and 15 m stand vertically upright on a plane ground. If the distance
between their tops is 10m, then find the distance between their feet.

29. The area of two similar triangles PQR and XYZ are 144 cm2 and 49 cm2 respectively. If the
shortest side of larger ∆PQR be 24 cm, then find the shortest side of the smaller triangle XYZ.

30. ∆ABC ~ ∆DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, then find perimeter
of ∆DEF.

Triangles 193

QQ Short Answer Questions-II: [3 marks each]

31. Prove that the area of an equilateral triangle described on one side of the square is equal to half
the area of the equilateral triangle described on one of its diagonal. [CBSE 2018 (C) (30/1)]

32. If the area of two similar triangles are equal, prove that they are congruent.

[CBSE 2018 (C) (30/1)]

33. In Fig. 7.73, ∠ACB = 90° and CD ⊥ AB, prove that CD2 = BD × AD. [CBSE 2019 (30/1/1)]

C

AD B

Fig. 7.73

34. If P and Q are the points on side CA and CB respectively of DABC, right angled at C, prove that
(AQ2 + BP2) = (AB2 + PQ2)
[CBSE 2019 (30/1/1)]

35. Diagonals of a trapezium PQRS intersect each other at the point O, PQ  RS and PQ = 3RS. Find

the ratio of the areas of triangles POQ and ROS. [CBSE 2019 (30/2/1)]

36. The perpendicular from A on side BC of a DABC meets BC at D such that DB= 3CD. Prove that
2AB2 = 2AC2 + BC2.
[CBSE 2019 (30/3/1)]

37. AD and PM are medians of triangles ABC and PQR respectively where DABC ~DPQR. Prove that

AB = AD . [CBSE 2019 (30/3/1)]
PQ PM

38. ABC is a right triangle in which ∠B = 90°. If AB = 8 cm and BC = 6 cm, find the diameter of the

circle inscribed in the triangle. [CBSE 2019 (30/4/2)]

39. In Fig. 7.74, BL and CM are medians of a DABC right-angled at A.

Prove that 4(BL2 + CM2) = 5 BC2. [CBSE 2019 (30/4/2)]

C

L

BM A

Fig. 7.74

40. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of

its diagonals. [CBSE 2019 (30/4/2)]

41. In DABC, ∠B = 90° and D is the mid-point of BC. Prove that AC2 = AD2 + 3CD2.

[CBSE 2019 (30/5/1)]

42. ABCD is a trapezium with AB  DC. E and F are points on non-parallel sides AD and BC

respectively, such that EF  AB. Show that AE = BF . [CBSE 2019 (C)(30/1/1)]
ED FC

43. Prove that in a right-triangle the square of the hypotenuse is equal to the sum of the squares of

the other two sides. [CBSE 2019 (C)(30/1/1)]

44. In ∆ABC, DE||BC . If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.

45. If a line intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove

that AD = AE .
AB AC
194 Xam idea Mathematics–X

46. ABCD is a trapezium in which AB||DC and P and Q are points on AD and BC respectively such

that PQ ||DC . If PD = 12 cm, BQ = 42 cm and QC = 18 cm, find AD.

47. For going to a city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and
CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two
cities, A and B. Find how much distance will be saved in reaching city B from city A after the
construction of the highway.

48. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO = CO .
BO DO

Show that ABCD is a trapezium. [NCERT]

49. A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5m
casts a shadow of 3 m, find how far she is away from the base of the pole.

50. In Fig. 7.75, find ∠E.

51. In Fig. 7.76, ABC is a right triangle, right-angled
at C and D is the mid-point of BC. Prove that
AB2 = 4AD2 – 3AC2.

Fig. 7.75

Fig. 7.76

52. In Fig. 7.77, P is the mid-point of EF and Q is the mid-point of DP. If EQ when produced meets
1
DF at R, prove that RD = 3 DF.

Fig. 7.77

53. In Fig. 7.78, E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at

F. Prove that ∆ABE ~ ∆CFB. [NCERT Exemplar]

Fig. 7.78

Triangles 195

54. In ∆ABC (Fig. 7.79), DE is parallel to base BC, with D on AB and E on AC. If AD = 2 , find BC
DB 3 DE

Fig. 7.79

55. O is any point inside a rectangle ABCD. Prove that OB2 + OD2 = OA2 + OC2.
56. In Fig. 7.80, AB  CD. If OA = 3x – 19, OB = x – 4, OC = x – 3, and OD = 4, find x.

Fig. 7.80

57. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm,
PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one-sixteenth of the area of
∆ABC.

QQ Long Answer Questions: [5 marks each]

58. In an equilateral DABC, D is a point on side BC such that BD = 1 BC . Prove that
3
9(AD)2 = 7(AB)2.
[CBSE 2018 (30/1)]

59. Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares

on the other two sides. [CBSE 2018 (30/1)]

60. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their

corresponding sides. [CBSE 2018 (C) (30/1)]

61. In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then

prove that the angle opposite the first side is a right angle. [CBSE 2019 (30/2/3)]

62. In Fig. 7.81, DACB, AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC × BD. [CBSE 2019 (30/4/2)]

A

BD C

Fig. 7.81

63. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB
and FE of ∆ABC and ∆EFG. If ∆ABC ~ ∆FEG, show that

(i) CD = AC (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
GH FG

196 Xam idea Mathematics–X

64. In Fig. 7.82, OB is the perpendicular bisector of the line segment DE, FA ⊥ OB and FE intersects

OB at the point C. Prove that: 1 + 1 = 2 .
OA OB OC
65. If a perpendicular is drawn from the vertex containing the right angle

of a right triangle to the hypotenuse, then prove that the triangle on

each side of the perpendicular are similar to each other and to the

original triangle. Also, prove that the square of the perpendicular is

equal to the product of the lengths of the two parts of the hypotenuse.

66. In Fig. 7.83, PQR is a right triangle right-angled at Q and QS ⊥ PR. Fig. 7.82
If PQ = 6 cm and PS = 4 cm, find the QS, RS and QR.

Fig. 7.83

67. In Fig. 7.84, DEFG is a square and ∠BAC = 90°. Prove that:

(i) ∆AGF ~ ∆DBG (ii) ∆AGF ~ ∆EFC

(iii) ∆DBG ~ ∆EFC (iv) DE2 = BD × EC Fig. 7.84

68. In a ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that:
(a + b) (a – b) = (c + d) (c – d).

69. In a triangle ABC, AC > AB, D is the mid-point of BC and AE ⊥ BC. Prove that:

(i) AC2 = AD2 + BC.DE + 1 BC2 (ii) AB2 = AD2 – BC.DE + 1 BC2
4 4

(iii) AB2 + AC2 = 2AD2 + 1 BC2.
2

70. In an equilateral triangle with side a, prove that:

(i) Altitude = a 3 (ii) Area = 3 a2.
2 4

Answers

1. (i) (a) (ii) (c) (iii) (c) (iv) (a) (v) (c)

2. (i) 4 : 5 (ii) parallel (iii) 90° (iv) similar (v) hypotenuse

3. 1 4. PR = 10 cm 5. 9 : 1 6. 4 2 cm 7. AD = 2.4 cm
9

8. BC = 11.2 cm 9. 3 : 10 10. 16 : 9 11. 9 cm 12. Yes

13. No 14. Yes 15. Yes, by AA similarity 16. Yes, because DL = DM =3
17. No LE MF
22. 25 cm 4
18. Yes 19. No, it will be 25 20. 6 cm

24. (i) 13 cm (ii) It is not a right triangle 25. 4 : 1 26. BD = 7.5 cm

27. 15 cm 28. 8 m 29. 14 cm 30. 30 cm 35. 9 : 1 38. 4 cm

44. x = 1 46. AD = 40 cm 47. 8 km 49. 9 m 50. ∠E = 60°

54. 5 56. 11 or 8 66. 2 5 cm, 5 cm, 3 5 cm
2

Triangles 197

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) In DABC, AB = 6 7 cm, BC = 24 cm and CA = 18 cm. Then angle A is

(a) an acute angle (b) an obtuse angle

(c) a right angle (d) can't say

(ii) It is given that DABC ~ DPQR, with BC = 1 . Then, ar (PRQ) is equal to [NCERT Exemplar]
QR 3 ar (BCA)

(a) 9 (b) 3 (c) 1 (d) 1
3 9

(iii) If S is a point on side PQ of a DPQR such that PS = QS = RS, then [NCERT Exemplar]

(a) PR . QR = RS2 (b) QS2 + RS2 = QR2

(c) PR2 + QR2 = PQ2 (d) PS2 + RS2 = PR2

(iv) If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is

not true?

(a) EF = DF (b) DE = EF (c) DE = DF (d) EF = DE
PR PQ PQ RP QR PQ RP QR

2. Fill in the blanks. (3 × 1 = 3)

(i) All squares with edges of equal length are _______________ .

(ii) Two polygons of the same number of sides are similar, if their corresponding angles are
_______________ and their corresponding sides are _______________ .

(iii) Triangles whose corresponding angles are equal are called ______________ .

3. Solve the following questions. (3 × 1 = 3)

(i) If ABC and DEF are two similar triangles such that BC = 4 cm and EF = 5 cm and area of
triangle ABC is 64 cm2, then what is the area of ∆DEF?

(ii) The diagonals of a rhombus are 30 cm and 40 cm. What is the length of its each side?

(iii) A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower
casts a shadow of 50 m long on the ground. Find the height of the tower.

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. In Fig. 7.85, PQ ||BC and AP : PB = 1: 2 find area (∆APQ ) .
area (∆ABC)

Fig. 7.85

198 Xam idea Mathematics–X

5. In Fig. 7.86, ABC is an isosceles triangle in which AB = AC. E is a point on the side CB produced such
that FE ⊥ AC. If AD ⊥ CB, prove that AB × EF = AD × EC.

Fig. 7.86
6. If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA, then prove that

AB2 + BC2 + CA2 = 4 BE2.

QQ Solve the following questions. (3 × 3 = 9)

7. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE
is constructed. Prove that: area (∆ADE) : area (∆ABC) = 3 : 4.

8. In Fig. 7.87, PQR is a right triangle, right angled at Q. X and Y are the
points on PQ and QR such that PX : XQ = 1 : 2 and QY : YR = 2 : 1. Prove
that 9 (PY2 + XR2) = 13PR2.

9. In Fig. 7.88, D and E trisects BC. Prove that 8AE2 = 3AC2 + 5AD2.

Fig. 7.87

Fig. 7.88

QQ Solve the following questions. (3 × 5 = 15)

10. P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right-angled at C.

Prove that:

(i) 4AQ2 = 4AC2 + BC2 (ii) 4BP2 = 4BC2 + AC2 (iii) 4(AQ2 + BP2) = 5AB2.

11. ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of

perpendicular from C on AB. Prove that:

(i) cp = ab (ii) 1 = 1 + 1 .
p2 a2 b2

12. State and prove Basic Proportionality Theorem.

Answers

1. (i) (c) (ii) (a) (iii) (c) (iv) (b)

2. (i) congruent (ii) equal, proportional (iii) similar

3. (i) 100 cm2 (ii) 25 cm (iii) 100 m

4. 1:9

zzz

Triangles 199

8 Circles

BASIC CONCEPTS – A FLOW CHART

Alternate Segment Theorem

An angle between a tangent and a chord E
CB
through the point of contact is equal to the D
angle in the alternate segment.
Hence, ‘BCE = ‘CDE
and, ‘ACD = ‘DEC

are the alternate segment angles. A

200 Xam idea Mathematics–X

MORE POINTS TO REMEMBER

 A tangent to a circle is perpendicular to the radius through the point of contact.
 A line drawn through the end point of a radius and

perpendicular to it is a tangent to the circle.
 The length of two tangents drawn from an external point

to a circle are equal. For example in Fig 8.2, PA = PB.

Fig. 8.1

Fig. 8.2 Fig. 8.3

 If two tangents are drawn to a circle from an external point
(Fig. 8.3), then

(i) they subtend equal angle at the centre.
(ii) they are equally inclined to the line segment joining the

centre and the external point.
i.e., ∠AOP = ∠BOP and ∠APO = ∠BPO

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the

radii is [NCERT Exemplar]

(a) 90° (b) 50° (c) 70° (d) 40°

2. In Fig. 8.4, the pair of tangents AP and AQ drawn from an external point

A to a circle with centre O are perpendicular to each other and length of

each tangent is 5 cm. Then radius of the circle is [NCERT Exemplar]

(a) 10 cm (b) 7.5 cm

(c) 5 cm (d) 2.5 cm

3. In Fig. 8.5, PQ is a chord of a circle and PT is the tangent at P such that Fig. 8.4

∠QPT = 60°. Then ∠PRQ is equal to [NCERT Exemplar]

Q

O
R

P T

Fig. 8.5

(a) 135° (b) 150° (c) 120° (d) 110°

Circles 201

4. At point A on a diameter AB of a circle of radius 10 cm, tangent XAY is drawn to the circle. The

length of the chord CD parallel to XY at a distance 16 cm from A is

(a) 8 cm (b) 10 cm (c) 16 cm (d) 18 cm

5. In Fig. 8.6, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then
∠PAB is equal to

A

OP

B

(a) 35° (b) 65° Fig. 8.6 (d) 70°
(c) 40°

6. Two circles touch each other externally at C and AB is common tangent of circles, then ∠ACB
is

(a) 70° (b) 60° (c) 100° (d) 90°

C

AB

Fig. 8.7

7. The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius
6 cm is

(a) 10 cm (b) 5 cm (c) 7 cm (d) 2 7 cm

8. In Fig. 8.8, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then
∠OAB is equal to [NCERT Exemplar]

(a) 25° (b) 30° Fig. 8.8 (d) 50°
(c) 40°

202 Xam idea Mathematics–X

9. In Fig. 8.9, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]

(a) 62.5° (b) 45° Fig. 8.9 (d) 55°
(c) 35°

10. In Fig. 8.10, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is
the tangent to the circle at the point A, then ∠BAT is equal to [NCERT Exemplar]

C

T

(a) 65° (b) 60° Fig. 8.10 (d) 40°
(c) 50°

11. If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle

which is tangent to the other circle is [NCERT Exemplar]

(a) 3 cm (b) 6 cm (c) 9 cm (d) 1 cm

12. In Fig. 8.11, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle

of 50° with PQ, then ∠POQ is equal to [NCERT Exemplar]

Fig. 8.11 A

(a) 100° (b) 80° 12 cm
(c) 90° (d) 75° FE

13. If ∆ABC is circumscribing a circle in the Fig. 8.12. The length B 6 cm C 4 cm C
of AB is
Fig. 8.12
(a) 12 cm (b) 13 cm

(c) 9 cm (d) 14 cm

Circles 203

14. From a point X, the length of the tangent to a circle is 20 cm and the distance of X from the centre
is 25 cm. The radius of the circle is

(a) 10 cm (b) 5 41 cm (c) 15 cm (d) 20 cm

15. If angle between two radii of a circle is 125°, then the angle between the tangents at the ends
of the radii is

(a) 90° (b) 75° (c) 55° (d) 125°

16. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the
pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(a) 60 cm2 (b) 65 cm2 (c) 30 cm2 (d) 32.5 cm2

17. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each

tangent is equal to [NCERT Exemplar]

(a) 3 3 cm (b) 6 cm (c) 3 cm (d) 3 3 cm
2

18. In the given figure, ∠CBD = 56° and ∠ABC = 65°. What is the measure of ∠ACB?

C

D

Fig. 8.13

(a) 56° (b) 90° (c) 59° (d) 121°

19. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle.
The length of the chord CD parallel to XY and at a distance 8 cm from A is

X
C

5 cm

A O 3 cm B
5 cm

D

Y

(a) 4 cm (b) 5 cm Fig. 8.14 (d) 8 cm
(c) 6 cm

20. The number of tangents that can be drawn to a circle from a point inside it is

(a) one (b) two (c) infinite (d) none

Answers

1. (b) 2. (c) 3. (c) 4. (c) 5. (b) 6. (d)
7. (d) 8. (a) 9. (d) 10. (c) 11. (b) 12. (a)
13. (d) 14. (c) 15. (c) 16. (a) 17. (d) 18. (c)
19. (d) 20. ( d)

204 Xam idea Mathematics–X

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Tangent is perpendicular to the ________________ through the point of contact.

2. Length of two tangents drawn from an external point are _______________.

3. _______________ is the Latin word from which the word tangent has been derived.

4. Number of tangents that can be drawn from an external point to a circle is _______________.

5. A line that intersects a circle in two distinct points is called ________________.

6. A line that intersects a circle in one point only is called _______________.

7. If a chord AB subtends an angle of 60° at the centre of a circle, the angle between tangents at
A and B is _______________.

8. The length of tangent from an external point to a circle is always _______________ than the
radius of the circle.

9. The length of tangent from an external point P to a circle with centre O is always _______________
than OP.

10. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°
then OP = _______________.

11. The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel
to side _______________.

12. If tangent at C, intersects extended diameter AB, of a circle at D, if BD = BC then ∠BAC is equal
to _______________.

13. A parallelogram circumscribing a circle is a _______________.

14. Perpendicular from centre of a circle to its chord _______________ it.

15. Two tangents are drawn to a circle from an external point then they subtend equal angles at the
_______________.

Answers

1. radius 2. equal 3. Tangere 4. 2 5. secant 6. tangent
11. BC 12. 30°
7. 120° 8. greater 9. less 10. a 2

13. rhombus 14. bisects 15. centre

Very Short Answer Questions [1 mark]

1. If a point P is 17 cm from the centre of a circle of radius 8 cm,
then find the length of the tangent drawn to the circle from point P.

Sol. OA ⊥ PA ( Radius is ⊥ to tangent at point of contact)

\ In ∆OAP, we have

PO2 = PA2 + AO2

⇒ (17)2 = (PA)2 + (8)2 Fig. 8.15

(PA)2= 289 – 64 = 225

⇒ PA = 225 = 15 O
17 cm
Hence, the length of the tangent from point P is 15 cm.
Q 15 cm P
2. The length of the tangent to a circle from a point P, which is 17 cm
away from the centre, is 15 cm. What is the radius of the circle?

Fig. 8.1C6ircles 205

Sol.  OQ ⊥ PQ ( Radius is perpendicular to tangent at

point of contact)

\ PQ2 + QO2 = OP2

⇒ OQ2 + (15)2 = (17)2

or OQ = 289 – 225

= 64 = 8 cm

3. In Fig. 8.17, ABCD is a cyclic quadrilateral. If ∠BAC = 50° and
∠DBC = 60° then find ∠BCD.

Sol. Here ∠BDC = ∠BAC = 50° (Angles in same segment are equal)

In DBCD, we have

∠BCD = 180° – (∠BDC + ∠DBC)

= 180° – (50° + 60°) Fig. 8.17
= 70°

4. In Fig. 8.18, the quadrilateral ABCD circumscribes a circle with centre O. If ∠AOB = 115°,

then find ∠COD.

Sol.  ∠AOB = ∠COD (Vertically opposite angles)

\ ∠COD = 115°

Fig. 8.18

5. In Fig. 8.19, ∆ABC is circumscribing a circle. Find the length of BC.

A

3 cm

N M 9 cm

4 cm

BL C

Fig. 8.19

Sol. AN = AM = 3 cm (Tangents drawn from an external point)

BN = BL = 4 cm (Tangents drawn from an external point)

CL = CM = AC – AM = 9 – 3 = 6 cm P R
40°
⇒ BC = BL + CL = 4 + 6 = 10 cm.

6. In Fig. 8.20, O is the centre of a circle, PQ is a chord and the
tangent PR at P makes an angle of 40° with PQ. Find ∠POQ.

Sol. ∠OPQ = 90° – 40° = 50° O Q

OP = OQ (Radii of a circle)

∠OPQ = ∠OQP = 50°

(Equal sides have equal opposite angles) Fig. 8.20

206 Xam idea Mathematics–X

∠POQ = 180° – ∠OPQ – ∠OQP

= 180° – 50° – 50° = 80°

7. If two tangents inclined at an angle 60° are drawn to a circle of radius 5 cm, then find the
length of each tangent.

Sol. In Fig. 8.21

∆AOP  ∆BOP (By SSS congruence criterion) 5 cm

⇒ ∠APO = ∠BPO = 60° = 30°
2
In ∆AOP, OA ⊥ AP
5 cm

∴ tan 30° = OA
AP
1 5 Fig. 8.21
3 = AP

⇒ AP = 5 3 cm

8. If radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of
one circle which is tangent to the other circle.

Sol. OA = 4 cm, OB = 5 cm

Also, OA ⊥ BC

\ OB2 = OA2 + AB2

⇒ 52 = 42 + AB2

⇒ AB = 25 – 16 = 3 cm

⇒ BC = 2 AB Fig. 8.22

= 2 × 3 = 6 cm

9. PQ is a tangent drawn from a point P to a circle with centre O and QOR is
a diameter of the circle such that ∠POR = 120° then find ∠OPQ.

Sol. ∠OQP = 90°

∠QOP = 180° – 120° = 60°

∠OPQ = 180° – ∠OQP – ∠QOP

= 180° – 90° – 60°

= 30° Fig. 8.23

10. From an external point P, tangents PA and PB are drawn to a circle with centre O. If

∠PAB = 50°, then find ∠AOB. [CBSE Delhi 2016]

Fig. 8.24

Sol. a PA = PB ⇒ ∠BAP = ∠ABP = 50°
∴ ∠APB = 180° – 50° – 50° = 80°
and ∠AOB = 180° – 80° = 100°

Circles 207

11. In Fig. 8.25, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and

∠CAB = 30°, find ∠PCA. [CBSE (AI) 2016]

Fig. 8.25

Sol. ∠ACB = 90° (Angle in the semicircle)

∠CAB = 30° (given)

In ∆ABC,

90° + 30° + ∠ABC = 180°

⇒ ∠ABC = 60°

Now, ∠PCA = ∠ABC (Angles in the alternate segment)

\ ∠PCA = 60° Fig. 8.26

OR

Construction: Join O to C.

∠PCO = 90° (Line joining centre to point of contact is perpendicular to PQ)

In ∆AOC, OA = OC (Radii of circle)

\ ∠OAC =∠OCA = 30° (Equal sides have equal opp. angles)

Now, ∠PCA = ∠PCO – ∠ACO

  = 90° – 30° = 60°

12. In fig. 8.27, there are two concentric circles with centre O. PRT and PQS are S
O
tangents to the inner circle from a point P lying on the outer circle. If Q

PR = 5 cm, find the length of PS. [CBSE Delhi 2017 (C)]

Sol. PQ = PR = 5 cm ( a Tangents drawn from external point are equal) P

∴ PS = 2PQ = 10 cm ( a Perpendicular drawn from centre to the chord R
bisects the chord)
T

Fig. 8.27

Short Answer Questions-I [2 marks]

1. In Fig. 8.28, PA and PB are tangents to the circle drawn from an external point P. CD is the

third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of ∆PCD.

Sol.  PA and PB are tangent from same external point

\ PA = PB = 15 cm

Now, perimeter of ∆PCD = P C + CD + DP

= PC + CQ + QD+ DP

= PC + CA + DB + DP Fig. 8.28
= PA + PB = 15 + 15 = 30 cm

208 Xam idea Mathematics–X

2. In Fig. 8.29, PA and PB are tangents to the circle from an external point P. CD is another tangent

touching the circle at Q. If PA =12 cm, QC = QD = 3 cm, then find PC + PD.
[CBSE Delhi 2017]

Sol. PA = PC + CA = PC + CQ [ a CA = CQ (tangents drawn A
from external point are equal] C

⇒ 12 = PC + 3 ⇒ PC = 9 cm OQ P
a PA = PB ⇒ PA – AC = PB – BD ⇒ PC = PD
∴ PD = 9 cm D
Hence, PC + PD = 18 cm B

Fig. 8.29

3. Prove that the line segment joining the points of contact of two parallel tangents of a circle,

passes through its centre. [CBSE Delhi 2014]

Sol. Let the tangents to a circle with centre O be ABC and XYZ.

Construction : Join OB and OY.

Draw OP||AC

 AB||PO

∴ ∠ABO + ∠POB = 180° (Adjacent interior angles) Fig. 8.30

∠ABO = 90° (A tangent to a circle is perpendicular to the radius
through the point of contact)

90° + ∠POB = 180°  ⇒ ∠POB = 90°

Similarly ∠POY = 90°

∠POB + ∠POY = 90° + 90° = 180°

Hence, BOY is a straight line passing through the centre of the circle.

4. If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that

QPR = 120°, prove that 2PQ = PO. [CBSE Delhi 2014, (F) 2016]

Sol. Given, ∠QPR = 120°

 Radius is perpendicular to the tangent at the point of contact.

∴ ∠OQP = 90° and

∠QPO = 60° (Tangents drawn to a circle from an external point are
equally inclined to the segment, joining the centre to that point.)

PQ ⇒  1 = PQ   ⇒ 2PQ = PO
In ∆QPO, cos 60° = PO 2 PO

5. In Fig. 8.32, common tangents AB and CD to two circles with Fig. 8.31
centres O1 and O2 intersect at E. Prove that AB = CD. [CBSE (AI) 2014]

Fig. 8.32

Sol. AE = CE and BE = ED (Tangents drawn from an external point are equal)
⇒ AB = CD
On addition, we get

AE + BE = CE + ED

Circles 209

6. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB

at D, E and F respectively. Prove that BD = DC. [CBSE (AI) 2014]

OR

In Fig. 8.33, if AB = AC, prove that BE = EC. [CBSE (F) 2017]

[Note: D, E, F replace by F, D, E]

Sol. Given, AB = AC (In Fig 8.34)
We have, BF + AF = AE + CE Fig. 8.33 ....(i)

AB, BC and CA are tangents to the circle at F, D and E respectively.

\   BF = BD, AE = AF and CE = CD ....(ii)

From (i) and (ii)

BD + AE = AE + CD ( AF = AE)

⇒ BD = CD Fig. 8.34

7. In Fig. 8.35, XP and XQ are two tangents to the circle with centre

O, drawn from an external point X. ARB is another tangent,

touching the circle at R. Prove that XA + AR = XB + BR.

[CBSE (F) 2014]

Sol. In the given figure,

AP = AR

BR = BQ Fig. 8.35
XP = XQ (Tangent to a circle from an external point are equal)

XA + AP = XB + BQ

XA + AR = XB + BR [AP = AR, BQ = BR]

8. In Fig. 8.36, a circle is inscribed in a ∆ABC, such that it touches the

sides AB, BC and CA at points D, E and F respectively. If the lengths
of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively,
find the lengths of AD, BE and CF.       [CBSE Delhi 2016]

OR Fig. 8.36
[CBSE 2019]
In Fig. 8.37, a circle inscribed in a ∆ ABC having sides BC = 8 cm,
AB = 10 cm and AC=12 cm. Find the lengths of BL, CM and AN.

Sol. Let AD = AF = x (In Fig. 8.36) A
\ DB = BE = 12 – x NM
and CF = CE = 10 – x
BC = BE + EC

⇒ 8 = 12 – x + 10 – x ⇒ x = 7

\ AD = 7 cm, BE = 12 – 7 = 5 cm, CF = 10 – 7 = 3 cm B LC
OR Fig. 8.37

Similar solution as above. Only vertices are changed with their values.

9. In Fig. 8.38, AP and BP are tangents to a circle with centre O, such that
AP = 5 cm and ∠APB = 60°. Find the length of chord AB. [CBSE Delhi 2016]

Sol. PA = PB (Tangents from an external point are equal)

and ∠APB = 60° Fig. 8.38
⇒ ∠PAB = ∠PBA = 60°
\ ∆PAB is an equilateral triangle.
Hence AB = PA = 5 cm.

210 Xam idea Mathematics–X

10. In Fig. 8.39 from an external point P, two tangents PT and PS are drawn T
Q
to a circle with centre O and radius r. If OP = 2r, show that O

∠OTS = ∠OST = 30°. [CBSE (AI) 2016] S P

Sol. Let ∠TOP = q Fig. 8.39

\ cos q = OO=TP 2=rr 1 ⇒ cos q = cos 60° ⇒ q = 60°
2

Hence, ∠TOS = 120°

In DOTS, OT = OS (Radii of circle)

⇒ ∠OTS = ∠OST = 60° = 30°
2

11. In Fig. 8.40, are two concentric circles of radii 6 cm and 4 cm with
centre O. If AP is a tangent to the larger circle and BP to the smaller
circle and length of AP is 8 cm, find the length of BP. [CBSE (F) 2016]

Sol. OA = 6 cm, OB = 4 cm, AP = 8 cm

OP2 = OA2 + AP2 = 36 + 64 = 100

⇒ OP = 10 cm Fig. 8.40
BP2 = OP2 – OB2 = 100 – 16 = 84
B
70°
⇒ BP = 2 21 cm
x° 66°
12. In the given Fig. 8.41, ∠ABC = 70°, ∠BCD = 66° then find the value of x. CD
Fig. 8.41
Sol. We have A

∠BCD = ∠CAB = 66° (Alternate Segment Angles)

Now, In ∆ABC, we have

∠ACB + ∠ABC + ∠CAB = 180°

⇒ x° + 70° + 66° = 180°

⇒ x = 180° – 136° = 44°

⇒ x = 44°

Short Answer Questions-II [3 marks]

1. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point

Q so that OQ = 12 cm. Find the length of PQ. [NCERT]

Sol. We have, ∠OPQ = 90°, OQ = 12 cm and OP = 5 cm

\ By Pythagoras Theorem

OQ2 = OP2 + QP2

⇒ 122 = 52 + QP2 Fig. 8.42
⇒ QP2 = 144 – 25 = 119

⇒ QP = 119 cm

2. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the

centre is 25 cm. Find the radius of the circle. [NCERT]

Sol. Let QT be the tangent and OT be the radius of circle. Therefore

OT ⊥ QT i.e., ∠OTQ = 90°

and OQ = 25 cm and QT = 24 cm

Circles 211

Now, by Pythagoras Theorem, we have

OQ2 = QT2 + OT2 ⇒ 252 = 242 + OT2

⇒ OT2 = 252 – 242 = 625 – 576

OT2 = 49 \ OT = 7 cm

3. In Fig. 8.44, if TP and TQ are the two tangents to a circle with centre Fig. 8.43
O so that ∠POQ = 110°, then find ∠PTQ .          [NCERT]

Sol. Since TP and TQ are the tangents to the circle with centre O

So, OP ⊥ PT and OQ ⊥ QT

⇒ ∠OPT = 90°, ∠OQT = 90° and ∠POQ = 110°

So, in quadrilateral OPTQ, we have

∠POQ + ∠OPT + ∠PTQ + ∠TQO = 360° Fig. 8.44
⇒ 110° + 90° + ∠PTQ + 90° = 360°  ⇒ ∠PTQ + 290° = 360°

\ ∠PTQ = 360° – 290° ⇒ ∠PTQ = 70°

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

[NCERT, CBSE (F) 2014, CBSE Delhi 2017, CBSE 2019 (30/5/2)]

Sol. Let AB be the diameter of the given circle with centre O, and two tangents PQ
and LM are drawn at the end of diameter AB respectively.

Now, since the tangent at a point to a circle is perpendicular to the radius
through the point of contact.

Therefore, OA ⊥ PQ and OB ⊥ LM Fig. 8.45
i.e., AB ⊥ PQ and also AB ⊥ LM

⇒ ∠BAQ = ∠ABL (Each 90°)

\ PQ||LM ( ∠BAQ and ∠ABL are alternate angles)

5. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at

angle of 80°, then find ∠POA. [NCERT]

Sol.  PA and PB are tangents to a circle with centre O,

\ OA ⊥ AP and OB ⊥ PB

i.e., ∠APB = 80°, ∠OAP = 90°, and ∠OBP = 90°

Now, in quadrilateral OAPB, we have

∠APB + ∠PBO + ∠BOA + ∠OAP = 360°

⇒ 80° + 90° + ∠BOA + 90° = 360°

⇒ 260° + ∠BOA = 360°

\ ∠BOA = 360° – 260° ⇒ ∠BOA = 100°

Now, in DPOA and DPOB we have

OP = OP (Common) Fig. 8.46
OA = OB (Radii of the same circle)

∠OAP = ∠OBP = 90°

\ DPOA  DPOB (RHS congruence condition)

⇒ ∠POA = ∠POB (CPCT)

Now, ∠POA = 12 × ∠BOA = 1 × 100° = 50°
2

212 Xam idea Mathematics–X

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm.

Find the radius of the circle. [NCERT]

Sol. Let O be the centre and P be the point of contact.

Since tangent to a circle is perpendicular to the radius through the
point of contact,

\ ∠OPA = 90°

Now, in right DOPA we have

OA2 = OP2 + PA2 (By Pythagoras Theorem) Fig. 8.47

52 = OP2 + 42

⇒ 25 = OP2 + 16

⇒ OP2 = 25 – 16 = 9 \ OP = 3cm

Hence, the radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger

circle which touches the smaller circle. [NCERT]

Sol. Let O be the common centre of two concentric circles and let AB be a chord
of larger circle touching the smaller circle at P. Join OP.

Since OP is the radius of the smaller circle and AB is tangent to this circle at P,

\ OP ⊥ AB

We know that the perpendicular drawn from the centre of a circle to any Fig. 8.48
chord of the circle bisects the chord.

Therefore, AP = BP

In right DAPO we have

OA2 = AP2 + OP2

52 = AP2 + 32 ⇒   25 – 9 = AP2
AP2 = 16 ⇒   AP = 4

Now, AB = 2.AP = 2 × 4 = 8 ( AP = PB)

Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm.

8. Prove that the tangents drawn at the ends of a chord of circle make equal angles with the

chord. [NCERT Exemplar, CBSE (AI) 2017]

Sol. Given: A circle with centre O, PA and PB are tangents drawn at the
ends A and B on chord AB.

To prove: ∠PAB = ∠PBA

Construction: Join OA and OB.

Proof: In DOAB we have

OA = OB …(i) (Radii of the same circle) Fig. 8.49

∠2 = ∠1 …(ii) (Angles opposite to equal sides of a D)

Also (∠2 + ∠3 = ∠1 + ∠4) ...(iii) (Both 90° as Radius ⊥ Tangent)

Subtracting (ii) from (iii), we have

\ ∠3 = ∠4 ⇒ ∠PAB = ∠PBA

Circles 213

9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through

the centre. [NCERT]

Sol. Let LM be tangent drawn at the point P on the circle with centre
O. Join OP. If possible, let PQ be perpendicular to LM, not passing
through O.

Now, since tangent at a point to a circle is perpendicular to the radius
through the point.

\ OP ⊥ LM ⇒ ∠OPM = 90° Fig. 8.50

Also, ∠QPM = 90° (As assumed above)

\ ∠OPM = ∠QPM, which is possible only when points O and Q are coincide.

Hence, the perpendicular at the point of contact to tangent to a circle passes through the centre.

10. A quadrilateral ABCD is drawn to circumscribe a circle (Fig. 8.51). Prove

that AB + CD = AD + BC. [NCERT, CBSE (AI) 2016]

OR

A circle touches all the four sides of a quadrilateral ABCD. Prove that

AB + CD = BC + DA [CBSE (AI) 2017, Delhi 2017 (C)]

Sol. Since lengths of two tangents drawn from an external point of circle are Fig. 8.51
equal,

Therefore, AP = AS, BP = BQ, DR = DS and CR = CQ

(where P, Q, R and S are the points of contact)

Adding all these, we have

(AP + BP) + (CR + RD) = (BQ + CQ) + (DS + AS)

⇒ AB + CD = BC + DA

11. A circle is touching the side BC of DABC at P and touching AB and AC produced at Q and R

respectively. Prove that AQ = 1 (Perimeter of DABC).
2

Sol. Since tangents from an exterior point to a circle are equal in length.

\ BP = BQ (Tangents from B) …(i)

CP = CR (Tangents from C) …(ii)

and, AQ = AR (Tangents from A) …(iii)

From (iii), we have Fig. 8.52

AQ = AR ⇒ AB + BQ = AC + CR

⇒ AB + BP = AC + CP (Using (i) and (ii)) …(iv)

Now, perimeter of DABC = AB + BC + AC

= AB + (BP + PC) + AC

= (AB + BP) + (AC + PC)

= 2(AB + BP) (Using (iv))

= 2(AB + BQ) = 2AQ (Using (i))

\ AQ = 1 (Perimeter of DABC)
2

214 Xam idea Mathematics–X

12. If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of
a+b–c
the circle which touches the sides of the triangle is given by r= 2 . [NCERT Exemplar]

Sol. Let the circle touches the sides BC, CA and AB of the right A
triangle ABC at D, E and F respectively,

where BC = a, CA = b, and AB = c

(See fig. alongside) b
Then AE = AF F

and BD = BF c r
Also CE = CD = r OE
i.e, AF = b – r,  a – r = BF
r

or AB = c = AF + BF = b – r + a – r B aD C

This gives, r= a+b– c Fig. 8.53
2

Long Answer Questions [5 marks]

1. Prove that the tangent to a circle is perpendicular to the radius through the point of contact.
[CBSE Delhi 2014, (AI) 2014, (F) 2014, 2020 (30/3/1)]

Sol. Given: A circle C(O, r) and a tangent AB at a point P.

To Prove: OP ⊥ AB.

Construction: Take any point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets
the circle at R.

Proof: We know that among all line segments joining the point O to a point on AB, the shortest
one is perpendicular to AB. So, to prove that OP ⊥ AB it is sufficient to prove that OP is shorter
than any other segment joining O to any point of AB.

Clearly, OP = OR (Radii of the same circle)

Now, OQ = OR + RQ

⇒ OQ > OR

⇒ OQ > OP (OP = OR)

Thus, OP is shorter than any other segment joining O to any point on AB. Fig. 8.54
Hence, OP ⊥ AB.

2. Prove that the lengths of two tangents drawn from an external point to a circle are equal.

[CBSE, Delhi 2014, (F) 2014, Delhi 2016, (AI) 2016, (F) 2016,

CBSE Delhi 2017, (AI) 2017, (F) 2017, Delhi 2017 (C)]

Sol. Given: AP and AQ are two tangents from a point A to a circle C (O, r).

To Prove: AP = AQ

Construction: Join OP, OQ and OA.

Proof: In order to prove that AP = AQ, we shall first prove that DOPA  DOQA.

Since a tangent at any point of a circle is perpendicular to the radius
through the point of contact.

\ OP ⊥ AP and OQ ⊥ AQ

⇒ ∠OPA = ∠OQA = 90° ...(i)

Now, in right triangles OPA and OQA, we have

OP = OQ (Radii of a circle) Fig. 8.55

Circles 215

∠OPA = ∠OQA (Each 90°)

and OA = OA (Common)

So, by RHS-criterion of congruence, we get

DOPA  DOQA ⇒ AP = AQ (CPCT)

Hence, lengths of two tangents from an external point are equal.

3. Prove that the parallelogram circumscribing a circle is a rhombus.

[CBSE Delhi 2014; CBSE 2019 (30/5/1)]

Sol. Let ABCD be a parallelogram such that its sides touch a circle with centre O.

We know that the tangents to a circle from an exterior point are equal in length.

Therefore, we have

AP = AS (Tangents from A) ... (i)
BP = BQ (Tangents from B) ... (ii)
CR = CQ (Tangents from C) ... (iii)

And DR = DS (Tangents from D) ... (iv)

Adding (i), (ii), (iii) and (iv), we have

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) Fig. 8.56
AB + CD = AD + BC

AB + AB = BC + BC      ( ABCD is a parallelogram \ AB = CD, BC = DA)

2AB = 2BC ⇒ AB = BC

Thus, AB = BC = CD = AD

Hence, ABCD is a rhombus.

4. In Fig. 8.57, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q

intersect at a point T. Find the length of TP. [CBSE (AI) 2014]

Sol. Given: PQ = 16 cm and PO = 10 cm

To find: TP 16 = 8 cm
PR = RQ = 2

(Perpendicular from the centre bisects the chord)

In DOPR,

OR = OP2 − PR2 = 102 − 82 = 100 − 64 Fig. 8.57
= 36 = 6 cm

Let ∠POR be q = PR = 8 = 4
In DPOR, tan q RO 6 3

We know, OP ⊥ TP (Point of contact of a tangent is perpendicular to the line from the centre)

In DOTP, tan q = OP ⇒ 4 = 10
TP 3 TP

TP = 10 × 3 = 15 = 7.5 cm.
4 2

5. If PQ is a tangent drawn from an external point P to a circle with centre O and QOR is a

diameter where length of QOR is 8 cm such that ∠POR = 120°, then find OP and PQ.

Sol. Let O be the centre and QOR = 8 cm is diameter of a circle. PQ is tangent such that ∠POR = 120°.

Now, OQ = OR = 8 = 4 cm
2

∠POQ = 180° – 120° = 60° (Linear pair)

Also OQ ⊥ PQ

216 Xam idea Mathematics–X

Now, in right ∆POQ, OQ
PO
cos 60° =

⇒ 1 = OQ ⇒ 1 = 4
2 PO 2 PO

⇒ PO = 8 cm

PQ 3 = PQ ⇒ PQ = 4 3 cm. Fig. 8.58
Again, tan 60° = OQ ⇒ 4

6. In Fig. 8.59, two equal circles, with centres O and O', touch each other at X.OO' produced

meets the circle with centre O' at A. AC is tangent to the circle with centre O, at the point C.

O'D is perpendicular to AC. Find the value of DO′ . [CBSE (AI) 2016]
Sol. AC is tangent to circle with centre O. CO


Thus ∠ACO = 90°

In DAO'D and DAOC

∠ADO' = ∠ACO = 90°

∠A = ∠A (Common)

\ DAO'D ~ DAOC (By AA similarity)

⇒ AO′ = DO′ Fig. 8.59
AO CO

\ DO′ = r = 1 ( AO = AO' + O' X+ XO = 3r)
CO 3r 3

7. In Fig. 8.60, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and
OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP

and TQ are two tangents to the circle. [CBSE Delhi 2016]

Sol. In right DPOT

PT = OT 2 − OP2

PT = 169 − 25 = 12 cm and

TE = 8 cm

Let PA = AE = x

(Tangents from an external point to a circle are equal)

In right ∆AET, Fig. 8.60

TA2 = TE2 + EA2

⇒ ( 12 – x)2 = 64 + x2 ⇒ 1 44 + x2 – 24x = 64 + x2
⇒ x = 3.3 cm
⇒ 80
x = 24

Thus, AB = 6.6 cm [ a D ATE @ DBTE, ⇒ AE = BE]

HOTS [Higher Order Thinking Skills]

1. Prove that opposite sides of a quadrilateral circumscribing a circle
subtend supplementary angles at the centre of the circle.
      [CBSE (AI) 2014, (F) 2017, 2019 (C) (30/1/1)]

Sol. Let a circle with centre O touches the sides AB, BC, CD and DA of a
quadrilateral ABCD at the points P, Q, R and S respectively. Then, we
have to prove that

∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180° Fig. 8.61

Circles 217

Now, join OP, OQ, OR and OS.

Since the two tangents drawn from an external point to a circle subtend equal angles at the
centre.

\ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8 …(i)

Now, ∠1 + ∠2 +∠3 + ∠4 + ∠5 +∠6 + ∠7 + ∠8 =360° …(ii)

(sum of all the angles subtended at a point is 360°)

⇒ 2(∠2 + ∠3 + ∠6 + ∠7) = 360° (using equation (i) and (ii))

⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°

⇒ ∠AOB + ∠COD = 180°

again 2(∠1 + ∠8 + ∠4 + ∠5) = 360° (from (i) and (ii))

\ (∠1 + ∠8) + (∠4 + ∠5) = 180°

⇒ ∠AOD + ∠BOC = 180°

2. A triangle ABC [Fig. 8.62] is drawn to circumscribe a circle of

radius 4 cm such that the segments BD and DC into which BC is

divided by the point of contact D are of lengths 8 cm and 6 cm

respectively. Find the sides AB and AC. [CBSE (AI) 2014]

Sol. Let DABC be drawn to circumscribe a circle with centre O and
radius 4 cm and circle touches the sides BC, CA and AB at D, E and
F respectively.

We have given that CD = 6 cm and BD = 8 cm Fig. 8.62
\ BF = BD = 8 cm and CE = CD = 6 cm

(Length of two tangents drawn from an external point of circle are equal)

Now, let AF = AE = x cm

Then, AB = c = (x + 8) cm, BC = a = 14 cm, CA = b = (x + 6) cm

\ 2s = (x + 8) + 14 + (x + 6)

⇒ 2s = 2x + 28 or   s = x + 14

s – a = (x + 14) – 14 = x

s – b = (x + 14) – (x + 6) = 8

s – c = (x + 14) – (x + 8) = 6

\ Area of ∆ABC = s( s − a)( s − b)( s − c)

= ( x +14)( x)(8)(6) = 48x ( x +14)

Also, area (∆ABC) = area (∆OBC) + area (∆OCA) + area (∆OAB)

= 1 × BC × OD + 1 × CA × OE + 1 × AB × OF
2 2 2

= 1 × 14 × 4 + 1 × (x + 6) × 4 + 1 × (x + 8) × 4
2 2 2

= 28 + 2x + 12 + 2x + 16 = 4x + 56

\ 48x ( x + 14) = 4x + 56 ⇒ 48x ( x + 14) = 4 (x + 14)

Squaring both sides, we have

48x (x + 14) = 16(x + 14)2  ⇒ 48x (x + 14) – 16 (x + 14)2 = 0

⇒ 16 (x + 14) [3x – (x + 14)] = 0

218 Xam idea Mathematics–X

⇒ 16(x + 14)(2x – 14) = 0

either 16(x + 14) = 0 or 2x – 14 = 0

⇒ x = –14 or 2x = 14

⇒ x = –14 or x = 7

But x cannot be negative so x ≠ – 14

\ x = 7 cm

Hence, the sides AB = x + 8 = 7 + 8 = 15 cm and

AC = x + 6 = 7 + 6 = 13 cm.

3. In Fig. 8.63, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent
AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.

[CBSE (AI) 2017, 2019 (30/2/1)]

Sol. Join OC.

In DAPO and DACO, we have

AP = AC (Tangents drawn from external point A)

AO = OA (Common)

PO = OC (Radii of the same circle)

\ DAPO  DACO (By SSS criterion of congruence)

\ ∠PAO = ∠CAO (CPCT)

⇒ ∠PAC = 2∠CAO Fig. 8.63

Similarly, we can prove that DOQB  DOCB.

\ ∠QBO = ∠CBO ⇒ ∠CBQ = 2∠CBO

Now, ∠PAC + ∠CBQ = 180° (Sum of interior angles on the same side of transversal is 180°)

⇒ 2∠CAO + 2∠CBO = 180°

⇒ ∠CAO + ∠CBO = 90°

⇒ 180° – ∠AOB = 90°   ( ∠CAO + ∠CBO + ∠AOB = 180°)

⇒ 180° – 90° = ∠AOB ⇒ ∠AOB = 90°

4. Let A be one point of intersection of two intersecting circles with
centres O and Q. The tangents at A to the two circles meet the circles
again at B and C respectively. Let the point P be located so that AOPQ
is a parallelogram. Prove that P is the circumcentre of the triangle
ABC.

Sol. In order to prove that P is the circumcentre of DABC it is sufficient to

show that P is the point of intersection of perpendicular bisectors of

the sides of DABC i.e., OP and PQ are perpendicular bisectors of sides Fig. 8.64

AB and AC respectively. Now, AC is tangent at A to the circle with centre at O and OA is its radius.

\ OA ⊥ AC
⇒ PQ ⊥ AC ( OAQP is a parallelogram so, OA||PQ)
Also, Q is the centre of the circle

QP bisects AC (Perpendicular from the centre to the chord bisects the chord)

⇒ PQ is the perpendicular bisector of AC.
Similarly, BA is the tangent to the circle at A and AQ is its radius through A.

\ BA ⊥ AQ ∵ AQPO is parallelogram 
\ BA ⊥ OP ∴ OP AQ 


Circles 219

Also, OP bisects AB ( O is the centre of the circle)
⇒ OP is the perpendicular bisector of AB.
Thus, P is the point of intersection of perpendicular bisectors PQ and PO of sides AC and AB

respectively.

Hence, P is the circumcentre of DABC.

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) The distance between two parallel tangents of a circle of radius 4 cm is

(a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm

(ii) In Fig. 8.65, ∠RPS = 25°, ∠ROS is

(a) 135° (b) 145° (c) 165° (d) 155°

R

O 25° P

S

Fig. 8.65
(iii) A tangent is drawn from a point at a distance of 17 cm of circle (o, r) of radius 8 cm. The

length of tangent is

(a) 5 cm (b) 9 cm (c) 15 cm (d) 23 cm

(iv) The lengths of tangents drawn from an external point to the circle

(a) are equal (b) are not equal

(c) some times are equal (d) are not defined

(v) Tangents drawn at the extremities of the diameter of a circle are

(a) parallel (b) perpendicular (c) equal (d) none

2. Fill in the blanks.
(i) A parallelogram circumscribing a circle is _____________ .

(ii) The two tangents are drawn to a circle from an external point, then they subtend
_____________ angle at the centre.

(iii) A tangent is perpendicular to _____________ through the point of contact.

(iv) Two concentric circles are of radii 13 cm and 5 cm. The length of the chord of larger circle
which touches the smaller circle is _____________ .

(v) A line which intersects a circle in two distinct points is called _____________ of circle.

QQ Very Short Answer Questions : [1 mark each]

3. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger

circle which touches the smaller circle. [CBSE 2019 (30/3/1)]

4. From a point A, the length of the tangent to a circle is 24 cm and the distance of A from the
centre is 26 cm. What will be the radius of the circle?

220 Xam idea Mathematics–X

5. At one end A of a diameter AB of a circle of radius 13 cm, tangent XAY is drawn to the circle. A
chord CD is parallel to XY and is at a distance of 18 cm from A. What will be the length of CD?

6. From a point A which is at a distance of 10 cm from the centre O of a circle of radius 6 cm, the pair
of tangents AB and AC to the circle are drawn. What will be the area of the quadrilateral ABOC?

7. In Fig 8.66, DABC is circumscribing a circle. Find the length of AB.

Fig. 8.66
8. In Fig. 8.67, PQ is a tangent of length 6 cm to the circle with centre O and ∠OQP = 60°. Find OQ.

Fig. 8.67
9. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an

angle of 110°, find ∠POA.
10. In Fig. 8.68, AB is a chord of a circle with centre O and AP is the tangent at A such that

∠BAP = 75°. Find ∠ACB.

Fig. 8.68
11. In Fig. 8.69, if ∠AOB = 125°, then find ∠COD.

Fig. 8.69

Circles 221

12. In Fig. 8.70, RQ is a chord of the circle and POQ is its diameter such that ∠RPQ = 30°. If QT is
the tangent to the circle at the point Q, then find ∠RQT.

Fig. 8.70

13. In Fig. 8.71, AOB is a diameter of a circle with centre O and AC is a tangent to the circle at A. If
∠BOC = 130°, then find ∠ACO.

Fig. 8.71

14. PQ is a tangent drawn from an external point P to a circle with centre O, QOR is the diameter of

the circle. If ∠POR = 120°, what is the measure of ∠OPQ? [CBSE (F) 2017]

QQ Short Answer Questions-I: [2 marks each]

15. In Fig. 8.72, ABC is a triangle in which ∠B = 90°, BC = 48 cm and AB = 14 cm. A circle is

inscribed in the triangle, whose centre is O. Find radius r of in-circle. [CBSE 2018 (30/1)]

C

48 cm rr A
B O

14 cm

Fig. 8.72
16. In Fig. 8.73, BOA is a diameter of a circle with centre O and the tangent at a point P meets BA

extended at T. If ∠ABP = 40°, then find the value of ∠PTA

Fig. 8.73 Fig. 8.74

17. In Fig. 8.74, CP and CQ are tangents from an external point C to
a circle with centre O. AB is another tangent which touches the
circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC.

222 Xam idea Mathematics–X

18. Find the length of the tangent from an external point P at a distance of 20 cm from the centre
of a circle of radius 12 cm.

19. If the angle between two tangents drawn from an external point P to a circle of radius a and

centre O, is 60°, then find the length of OP. [CBSE (AI) 2017]

20. In Fig. 8.75, BA and BC are tangents to the circle drawn from an external point B. PQ is a third
tangent touching the circle at R. If BC = 12 cm and PR = 3 cm, what is the perimeter of DBPQ?

Fig. 8.75

21. In Fig. 8.76, there are two concentric circles with centre O and of radii 5 cm and 3 cm. From an
external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length
of BP.

Fig. 8.76

QQ Short Answer Questions-II: [3 marks each]

22. Prove that the lengths of tangents drawn from an external point to a circle are equal.
[CBSE 2018, (30/1)]

23. In Fig. 8.77, AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at

A and B intersect at P. Find the length of AP. [CBSE 2018, (C) (30/1)]

A

5 cm P
O 8 cm

B

Fig. 8.77

24. Two concentric circles are of radii 8 cm and 5 cm. Find the length of the chord of the larger circle
which touches the smaller circle.

25. In Fig. 8.78, DABC is an isosceles triangle in which AB = AC, circumscribed about a circle, as
shown in the figure. Prove that the base is bisected by the point of contact.

Fig. 8.78

Circles 223

26. In the figure shown below, point C is the centre of the circle with a radius of 8 cm and
∠QRS = 80°. Find the length of arc QTR. (Use π = 3.14)

Q

T
CS
P 80°

R

Fig. 8.79
27. In Fig. 8.80, ABC is a right triangle, right angled at B such that BC = 6 cm and AB = 8 cm. Find

the radius of its incircle.

Fig. 8.80
28. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that

QORP is a cyclic quadrilateral.
29. In Fig. 8.81, PQL and PRM are tangents to the circle with centre O at the points Q and R,

respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Find ∠QSR.

Fig. 8.81
30. In Fig. 8.82, ‘O’ is the centre of the circle. Determine ∠AQB and ∠AMB, if PA and PB are

tangents and ∠APB = 75°.

Fig. 8.82
31. In Fig. 8.83, from a point P two tangents PT and PS are drawn to a circle with centre O and

radius r. If OP = 2r, then show that DTPS is an equilateral triangle.

Fig. 8.83

224 Xam idea Mathematics–X

32. Find the actual lengths of sides of DOTP. (Fig. 8.84.)

Fig. 8.84
33. Find the perimeter of DEFG. (Fig. 8.85).

Fig. 8.85
34. If d1, d2, (d2 > d1) be the diameters of two concentric circles and c be the length of a chord of a

circle which is tangent to the other circle, prove that d22 = c2 + d12 .

35. Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle
touches the sides BC, CA, AB, at D, E, F, respectively, prove that BD = s – b.

QQ Long Answer Questions: [5 marks each]

36. If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
[NCERT Exemplar]

37. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the
hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

38. A is a point at a distance 10 cm from the centre O of a circle of radius 6 cm. AP and AQ are the
tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ
to intersect AP at B and AQ at C, find the perimeter of the DABC.

39. If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find

the area of the triangle. [NCERT Exemplar]

40. In Fig. 8.86, the tangent at a point C of a circle and a diameter AB when extended intersect at
P. If ∠PCA = 130°, find ∠CBA.

Fig. 8.86

Answers

1. (i) (d) (ii) (d) (iii) (c) (iv) (a) (v) (a)

2. (i) rhombus (ii) equal (iii) radius (iv) 24 cm (v) secant

3. 2 a2 – b2 4. 10 cm 5. 24 cm 6. 48 cm2 7. 14 cm 8. 12 cm

Circles 225

9. 35° 10. 105° 11. 55° 12. 30° 13. ∠ACO = 40°

14. ∠OPQ = 30° 15. r = 6 cm 16. ∠PTA = 10° 17. 7 cm

18. 16 cm 19. 2a 20. 24 cm 21. 4 10 cm 23. AP = 20 cm 24. 2 39 cm
29. 70° 3

26. 22.33 cm 27. 2 cm 30. 52.5°, 127.5°

32. 6 cm, 8 cm, 10 cm 33. 35 cm 38. 16 cm 39. 8 2 cm2 40. 50°

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) In Fig. 8.87, if PA and PB are tangents to the circle with centre O such that ∠APB = 80°,
then ∠OAB is equal to

A O
P 80°

B

Fig. 8.87

(a) 25° (b) 30° (c) 40° (d) 50°

(ii) In Fig. 8.88, if PQR is tangent to a circle at Q whose centre is O, AB is a chord parallel to
PR and ∠BQR = 70°, then ∠AQB is equal to

AB

O

70°

PQR

Fig. 8.88

(a) 20° (b) 40° (c) 35° (d) 45°

(iii) If tangents AB and AC from a point A to a circle with centre O are inclined to each other at
angle of 70°, then ∠AOB is equal to

(a) 110° (b) 55° (c) 70° (d) 60°

(iv) In a cyclic quadrilateral one angle is 40°, then the opposite angle is equal to

(a) 140° (b) 50° (c) 60° (d) Cannot say

226 Xam idea Mathematics–X

2. Fill in the blanks. (3 × 1 = 3)

(i) In Fig. 8.89, AB is the diameter and AC is a chord of the circle such that ∠CAB = 40°. If BD
is the tangent to the circle, then ∠CBD is _______________ .

A B
40°

CD

Fig. 8.89

(ii) If radii of two concentric circles are 6 cm and 10 cm, then length of each chord of one circle
which is tangent to the other circle is ______________ .

(iii) Sum of opposite angles of a cyclic quadrilateral is ______________ .

3. Solve the following questions. (3 × 1 = 3)

(i) In a circle of radius 3 cm, a point lies 5 cm away from its centre. What is the length of the
tangent to the circle through this point?

(ii) Radii of two concentric circles are 5 cm and x cm, if x is radius of smaller circle and length of
each chord of one circle which is tangent to other, is 6 cm, find x.

(iii) If the angle between two radii of circle is 125°, then find the angle between tangents at the
ends of the radii.

Section B

QQ Solve the following questions. [3 × 2 = 6]

4. If the angle between two tangents drawn from an external point P to a circle of radius a and

centre O is 60°, then find the length of OP. [CBSE (AI) 2017]

5. In Fig. 8.90, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58° find

∠ATQ. [CBSE Delhi 2015]

Fig. 8.90
6. Two tangents TP and TQ are drawn to a circle with centre O from an external point T (Fig. 8.91).

Prove that ∠PTQ = 2 ∠OPQ.       [CBSE Delhi 2017, Delhi 2017 (C)]

Q

T

P
Fig. 8.91

Circles 227

QQ Solve the following questions. [3 × 3 = 9]

7. In Fig. 8.92, PA and PB are tangents to a circle from an external point P such that PA = 4 cm

and ∠BAC = 135°. Find the length of chord AB. [CBSE (F) 2017]

P

4 cm

A B
135°

C

Fig. 8.92
8. In Fig. 8.93, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB

= 29 cm and DS = 5 cm, find the radius (r) of the circle.

Fig. 8.93

9. In Fig. 8.94, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is

drawn parallel to the tangent PQ. Find ∠RQS. [NCERT Exemplar]

Fig. 8.94

QQ Solve the following questions. [3 × 5 = 15]

10. In Fig. 8.95, O is the centre of the circle and TP is the tangent to the circle from an external point

T. If ∠PBT = 30°, prove that BA : AT = 2 : 1. [CBSE (F) 2015]

Fig. 8.95

228 Xam idea Mathematics–X

11. In Fig. 8.96, AB is a chord of a circle, with centre O, such that AB = 16 cm and radius of circle is
10 cm. Tangents at A and B intersect each other at P. Find the length of PA. [CBSE (F) 2015]

Fig. 8.96

12. In Fig. 8.97, the common tangent, AB and CD of two circles with centres O and O’ intersect at E.

Prove that the points O, E, O′ are collinear. [NCERT Exemplar]

Fig. 8.97

Answers

1. (i) (c) (ii) (b) (iii) (b) (iv) (a)

2. (i) 40° (ii) 16 cm (iii) 180°

3. (i) 4 cm (ii) 4 cm (iii) 55°

4. 2a 5. 61° 7. 4 2 cm 8. 11 cm 9. 30°

11. 40 cm
3
zzz

Circles 229

9 Constructions

BASIC CONCEPTS – A FLOW CHART

230 Xam idea Mathematics–X

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. To divide a line segment AB in the ratio m : n (m, n are positive integers), draw a ray AX, so
that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the
minimum number of these points is

(a) mn (b) m + n – 1 (c) m + n (d) reater of m and n

2. To draw a pair of tangents to a circle which are inclined to each other at an angle of 30°, it
is required to draw tangents at end points of those two radii of the circle, the angle between
which should be

(a) 60° (b) 90° (c) 120° (d) 150°

3. To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ∠BAX is an acute

angle and then at equal distances points are marked on the ray AX such that the minimum

number of these points is [NCERT Exemplar]

(a) 8 (b) 10 (c) 11 (d) 12

4. The construction of triangle is not possible when the three sides of triangle are

(a) 6 cm, 7 cm, 8 cm (b) 8 cm, 3 cm, 4 cm

(c) 4 cm, 4 cm, 4 cm (d) None of these

5. To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute

angle and then points A1, A2, A3,.. are located at equal distances on ray AX and the point B is

joined to [NCERT Exemplar]

(a) A12 (b) A11 (c) A10 (d) A9

6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is

required to draw tangents at end points of those two radii of the circle, the angle between them

should be [NCERT Exemplar]

(a) 135° (b) 90° (c) 60° (d) 120°

7. To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle,

then draw a ray BY parallel to AX and the points A1, A2, A3, ... and B1, B2, B3, ... are located at
equal distances on ray AX and BY, respectively. Then the point joined are [NCERT Exemplar]

(a) A5 and B6 (b) A6 and B5 (c) A4 and B5 (d) A5 and B4

8. To construct a triangle similar to a given ∆ABC, with its sides 3 of the corresponding sides of
7
∆ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of

A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and next step
is to join

(a) B10 to C (b) B3 to C (c) B7 to C (d) B4 to C

Answers

1. (c) 2. (d) 3. (d) 4. (b) 5. (b) 6. (d)
7. (a) 8. (c)

Constructions 231

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Two points on a line segment are marked such that the three parts they make are equal then we
say that the two points _________________ the line segment.

2. Two circles are drawn with same centre then the _________________ circle have bigger radius.

3. Only two _________________ can be drawn to a circle from an external point.

4. A curve made by moving one point at a fixed distance from another is called ________________.
8
5. To construct a triangle similar to a given ∆ABC, with its sides 5 of the corresponding sides of

∆ABC, we draw a ray BX, on opposite sides of A, such that ∠CBX is an acute angle, the minimum

number of points to be located at equal distances on ray BX is ________________.

6. To divide a line segment in the ratio 2 : 3, it is divided into ________________ parts.

Answers 2. outer 3. tangents 4. circle 5. 8 6. 5

1. trisect

Very Short Answer Questions [1 mark]

1. Is construction of a triangle with sides 8 cm, 4 cm, 4 cm possible?
Sol. No, we know that in a triangle sum of any two sides of a triangle is greater than the third side.

Here the condition is not satisfied.

2. To divide the line segment AB in the ratio 3:7, draw a ray AX such that ∠BAX is an acute angle,
then draw a ray BY parallel to AX and the point A1, A2, A3... and B1, B2, B3... are located at equal
distances on ray AX and BY respectively. Then which points should be joined?

Sol. A3 and B7.

3. To draw a pair of tangents to a circle which are inclined to
each other at an angle of 80°, it is required to draw tangents
at end points of those two radii of the circle. What should
be the angle between them?

Sol. 100° P
4. In Fig. 9.1 by what ratio does P divide AB internally?

Sol. From Fig. 9.1, it is clear that there are 3 points at equal Fig. 9.1
distances on AX and 4 points at equal distances on BY. Here
P divides AB on joining A3 B4. So P divides AB internally by
3:4

5. Given a triangle with side AB = 8 cm. To get a line segment
Sol. AGBiv'e=n 43 ofAABB =, in8wcmhat ratio will line segment AB be divided?
3 of AB = 43
AB' = 4 × 8 = 6 cm.

BB' = AB – AB' = 8 – 6 = 2 cm.

⇒ AB' : BB' = 6 : 2 = 3 : 1

Hence the required ratio is 3 : 1.

Fig. 9.2

232 Xam idea Mathematics–X

Short Answer Questions-I and II [2, 3 marks]

1. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides

are 2 of the corresponding sides of the first triangle. [NCERT]
3
OR

Draw a triangle with sides 4 cm, 5 cm and 6 cm. Then construct another triangle whose sides

are 2 of the corresponding sides of first triangle. [CBSE Delhi (C) 2017]
3

Sol. Steps of Construction:

Step I: Draw a line segment BC = 6 cm.

Step II: Draw an arc with B as centre and radius equal to 5 cm.

Step III: Draw an arc, with C as centre and radius equal to 4 cm
intersecting the previous drawn arc at A.

Step IV: Join AB and AC, then DABC is the required triangle.

Step V: Below BC make an acute angle CBX

Step VI: Along BX mark off three points at equal distance: B1, B2, B3, Fig. 9.3
such that BB1 = B1 B2 = B2 B3.

Step VII: Join B3C.

Step VIII: From B2, draw B2D ||B3C, meeting BC at D.

Step IX: From D draw ED ||AC meeting BA at E. Then we have DEDB which is the required

triangle.

Justification:

Since DE ||CA

\ DABC ~ DEBD and EA=BB BB=DC DC=AE 2
3
2
Hence, we have the new DEBD similar to the given DABC, whose sides are equal to 3 rd of the
corresponding sides of DABC.

2. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

[NCERT]

Sol. Steps of Construction:

Step I: Draw a line segment AB = 7.6 cm.

Step II: Draw any ray AX making an acute angle ∠BAX with AB.

Step III: On ray AX starting from A, mark 5 + 8 = 13 equal arcs. AA1, A1A2, A2A3, A3A4, ...
A11A12, and A12A13.

Step IV: Join A13B.
Step V: From A5, draw A5P || A13B, meeting AB at P.
Thus, P divides AB in the ratio 5 : 8. On measuring the two

parts, we find AP = 2.9 cm and PB = 4.7 cm (approx).

Justification:

In DABA13, PA5 ||BA13

\ By Basic Proportionality Theorem

=APPB A=A5 AA153 5 Fig. 9.4
8 AP : PB = 5:8
⇒ AP = 5
PB 8 ⇒

Constructions 233

3. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then draw another
1
triangle whose sides are 1 2 times the corresponding sides of the isosceles triangle.

[NCERT, CBSE Delhi 2017]

Sol. Steps of Construction:

Step I: Draw BC = 8 cm.

Step II: Construct XY, the perpendicular bisector of line
segment BC, meeting BC at M.

Step III: Along MP, cut-off MA = 4 cm.

Step IV: Join BA and CA. Then DABC so obtained is the required

DABC. that BD = 12 cm  = 3 × 8 cm  . Fig. 9.5
Step V: Extend BC to D, such  2 

Step VI: Draw DE  CA meeting BA produced at E. Then DEBD is the required triangle.

Justification:

Since, DE  CA EA=BB DC=AE BB=DC 12 = 3 1 3
\ DABC ~ DEBD and 8 2 2 2

Hence, we have the new triangle similar to the given triangle whose sides are 1 i.e, times the

corresponding sides of the isosceles DABC.

4. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a
3
triangle whose sides are 4 th of the corresponding sides of the triangle ABC.

[NCERT, CBSE Delhi 2017 (C)]
Sol. Steps of Construction:

Step I: Construct a DABC in which BC = 6 cm and, AB = 5 cm and
∠ABC = 60°.

Step II: Below BC make an acute ∠CBX.

Step III: Along BX mark off four arcs: B1, B2, B3 and B4 such that
BB1 = B1B2 = B2B3 = B3B4.
Step IV: Join B4C.
Step V: From B3, draw B3D  B4C, meeting BC at D.

Step VI: From D, draw ED  AC, meeting BA at E.

Fig. 9.6

Now, we have DEBD which is the required triangle whose sides are 3 th
of the corresponding sides of DABC. 4

Justification:

Here, DE  CA

\ DABC ~ DEBD

and EA=BB BB=DC DC=AE 3 3
4 4
Hence, we get the new triangle similar to the given triangle whose sides are equal to th of the

corresponding sides of DABC.

234 Xam idea Mathematics–X

5. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of

tangents to the circle and measure their lengths. [NCERT]

Sol. Steps of Construction:

Step I: Take a point O and with O as centre draw a circle of radius 6 cm.

Step II: Take a point P at a distance of 10 cm from the centre O.

Step III: Join OP and bisect it. Let M be the mid-point.

Step IV: With M as centre and MP as radius, draw a circle to
intersect the previous circle at Q and R.

Step V: Join PQ and PR. Then, PQ and PR are the required
tangents. On measuring, we find, PQ = PR = 8 cm.

Justification: Fig. 9.7

On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in the semicircle.

\ PQ ⊥ OQ

Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is
also a tangent to the circle.

6. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius
6 cm and measure its length. Also, verify the measurement by actual calculation.

Sol. Steps of Construction:
Step I: Take a point O and draw a circle of radius OA = 4 cm. Also, draw a concentric circle of

radius OB = 6 cm

Step II: Find the mid-point C of OB and draw a circle of radius
OC = BC. Suppose this circle intersects the circle of
radius 4 cm at P and Q.

Step III: Join BP and BQ to get the desired tangents from a
point B on the circle of radius 6 cm.

By actual measurement, we find BP = BQ = 4.5 cm.

Justification:

In DBPO, we have

∠BPO = 90°, OB = 6 cm and OP = 4 cm Fig. 9.8

\ OB2 = BP2 + OP2 [Using Pythagoras theorem]

⇒ BP = OB2 − OP2 ⇒ BP = 36 − 16 = 20 cm = 4.47 cm

Similarly, BQ = 4.47 cm

7. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and

taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from

the centre of the other circle. [CBSE (F) 2017]

Sol. Steps of Construction:

Step I: Draw a line segment AB = 8 cm.

Step II: With A as centre, draw a circle of radius 4 cm
and let it intersect the line segment AB in M.

Step III: With B as centre, draw a circle of radius 3 cm.

Step IV: With M as centre, draw a circle of radius AM
and let it intersect the given two circles in P, Q
and R, S.

Step V: Join AP, AQ, BR and BS.

These are the required tangents.

Fig. 9.9

Constructions 235

Justification:

On joining BP, we have ∠BPA = 90°, as ∠BPA is the angle in the semicircle.

\ AP ⊥ PB.

Since BP is the radius of given circle, so AP has to be a tangent to the circle.

Similarly, AQ, BR and BS are also the tangents to the given circles.

8. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then draw another
3
triangle whose sides are 4 times the corresponding sides of the isosceles triangle.

[CBSE 2019 (30/5/1)]
Sol. Steps of Construction:

Step I: Draw base BC = 8 cm.

Step II: Draw perpendicular bisector of BC.

Step III: Let perpendicular bisector intersect BC at D.

Step IV: Taking D as centre and 4 cm as radius, draw an arc
intersecting perpendicular bisector at A.

Step V: Join AB and AC. Then ∆ABC is an isosceles triangle.

Now,

Step VI: Draw any ray BX making an acute angle with BC on the Fig. 9.10
side opposite to the vertex A.

Step VII: Mark four points B1, B2, B3, B4 on BX such that BB1 = B1B2
= B2B3 = B3B4.

Step VIII: Join B4C and draw a line through B3 parallel to B4C which intersect BC at C'.
Step IX: Draw a line through C' parallel to CA to intersect AB at A'

Thus, A' BC' is the required triangle.

Long Answer Questions [5 marks]

1. Draw a circle of radius of 3 cm. Take two points P and Q on one of its diameters extended

on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the

circle from these two points P and Q. [CBSE (F) 2017]

Sol. Steps of Construction:

Step I: Taking a point O as centre, draw a circle of radius 3 cm.

Step II: Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.

Step III: Bisect OP and OQ and let M1 and M2 be the mid-points of OP and OQ respectively.
Step IV: Draw a circle with M1 as centre and M1 P as radius to intersect the circle at T1 and T2.
Step V: Join PT1 and PT2.

Fig. 9.11

236 Xam idea Mathematics–X

Then, PT1 and PT2 are the required tangents. Similarly, the tangents QT3 and QT4 can be
obtained.

Justification:

On joining OT1, we find ∠PT1O = 90°, as it is an angle in the semicircle.

PT1 ⊥ OT1

Since OT1 is a radius of the given circle, so PT1 has to be a tangent to the circle.

Similarly, PT2, QT3 and QT4 are also tangents to the circle.

2. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the

perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from

A to this circle. [CBSE Delhi 2014]

Sol. Steps of Construction:

Step I: Draw DABC and perpendicular BD from B on AC.

Step II: Draw a circle with BC as a diameter. This circle will pass
through D.

Step III: Let O be the mid-point of BC. Join AO.

Step IV: Draw a circle with AO as diameter. This circle cuts the circle
drawn in step II at B and E.

Step V: Join AE. AE and AB are desired tangents drawn from A to the Fig. 9.12
circle passing through B, C and D.

3. Construct a right triangle in which sides (other than the hypotenuse) are 8 cm and 6cm.
3
Then construct another triangle whose sides are 5 times the corresponding sides of the right

triangle. [CBSE 2019 (30/5/2)]

Sol. Steps of Construction: Y
Step I: Draw the base, BC = 8 cm. A

Step II: At point B draw an ∠YBC = 90°. 6 cm
Step III: Taking B as centre and radius 6 cm draw an arc A'
intersecting BY at A.

Step IV: Join AC.

Thus, ∆ABC is constructed (obtained) 3
5
Now, we construct another triangle whose sides are 90° C' C
times of the sides of ∆ABC. B
8 cm

B1

Steps of Construction: B2

B3

(i) Draw a line BX making acute angle with BC. B4
(ii) Take 5 points on BX such that B5
BB1 = B1B2 = B2B3 = B3B4 = B4B5
X
(iii) Join B5C.
Fig. 9.13

(iv) Draw a line through B3 parallel to B5C which cuts BC at C′ i.e., B5C  B3C′.

(v) Draw a line through C' parallel to AC which cuts AB at A'.

Thus, DA′BC′ is the required triangle.

4. Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle

similar to DABC with scale factor 3 . Justify the construction. Are the two triangles congruent?
2

Note that all the three angles and two sides of the two triangles are equal. [NCERT Exemplar]

Constructions 237

Sol. Steps of Construction
Step I: Draw a line segment BC = 6 cm.
Step II: With centre B and radius 4 cm draw an arc.
Step III: With centre C and radius 9 cm draw another arc which intersects the previous arc at A.
Step IV: Join BA and CA. ABC is the required triangle.
Step V: Through B, draw an acute angle CBX on the side opposite to vertex A.
Step VI: Locate three arcs B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3.
Step VII: Join B2C.
Step VIII: Draw B3C' || B2C intersecting the extended line segment BC at C'.
Step IX: Draw C'A' || CA intersecting the extended line segment BA at A'.

Fig. 9.14

Thus, DA'BC' is the required triangle (DA'BC' ~ DABC).

The two triangles are not congruent because, if two triangles are congruent, then they have
same shape and size. Here, all three angles are equal but all three sides are not equal.

Justification:

 B2C || B3C'

BC = 2
CC ' 1

Now, BC ' = BC + CC ' = 1 + CC ' =1 + 1 = 3
BC BC BC 2 2

Again, CC' || AC

DABC ~ DA'BC'
\
A'B = BC' = A'C' = 3
AB BC AC 2

238 Xam idea Mathematics–X

HOTS [Higher Order Thinking Skills]

1. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of

tangents from this point to the circle. [NCERT]

Sol. Steps of Construction:

Step I: Draw a circle with the help of a bangle.

Step II: Let P be the external point from where the tangents are
to be drawn to the given circle. Through P, draw a secant
PAB to intersect the circle at A and B (say).

Step III: Produce AP to a point C, such that AP = PC, i.e., P, is the
mid-point of AC.

Step IV: Draw a semicircle with BC as diameter. Fig. 9.15

Step V: Draw PD ⊥ CB, intersecting the semicircle at D.

Step VI: With P as centre and PD as radius, draw arcs to intersect the given circle at T and T1.

Step VII: Join PT and PT1. Then, PT and PT1 are the required tangents.

2. Draw a DABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose

sides are 3 times the corresponding sides of DABC. [CBSE Delhi 2017, (AI) 2017]
4 OR

Construct a triangle ABC with side BC = 6 cm, ∠B = 45°, ∠A = 105°. Then construct another

triangle whose sides are 3 times the corresponding sides of the DABC. [CBSE 2019 (30/1/2)]
4

Sol. Steps of Construction:

Step I: Construct a DABC in which BC = 7 cm,

∠ B = 45°, ∠C = 180° – (∠A + ∠B)
= 180° – (105° + 45°) = 180° – 150° = 30°.

Step II: Below BC, make an acute angle ∠CBX. Fig. 9.16

Step III: Along BX, mark off four arcs: B1, B2, B3 and B4 such that
BB1 = B1B2 = B2B3 =B3B4.
Step IV: Join B4C
Step V: From B3, draw B3D || B4C meeting BC at D.

Step VI: From D, draw ED || AC, meeting BA at E. Then EBD is the required triangle whose

sides are 3 times the corresponding sides of DABC.
4
Justification:

a DE || CA EB BD DE 3
AB BC CA 4
\ DABC ~ DEBD and = = =

Hence, we have the new triangle similar to the given triangle whose sides are equal to 3 times
the corresponding sides of DABC. 4

OR
Similar solution as above, only length of BC change from 7 cm to 6 cm.

Constructions 239

3. Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle
of 60°.

OR

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°.

Also justify the construction. Measure the distance between the centre of the circle and the

point of intersection of tangents. [NCERT Exemplar, CBSE (AI) 2016]

Sol. Steps of Construction:

Step I: Draw a circle with centre O and radius 4 cm.

Step II: Draw any diameter AOB.

Step III: Draw a radius OC such that ∠BOC = 60°.

Step IV: At C, we draw CM ⊥ OC and at A, we draw AN ⊥ OA.

Step V: Let the two perpendiculars intersect each other at P.

Then, PA and PC are required tangents.

Justification:

Since OA is the radius, so PA has to be a tangent to the circle.
Similarly, PC is also tangent to the circle.

∠APC = 360° – (∠OAP + ∠OCP + ∠AOC) Fig. 9.17

= 360° – (90° + 90° + 120°) = 360° – 300° = 60°

Hence, tangents PA and PC are inclined to each other at an angle of 60°.

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) To divide a line segment AB in the ratio 5 : 7, first a ray PX is drawn so that ∠QPX is an
acute angle, then at equal distances points are marked on ray PX, such that minimum
number of these points is

(a) 5 (b) 7 (c) 12 (d) 10

(ii) To divide a line segment AB in the ratio 4 : 7 a ray AX is drawn first such that ∠BAX is an
acute angle and then points A1, A2, A3 .... are located at equal distances on ray AX, and point
B is joined to

(a) A4 (b) A11 (c) A10 (d) A7

(iii) To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it

is required to draw tangents at the end points of those two radii of circle, the angle between

which is

(a) 145° (b) 135° (c) 120° (d) 110°

(iv) When a line segment is divided in the ratio 2 : 3, how many parts is it divided into?

(a) 2 (b) 2 (c) 3 (d) 5
3
2
(v) To construct a triangle similar to a given ∆ABC, with its sides 5 of the corresponding sides

of ∆ABC, first draw an acute angle CBX below BC, then after locating points A1, A2 ..... on
BX at equal distances, next step is to join

(a) A7 to C (b) A5 to C (c) A2 to C (d) A4 to C

240 Xam idea Mathematics–X

QQ Very Short Answer Questions: [1 mark each]

2. To divide a line segment AB in the ratio a : b (a, b are positive integers), a ray AX is drawn so that
∠BAX is an acute angle and points on ray AX are marked at equal distances. What should be the
minimum number of these points?

3. In order to divide a line segment PQ in the ratio 3 : 7, a ray PX is drawn such that ∠QPX is an

acute angle and then points P1, P2, ... are located at equal distances on the ray PX. With which

point is the point Q joined? 3
5
4. To construct a triangle similar to given DABC with its sides of the corresponding sides of DABC,

we draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect

to BC. Then we locate points B1, B2, ... on BX at equal distances. What will be the next step?

5. To construct a triangle similar to a given DABC with its sides 7 of the corresponding sides of
5

DABC, we draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with

respect to BC. Then we locate points B1, B2, ... on BX at equal distances. What will be the next step?

6. To construct a triangle similar to a given DPQR with its sides 9 of the corresponding sides of
4
DPQR, we draw a ray QX such that ∠RQX is an acute angle and X is on the opposite side of P

with respect to QR. What will be the minimum number of points to be located at equal distances

on ray QX?

7. To draw a pair of tangents to a circle which are inclined to each other at an angle of 80°, it is
required to draw tangents at end points of those two radii of the circle. What will be the angle
between the two radii?

QQ Short Answer Questions–I and II: [2 and 3 marks each]

8. Construct a right triangle ABC in which AC = 13 cm, AB = 5 cm and ∠B = 90°. Construct a
3
triangle similar to it and of scale factor 4 . Is the new triangle also a right triangle?

9. Draw a line segment of length 6.5 cm. Find a point P on it which divides it in the ratio 3 : 2.

10. Draw a line segment of length 8 cm and divide it internally in the ratio 4:5. [CBSE Delhi 2017]

11. Draw a line segment of length 7 cm and divide it internally in the ratio 2 : 3. [CBSE Delhi 2017]

12. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5cm and 4 cm.
5
Then, construct another triangle whose sides are 3 times the corresponding sides of the given
triangle.

13. Construct a triangle ABC in which BC = 8 cm, ∠B = 60° and ∠C = 45°. Then, construct another
3
triangle whose sides are 4 of the corresponding sides of DABC.

14. Draw a circle of radius 7 cm. From a point 12 cm away from its centre, construct the pair of

tangents to the circle and measure their lengths.

15. Construct a tangent to a circle of radius 5 cm from a point which is at a distance of 6.2 cm from
its centre.

Constructions 241

QQ Long Answer Questions: [5 marks each]

16. Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle
3
whose sides are 4 of the corresponding sides of the ∆ ABC. [CBSE 2018 (30/1)]

17. Construct a ∆ ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°. Then construct a triangle
3
whose sides are 5 of the corresponding sides of ∆ABC. [CBSE 2019 (30/1/1)]

18. Construct an equilateral ∆ ABC with each side 5 cm. Then construct another triangle whose sides
2
are 3 times the corresponding sides of ∆ABC. [CBSE 2019 (30/2/1)]

19. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and

construct a pair of tangents PA and PB to the smaller circle. Measure PA. [CBSE 2019 (30/2/1)]

20. Construct a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct
3
another triangle whose sides are 4 of the corresponding sides of triangle ∆ABC.

[CBSE 2019 (30/3/1)]

21. Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another
5
triangle whose sides are 7 times the corresponding sides of the first triangle.

[CBSE 2019 (30/3/3)]

22. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
3
5 of the corresponding sides of the first triangle. [CBSE 2019 (30/4/2)]

23. Draw a circle of radius 4 cm. From a point 6 cm away from its centre, construct a pair of tangents

to the circle and measure their lengths. [CBSE 2019 (30/4/3)]

24. Construct a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle

of 60°. [CBSE 2019 (30/5/3)]

25. Draw a circle of radius 3 cm. Take a point A on its extended diameter at a distance of 7 cm from

its centre. Draw two tangents to the circle from A. [CBSE 2019 (C) (30/1/1)]

26. Construct tangents to a circle of radius 4 cm from a point on the concentric circle of radius

7 cm. [CBSE 2019 (30/1/2)]

27. Draw a triangle ABC in which AB = 6 cm, BC = 5.8 cm and ∠ABC = 75°. Construct a triangle
7
similar to DABC with scale factor 5 . Justify the construction.

28. Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle
PQR similar to DABC in which PQ = 8 cm. Also, justify the construction. [NCERT Exemplar]

29. Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ∠ABC = 60°, divide it into

triangles BCD and ABD by the diagonal BD. Construct the triangle BD' C' similar to DBDC with
∠scAal'eBCfa'Dctoarpa43r.aDllerlaowgrtahme ?li ne seg ment ∠D'A ' parallel
to DA where A' lies on extended side BA. Is
[NCERT Exemplar]

242 Xam idea Mathematics–X