Lines, angles and circles7. Two complementary angles are in the ratio 7 : 1 1. Find the angles. Ans. Let first angle be x. Then its complementary angle is 90 – x According to the

Math Class VIII 1 Question Bank

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1. Two adjacent angles on a straight line are (3x – 2)° and 4(x + 7)°.

Find :

(i) the value of x (ii) the measure of each angle.

Ans. (i) Two adjacent angles on a straight line are (3x – 2)° and 4(x + 7)°

⇒ 3x – 2 + 4(x + 7) = 180°

⇒ 3x – 2 + 4x + 28 = 180°

⇒ 7x + 26 = 180° ⇒ 7x = 180° – 26°

⇒ 7x = 154° ⇒ 154

227

x°

= = °

(i) Hence, x = 22°

(ii) Measure of first angle = (3x – 2)°

= (3 × 22 – 2)° = 66° – 2° = 64°

Measure of second angle = 4 (x + 7)° = 4(22 + 7)° = 116°.

2. In the adjoining figure, AOB is a straight line. Find the value of x.

Hence, find ∠AOC and ∠BOD.

(3 – 5)°x55°

( + 20)°x

OA B

C D

Ans. AOB is a straight line.

∴ ∠AOC + ∠COD + ∠DOB = 180°

⇒ (3x – 5)° + 55° + (x + 20)° = 180°

⇒ 3x – 5° + 55° + x + 20° = 180°

⇒ 4x + 70° = 180° ⇒ 4x = 180° – 70°

1LINES, ANGLES AND

CIRCLES


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⇒ 4x = 110° ⇒ 110 1

274 2

x° °

= =

Therefore, ∠AOC = 3x – 5° = 3 × 1

272

– 5° = 1

772

°

and 1 1

( 20) 27 20 472 2

BOD x ° °

∠ = + ° = + =

3. In the adjoining figure, AOB is a straight line.

If x : y : z = 6 : 5 : 4, find the values of x, y and z.

Ans. AOB is a straight line

z°

y°x°

OA B

C D

x : y : z = 6 : 5 : 4

Let x = 6, y = 5, z = 4

∴ 6 + 5 + 4 = 180°

⇒ 180

1215

°= = °

Hence, x = 6 = 6 × 12° = 72°

y = 5 = 5 × 12° = 60°

z = 4 = 4 × 12° = 48°

4. Find the angle which is 34° more than its complement.

Ans. Let required angle be x. Then its complement = 90° – x

According to the given problem

x + x = 90° + 34°

⇒ 2x = 124° ⇒124

622

x = = °

Hence, required angle is 62°.


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5. Find the angle which is 32° less than its supplement.

Ans.Let required angle be x

Then its supplement = 180° – x

According to the given problem, we have

x = 180° – x – 32°

⇒ x + x = 180° – 32°

⇒ 2x = 148° ⇒ 148

742

x°

= = °

Hence, the required angle is 74°.

6. Find the angle whose complement is one-third of its supplement.

Ans.Let the required angle be x.

Then its complement is 90° – x

and supplement is 180° – x


90° – x = 1

3(180° – x)

⇒ 270° – 3x = 180° – x ⇒ 270° – 180° = – x + 3x

⇒ 2x = 90° ⇒ 90

452

x°

= = °

Hence, required angle is 45°.

7. Two complementary angles are in the ratio 7 : 11. Find the angles.

Ans. Let first angle be x. Then its complementary angle is 90° – x


7

90 11

x

x=

° −

⇒ 11x = 630° – 7x ⇒ 11x + 7x = 630°

⇒ 18x = 630°

Thus, 630

3518

x = = °


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Hence, first angle = 35°

and second angle = 90° – x = 90° – 35° = 55°

8. Two supplementary angles are in the ratio 7 : 8. Find the angles.

Ans.Let first angle be x. Then its supplementary angle is 180° – x.


7

180 – 8

x

x=

°

⇒ 8x = 1260° – 7x [by cross multiplication]

⇒ 8x + 7x = 1260°

⇒1260

8415

x°

= = °

Hence, first angle = 84°. And

Its supplement = (180° – x) = 180° – 84° = 96°

9. Find the measure of an angle, if seven times its complement is 10°less than three times its supplement.

Ans. Let the required angle be x. Then its complement angle is 90° – x

and its supplement angle is 180° – x

According to given problem, we have

7(90° – x) = 3(180° – x) – 10°

⇒ 630° – 7x = 540° – 3x – 10°

⇒ 630° – 540° + 10° = – 3x + 7x

⇒ 4x = 640° – 540° = 100°

Thus, 100

254

x°

= = °

Hence, required angle = 25°

10. Prove that the bisectors of two adjacent supplementary angles in-clude a right angle.

Ans. Let AOC∠ and BOC∠ are adjacent supplementary angles

∴ 180AOC BOC∠ + ∠ = °


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OA B

E

CD

But OD and OE are the angle bisectors of AOC∠ and BOC∠

respectively.

∴1

2AOD COD AOC∠ = ∠ = ∠

And, 1

2BOE COE BOC∠ = ∠ = ∠

∴ 1 1

2 2COD COE AOC BOC∠ + ∠ = ∠ + ∠

⇒ 1

( )2

DOE AOC BOC∠ = ∠ + ∠ ⇒ 1

180 902

DOE∠ = × ° = °

Hence, DOE∠ = right angle

11. In the adjoining figure AB || CD. Find the values, of x and y.

E

DC

A BH

F

y

(2 )°x G

(3 )°x

Ans.∵ AB || CD and EF is the transversal which intersects them at Gand H

∴ CGH GHB∠ = ∠ (Alternate angles)

⇒ 3 , 3∠ = =CGH x y x


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But 180CGH CGE∠ + ∠ = ° (Linear pair)

∴ 3x + 2x = 180° ⇒ 5x = 180°

⇒ 180

365

x°

= = ° and y = 3x = 3 × 36° = 108°

Hence, x = 36°, y = 108°

12. In each of the following figures, AB || CD. Find the value of x . Givereasons.

E

DC

A BH

F

G

(3x – 20)°

(2x + 15)°

E

DC

A BH

F

(3 )°x

G

(4x – 23)°

(i) (ii)

Ans. In fig. (i)

AB || CD and a transversal EF intersects them at G and H

(2 15 )EGD x∠ = + °

and ) (3 20)GHB x∠ = − °

But ∠EGD = ∠GHB (corresponding angles)

∴ 2x + 15 = 3x – 20 ⇒ 15 + 20 = 3x – 2x

⇒ x = 35°

Hence x = 35°

In fig (ii) AB || CD and a transversal EF intersects them at G

and H.

(4 23)DGH x∠ = − ° and 3 )GHB x∠ = °

But 180DGH GHB∠ + ∠ = ° (sum of co-interior angles)


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⇒ 4x – 23° + 3x = 180° ⇒ 7x = 180° + 23°

∴203

297

x = = °

Hence, x = 29°.

13. In the adjoining figure, AB || CD are cut by a transversal t at E and F

respectively. If ∠2 : ∠1 = 5 : 4, find the measure of each one of

the marked angles.

t

BA

C D

F

E

5

43

21

87

6

Ans. In the figure, AB || CD and a transversal t intersects them at E and F

respectively

Also, ∠2 : ∠1 = 5 : 4 (given)

Let ∠2 = 5x and ∠1 = 4x

But ∠1 + ∠2 = 180° (Linear pair)

⇒ 4x + 5x = 180° ⇒ 9x = 180° ⇒ 180

209

x°

= = °

∴ ∠1 = 4x = 4 × 20° = 80° and ∠2 = 5x = 5 × 20° = 100°

Now ∠3 = ∠1 (Vertically opposite angles)

and ∠2 = ∠4 (Vertically opposite angles)

∴ ∠3 = 80° and ∠4 = 100°

∵ AB || CD

∴ ∠5 = ∠3 (alternate angles)


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and ∠6 = ∠4 (alternate angles)

∴ ∠5 = 80° and ∠6 = 100°

Now ∠7 = ∠5 (vertically opposite angles)

and ∠8 = ∠6 (vertically opposite angles)

Thus, ∠7 = 80° and ∠8 = 100°

14. In the adjoining figure AB || CD. Find the values of x, y and z.

A Q R B

D

P

C 1 (3 )°x(2 )°x

80° y° z°

Ans. 1 180DPR RPQ∠ + ∠ + ∠ = ° (angles at a point of a line)

⇒ 3x + 2x + 80° = 180° ( 1∠ = PQR∠ alternate angles)

⇒ 5x = 180° – 80° = 100° 100

205

x = = °

y = RPD∠ (alternate angles) = 3x = 3 × 20° = 60°

But y + z = 180° (linear pair)

∴ 60° + z = 180°

⇒ z = 180° – 60° = 120°

Hence, x = 20°, y = 60° and z = 120°

15. In each of the following figure, AB || CD. Find the value of x in each

case.

A

C

E

G

B

F

D

1

2

x°

104°

116°

C

A

G

D

F

B

1

2

35°

65°

x

E

(i) (ii)


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Ans. (i) In figure (i) AB || CD,

Through E, draw GF || AB or CD.

∴ 1 180BAE∠ + ∠ = ° (co-interior angles) ...(i)

Similarly 2 180ECD∠ + ∠ = ° (co-interior angles) ...(ii)

Adding (i) & (ii) we get

1 2 180 180 360BAE ECD∠ + ∠ + ∠ + ∠ = ° + ° = °

⇒ 104° + 116° + x = 360°

⇒ x = – 220° + 360° = 140°

Hence, x = 140°

(ii) In figure (ii), AB || CD

Through E, draw GF || AB or CD

Now DCE∠ = 1∠ ... (i) (alternate angles) ...(iii)

Similarly BAE∠ = 2∠ ... (ii) (alternate angles) ...(iv)

Adding (iii) and (iv) we have

1 2 DCE BAE∠ + ∠ = ∠ + ∠

⇒ x = 35° + 65° = 100°

Hence, x = 100°

16. In the adjoining figure, AB || CD || EF and BC || ED. Find the values

of x, y and z.

A B

x°40°

C D

E F78°

y

z

Ans. Since CD || EF and ED is the transversal

∴ y = 78° (alternate angles)

Now, BC || ED


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∴ (x + 40°) + y° = 180° (co-interior angles)

⇒ x + 40° + 78° = 180°

⇒ x = 180° – 118° = 62°

Also, AB || CD and BC is transversal

∴ (x + 40)° = z (alternate angles)

or z = 62° + 40° = 102°

Hence, x = 62°, y = 78° and z = 102°

17. Caculate the size of each lettered angles in the following figures:

d

c b

108° a20°b 68°

dc

a b

95°

ca

d62°

(i) (ii) (iii)

Ans. (i) c = 108° (alternate angles)

Also, c + d = 180° (linear pair)

⇒ 108° + d = 180°

⇒ d = 72°

Also, b + c = 180° (co-interior angles)

⇒ b + 108° = 180°

⇒ b = 72°

Now, b = a = 72° (alternate angles)

Hence, a = 72°, b = 72°, c = 108°, d = 72°

(ii) 20° + b = 68° (corresponding angles)

⇒ b = 68° – 20° = 48°

Now, a + (20° + b) = 180° (co-interior angles)

⇒ a + 20° + 48° = 180°

⇒ a = 180 – 68 =112°


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20°b 68°

dc

a

B

A D

C

In ABD∆ , a + 20° + ADB∠ = 180°

⇒ ADB∠ = 180° – 112° – 20° = 48°

∴ d = ADB∠ (vertically opposite angles) = 48°

Also c + d = 68° (corresponding angle)

c = 68 – 48 = 20°

Hence, a = 112°, b = 48°, c = 20°, d = 48°

(iii)b

95°

ca

d62°C

D

E

FBA

CDF∠ = 62° (vertically opposite angles)

and d + CDF∠ = 180° (co-interior angles)

⇒ d = 180° – 62° = 118°

Also, b + d = 180° (co-interior angles)

⇒ b = 180° – 118° = 62°

Also, c + ∠CDF = 180° (co-interior angles)

⇒ c + 62° = 180°

⇒ c = 180° – 62° = 118°

Now, FEC FEA∠ + ∠ = 180° (linear pair)

FEC∠ + 95° = 180°

⇒ FEC∠ = 85°


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∴ a = FEC∠ = 85° (corresponding angles)

Hence a = 85°, b = 62°, c = 118°, d = 118°

18. Calculate the size of each lettered angles in the following figure :

c

a

b 44°

c

b72°

a

65°

(i) (ii)

x

60°

(iii)

Ans. (i) 44° + a = 180° (co-interior angles)

⇒ a = 180° – 44° = 136°

Also c = 44° (alternate angles)

Now, 90° + (44° + b°) = 180° (co-interior angles)

⇒ b = 180° – 134° = 46°

Hence, a = 136°, b = 46 °, c = 44°

(ii)

65°

Gc E D

b

72° a

F C

BA

a = 72° (corresponding angles)


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Now, a + BAC∠ = 180° (Linear pair)

⇒ BAC∠ = 180° 72° = 108°

∴ b = 108° (corresponding to BAC∠ )

Now GEF∠ = 65° (corresponding angles)

Also, c + GEF∠ = 180° (linear pair)

⇒ c = 180° – 65° = 115°

Hence, a = 72°, b = 108°, c = 115°

(iii) A

F y

BE

Gx

90–y

60°

QP

C D

Through F draw a straight line PQ || AB || CD.

Let GFQ∠ = y then EFQ∠ = 90 – y°

Also, 60 + BEF∠ = 180° (linear pair)

∴ BEF∠ = 120°

Now, BEF∠ + (90 – y)° = 180° (co-interior angles)

⇒ 120° + 90° – y = 180°

or y = 210° – 180° = 30°

Hence, y = x = 30° (alternate angles)

19. Find the measure of each lettered angles in the following figures :

112°

l mn

a70°

b

ce

d

38°dc

a

b 70° n

m

l

(i) (ii)


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Ans. (i)112°

l mn

a70°

b

ce

d

B

A

C

D

Since m || n

∴ a = 70° (corresponding angles)

Also, a + b = 180° (linear pair)

⇒ b = 180° – 70° = 110°

l || n and AB is a transversal.

∴ e = 112° (corresponding angles)

Also, c + e = 180° (linear pair)

∴ c = 180° – 112° = 68°

Now, l || n and CD is a transversal

∴ d = 70° (alternate angles)

Hence, a = 70°, b = 110°, c = 68°, d = 70°, e = 112°

(ii) 38°

dc

a

b 70°

l

m

n

p q

Since l || m

∴ c = 38° (alternate angles)

Now, c + d = 180° (linear pair)


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∴ 38° + d = 180° ⇒ d = 142°

Also l || n,

∴ a + 70° = 180° (co-interior angles)

⇒ a = 110°

Now, p || q

∴ b = 70° (corresponding angles)

Hence, a = 110°, b = 70°, c = 38°, d = 142°

20. Find the measure of each lettered angles in the following figures :

80°x

m

l

x

z

y

B

A

D

C

Ans.

80°xx

z

y D

B Cm

l

80° + x + x = 180° (straight angle)

⇒ 80° + 2x = 180°

⇒ 2x = 100 ⇒ x = 50°

In ABC∆ , 90 + x + y = 180

∴ y = 180° – 90° – 50° = 40°

Since l || m

∴ x + z = 180° (co-interior angles)

⇒ 50° + z = 180° ⇒ z = 130°

Hence, x = 50°, y = 40°, z = 130°


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21. In the given figure, AB || CD and EF || GH. Find the values of x, y, z,

t and w.

E G

H

J

C D

110° z°

x°t°

y°

A B60°

w°

IF

Ans. In the figure,

E G

H

J

C D

110° z°

x°t°

y°

A B60°

w°

IF

P

M

L

N

AB || CD and EF || GH

CPL∠ = 110° and ILF∠ = 60°

PLN ILF∠ = ∠ (vertically opposite angles)

⇒ x = 60°

∵ AB || CD

∴ PLN LNM∠ = ∠ (alternate angles)

∴ x = y = 60°


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CPL LPN∠ + ∠ = 180° (linear pair)

⇒ 110° + LPN∠ = 180°

⇒ LPN∠ = 180° – 110° = 70°

∵ EF || GH

∴ DNM LPN∠ = ∠ (corresponding angles)

∴ z = 70°

∵ AB || CD

∴ DNM NML∠ = ∠ (alternate angles)

∴ t = z

∴ t = 70°

But NML NMB∠ + ∠ = 180° (linear pair)

∴ 70° + w = 180°

⇒ w = 180° – 70° = 110°

Hence, x = 60°, y = 60°, z = 70°, w = 110° and t = 70°

22. In the given figure, AB || CD, EF || GH and JK || LM. Find the values

of x, y, z, t and w.

C D

A B

P

J L EG

T

70°

R

F K H M

S

y°

z°

w°

45°x°

t°

Q

Ans. In the figure,

AB || CD, EF || GH and JK || LM

TRS∠ = 70° and RST∠ = 45°

∵ EF || GH

∴ QRS∠ = RST∠ (alternate angles)


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⇒ x = 45°

But 180PRQ QRS SRT∠ + ∠ + ∠ = ° (Angle on one sides of a line)

⇒ y + x + 70° = 180°

⇒ y + 45° + 70° = 180°

⇒ y + 115° = 180° ⇒ y = 180° – 115° = 65°

∵ AB || CD

∴ SRT∠ = RSQ∠ (alternate angles)

∴ 70° = w ⇒ w = 70°

and PRQ∠ = RSQ∠ (alternate angles)

⇒ y = t ⇒ t = 65°

∴ JK || LM

∴ QRS∠ = RQP∠ (alternate angles)

⇒ x = z ⇒ z = 45°

Hence, x = 45°, y = 65°, z = 45°, t = 65° and w = 70°

23. In the adjoining figure AB || CD and they cut the lines PQ and QR at

E, F and G, H respectively. Find the value of x.

R

85°x°P

B

E G

D

CA

F

H 125°

Q

Ans. In the figure, AB || CD and PQ, QR intersect them

Also, PEF∠ = 85° and CHQ∠ = 125°

∵ AB || CD and PQ intersects at E and G

∴ EGH PEF∠ = ∠ = 85° (corresponding angles)


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But EGH HGQ∠ + ∠ = 180° (Linear pair)

⇒ 85° + HGQ∠ = 180°

⇒ HGQ∠ = 180° – 85° = 95°

In ∆HGQ

Ext. GHQ HGQ GQH∠ = ∠ + ∠

⇒ 125° = 95° + x

⇒ x = 125° – 95° = 30°

Hence, x = 30°

24. In the adjoining figure, AB || CD || EF. Find the value of x.

C D

130°

FE

BA

70°

x°

Ans. In the figure AB || CD || EF.

ABC∠ = 70° and CEF∠ = 130°

∵ AB || CD

C D

130°

FE

BA

70°

x°

1

∴ ABC∠ = BCD∠ (alternate angles)


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⇒ 70° = x + 1∠ ... (i)

∵ EF || CD

∴ CEF ECD∠ + ∠ = 180° (co-interior angles)

⇒ 130° + 1∠ = 180° ⇒ 1∠ = 180° – 130° ⇒ 1∠ = 50°

From (i), we have

x + 50° = 70° ⇒ x = 70° – 50° = 20°

Hence, x = 20°

25. In the adjoining figure, AB || CD. Find the values of x and y.

A75°

B

P

E

F

20°

x°

25°y°C

D

G

Q

Ans.∵ AB || CD

∴ PEB EFD∠ = ∠ [corresponding angles]

⇒ 75° = 25° + y

⇒ y = 75° – 25° = 50°

A 75°B

20°

x°

25°y°C

D

G

P

Q

F

E 1


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But PEB BEF∠ + ∠ = 180° (Linear pair)

∴ 75° + 20° + 1∠ = 180° ⇒ 1∠ + 95° = 180°

⇒ 1∠ = 180° – 95° ⇒ 1∠ = 85°

In EFG∆ , GEF EFG EGF∠ + ∠ + ∠ = 180°

(Sum of angles of a triangles)

⇒ 1∠ + 25° + x = 180°

⇒ 85° + 25° + x = 180° ⇒ 110° + x = 180°

⇒ x = 180° – 110° = 70° Hence, x = 70°, y = 50°

26. In the adjoining figure, AB || CD and EF cuts them at G and H

respectively. GP amd HQ are bisectors of AGH∠ and GHD∠

respectively. Proved that GP || HQ.

A B

E

G

H

P

Q DC

F

Ans. Given : AB || CD and EF intersects them at G and H respectively.

PG and HQ are the bisectors of AGH∠ and GHD∠ respectively.

A B

E

G

P

Q DC

F

H

2

1

To prove : GP || HQ.


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Proof : AB || CD

∴ AGH∠ = GHD∠ (alternate angles)

∴ GP and HQ are the bisectors of ∠AGH and ∠GHD

respectively

∴ 1

12

∠ = AGH∠ and 1

22

∠ = GHD∠

∴ 1 2∠ = ∠ (Half of equal angles)

But, these are alternate angles.

Hence, GP || HQ

27. In the given figure, AB || CD and A C∠ = ∠ . Prove that AE || CF.

C

A

D

B

EP

F

Ans. Given : AB || CD and A C∠ = ∠

To prove : AE || CF

Proof : ∴ AB || CD given, and AE is its transversal.

C

AP

D

B

E

F

∴ A DPE∠ = ∠ (corresponding angles)

But A C∠ = ∠ (given)

∴ DPE C∠ = ∠


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But these are also corresponding angles.

Hence, AE || CF.

28. Two parallel lines are cut by a transversal at points P and Q. The

bisectors of interior angles, on the same side of the transversal,

intersect at point R. Prove that angle PRQ is a right angle.

Ans. Given : AB || CD and transversal PQ intersects them at X and S

respectively. XR and SR are bisectors of interior angles which meet

at R.

To prove : XRS∠ = 90°

P

SC D

BX

A

R

Q

Proof : ∵ AB || CD (Given)

BXS XSD∠ + ∠ = 180°

1 1 1

1802 2 2

BXS XSD∠ + ∠ = × °

∵ XR & RS are bisectors. (Given)

SXR RSX∠ + ∠ = 90° ... (i)

∴ In SXR∆ ,

SXR RSX XRS∠ + ∠ + ∠ = 180°

90° + XRS∠ = 180° [from (i)]

XRS∠ = 180° – 90°

Hence, XRS∠ = 90° ... (i)

29. Two straight lines are cut by a transversal. If the bisectors of a pairs

of co-interior angles are perpendicular to each other; prove that the

two straight lines are parallel to each other.


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Ans.

P

SC D

BA

T

Q

R

Given: Transversal PQ cuts two straight lines AB and CD at R & S

respectively. Bisectors RT and ST of co-interior angles are such

that RTS∠ = 90°.

To prove : AB || CD

Proof : RT is bisector(Given)

∴1

2SRT BRS∠ = ∠

∵ ST is bisector (Given)

∴1

2TSR RSD∠ = ∠

In RST∆ ,

180SRT TSR RTS∠ + ∠ + ∠ = °

∴1 1

90 1802 2

BRS RSD∠ + ∠ + ° = °

1( ) 180 90 90

2∠ + ∠ = ° − ° = °BRS RSD

BRS RSD∠ + ∠ = 180° (co-interior angles)

Hence, AB || CD

30. In a parallelogram ABCD, the bisectors of angles B∠ and C∠

intersect each other at point E. Prove that angle BEC is equal to a

right angle.


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Ans. Given: Parallelogram ABCD in which bisectors BE & CE of

B∠ and C∠ intersect at E.

To prove: BEC∠ = 90°

Proof : AB || CD (Given)

∴ 180ABC BCD∠ + ∠ = °

1 1 1

1802 2 2

ABC BCD∠ + ∠ = × °

∵ BE & CE are bisectors (Given)

∴ 90CBE BCE∠ + ∠ = ° ... (i)

In BEC∆ ,

180BEC CBE BCE∠ + ∠ + ∠ = °

⇒ 90 180BEC∠ + ° = ° [from (i)]

⇒ 180 90BEC∠ = ° − °

Hence, 180 90BEC∠ = ° − ° = 90°

31. Draw a line segment AB = 7 cm. Now draw a circle with AB as

diameter. In this circle draw a chord of length 5 cm. Now shade the

minor segment of the circle.

Ans. Steps of construction :

(i) Draw a line segment AB = 7 cm

(ii) Bisect it at O

(iii) With centre O and AB as diameter or

OA as radius, draw a circle.

(iv) Take any point P on the circle.

(v) With centre P and radius 5 cm. draw an

arc which intersects the circle at Q.

(vi) Join PQ,

D C

BA

E

A B7 cm

5 cm

O

Q P

C


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PQ is the chord whose length is 5 cm.

Then PCQ is the minor segment which has been shaded.

32. Find the length of the diameter of a circle of radius.

(i) 4.8 cm (ii) 5.6 cm (iii) 3.7 cm

Ans. We know that diameter of a circle is twice the length of its radius,

Now

(i) Radius = 4.8 cm, then diameter = 2 × 4.8 = 9.6 cm

(ii) Radius = 5.6 cm, then diameter = 2 × 5.6 = 11.2 cm

(iii) Radius = 3.7 cm, then diameter = 2 × 3. 7 = 7.4 cm

33. In the given figure, ABC∆ is inscribed in a circle with centre O. If

ACB∠ = 40°. Find ABC∠ .

Ans. We know that the angle in semi-circle is a right angle

∴ BAC∠ = 90°

Also BCA∠ = 40° (given)

But the sum of all the angles in a trianle is 180°

∴ 180BCA BAC ABC∠ + ∠ + ∠ = °

⇒ 90° + 40° + ABC∠ = 180°

⇒ ABC∠ = (180° – 130°) = 50°

Hence, ABC∠ = 50°

34. In the given figure, ABC∆ is inscribed in a circle with centre O. If

ABC∠ = 34°. Find ACB∠ .

34°O

A

C

B

40°O

A

B C


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Ans. We know that the angle in semi-circle is a right angle.

So, BAC∠ = 90°

Also, ABC∠ = 34° (given)

But the sum of all the angles of a triangle is 180°

∴ 180BAC BCA ACB∠ + ∠ + ∠ = °

⇒ 90° + 34° + ACB∠ = 180°

⇒ ACB∠ = 180° – 124° = 56°

35. In the given figure, O is the centre of a circle ABC∆ is inscribed in

this circle. If AB = AC, find ABC∠ and ACB∠ .

O

A

B C

Ans. Here, BAC∠ = 90° [Angle in semi-circle]

180 90 90ABC ACB∠ + ∠ = ° − ° = ° ... (i)

Let ACB∠ = x°

∴ x + x = 90° (AB = AC)

⇒ 2x = 90° ⇒90

452

x°

= = °

Hence, ACB∠ = 45°

Also, 45ABC ACB∠ = ∠ = °

36. In the given figure, O is the centre of a circle. If ABC∠ = 54°, find

ACB∠ . Also if BCD∠ = 43°, find ∠CBd

Ans. We know that the angle in semi-circle is a right angle.


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O

54°43°

D

A

B C

(i) ∴ BAC∠ = 90°

Also ABC∠ = 54° (given)

But the sum of all the angles of a triangle is 180°

180BAC ABC ACB∠ + ∠ + ∠ = °

⇒ 90° + 54° + ACB∠ = 180°

⇒ ACB∠ = 180° – 144° = 36°

(ii) BDC∠ = 90° (Angle in the semi circle)

Also BCD∠ = 43° (given)

But, the sum of all the angles in a triangle is 180°

180BDC BCD CBD∠ + ∠ + ∠ = °

⇒ 90° + 43° + CBD∠ = 180°

⇒ CBD∠ = 180° – 133° = 47°.

37. In the adjoining diagram, AB is a chord of a circle with centre C.If

CM ⊥ AB, prove that AM = MB. Hence, complete the

following staement : If a perpendicular is drawn from the centre of

a circle to a chord then it .... the chord.

A BM

C


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Ans. In AMC∆ and BMC∆

90CMA CMB∠ = ∠ = °

AC = BC (radii of circle)

MC = MC (common)

AMC BMC∆ ≅ ∆ (R.H.S. criterion of congruency)

∴ AM = MB (c.p.c.t.)

Hence, If a perpendicular is drawn from the centre of a circle to a

chord then it bisects the chord.

38. In the adjoining figure, AB is a diameter of the circle. If AC = BC,

find CAB∠ .

A B

C

Ans. We know that, angle in a semi-circle is 90°

Thus ACB∠ = 90°

So, in ∆ ABC

180ACB CAB CBA∠ + ∠ + ∠ = °

(sum of the angles of a triangle is 180°)

90° + CAB CBA∠ + ∠ = 180°

⇒ CAB CBA∠ + ∠ = 180° – 90° =90°

But CAB CBA∠ = ∠

[Being angles opposite to equal sides of an isosceles triangle]

∴ 2 CBA∠ = 90°

⇒ CBA∠ = 90

452

°= °

Hence, CAB CBA∠ = ∠ = 45°


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39. In the adjoining diagram, AB is a chord of a circle with centre C. If

M is mid-point of AB, prove that CM is perpendicular to AB.

A BM

C

Ans. In CAM∆ and CMB∆

CM = CM (common)

AC = BC (radii of same circle)

AM = MB (M is mid-point of AB)

∴ CAM CMB∆ ≅ ∆ (SSS criterion of congruency)

∴ CMA CMB∠ = ∠ = 90° (c.p.c.t.)

Hence, CM is perpendicular to AB.

40. Construct an angle of 45° and bisect it. Measure each part by pro-

tractor.

O M

N T22½

P R

L

A

Ans. (i) Construct 90AOP∠ = °

(ii) With M and N as centres and radius 1

2> MN, draw arcs to inter-

sect at R = join OR we get 45ROA∠ = °

(iii) Now, with M and T as centres and radius 1

2> MT, draw arcs to

intersect at L join OL. we have get 1

222

LOA°

∠ = .


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41. Draw a line segment of length 6.4 cm and divide it into equal parts.

Ans. Steps of construction:

(i) Draw AB = 6.4 cm

(ii) With A as centre and radius greater

than half of AB draw two arcs on

either side AB.

(iii) With B as centre and same radius draw

two more arcs, cutting the previous

arcs at P and Q.

(iv) Join PQ which divides the line seg-

ment AB into two equal parts.

42. Draw a line segment AB of length 7 cm and divide it into the ratio

2 : 3.


(i) Draw AB = 7cm.

(ii) With A and B as a centre we draw

two rays AX and BY making an acute

equal either side of AB

(iii) From AX and AY cut of 5 equal dis-

tance C, D, E, F, G.

(iv) With same radius, cut off 5 equal

distances along BY, at points H, I,

J, K, L,

(v) Join AL, DJ and GB.

(vi) Line segment DJ divides the AB into

2 : 3.

43. Draw angle ABC of any suitable measure.

(i) Draw BP, the bisector of angle ABC.

(ii) Draw BR, the bisector of angle PBC and draw BQ, the bisector

of angle ABP.

A B

3.2 cm 3.2 cmM

6.4 cm

Q

P

A

C

D

E

F

GX

B

L

K

J

I

H7 cm

Y


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(iii) Are the angles ABQ, QBP, PBR and RBC equal ?

(iv) Are the angles ABR and QBC equal ?

Ans. Steps of Construction :

(1) Construct any angle ABC

(2) With B as centre, draw an arc EF meet-

ing BC at E and AB at F.

(3) With E, F as centres draw two arc of

equal radii meeting each other at the

point P.

(4) Join BP. then BP is the bisector of ABC∠ . ABP∠

1

2PBC ABC∠ = ∠

(5) Similarly draw BR, the bisector of PBC∠ and draw BQ as the

bisector of ABP∠ .

(6) Then ∠ABQ = ∠QBP = ∠PBR = ∠RBC.

(7) Then 3

4∠ = ∠ABR ABC and

3

4QBC ABC∠ = ∠

∴ .ABR QBC∠ = ∠

44. Draw an angle ABC = 60°. Draw the bisector of it. Also, draw a

line parallel to BC a distance of 2.5 cm from it.

Let this parallel line meet AB at point P and angle bisector at point

Q. Measure the length of BP and PQ. Is BP = PQ ?


(i) Draw an 60ABC∠ = °

(ii) Draw BD, the bisector of .ABC∠

(iii) Taking B as centre, draw an arc

of radius 2.5 cm.

(iv) Taking C as centre, draw another

B E C

FHGI

R

P

Q

A

MP Q

A

D

N

B C


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arc of radius 2.5 cm.

(v) Draw a line MN which touches these two arcs drawn.

Then, MN is the required line parallel to BC.

(vi) Let this line MN meets AB at P and bisector BD at Q.

(vii) Measure BP and PQ.

By measurement we see BP = PQ.

45. Draw a line segment PQ = 5.4 cm and draw its perpendicular bisec-

tor.


(i) We draw a line segment PQ =

5.4 cm

(ii) With centres P and Q, and radius more

than half of PQ, draw two arcs inter-

secting each other at L and M.

(iii) Join LM intersecting PQ at N.

Then LM is the required perpendicular

bisector of line segment PQ.

46. Draw a line segment of length 7.2 cm and divide it into 6 equal

parts, using ruler and compasses.


(i) We draw a line segment AB = 7.2 cm

(ii) At A adn B as a centre we draw two

rays AX and BY making an acute angle

either of AB

(iii) Cut off 6 equal parts from AX and AY

(iv) Join A, 6; 1, 5; 2, 4; 3, 3; 4, 2; 5, 1;6,

B intersecting AB at L, M, N, P, Q re-

spectively.

Then, AL, LM, NP, PQ, are six equal

parts.

L

N

M

P Q5.4 cm

A B

1

2

3

4

5

6X

6

5

4

3

2

1

L M N Q

P

7.2 cm

Y


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47. Draw a line segment AB = 4.8 cm. Take point P on it such that PA

= 2.5 cm. Draw a line passing through P and perpendiclar to AB.


(i) Draw a line segment AB = 4.8 cm

(ii) Take a point P on AB such that AP = 2.5 cm

(iii) P as centre and with a suitable

radius, draw an arc meeting AB at E

and F.

(iv) With centres E and F, and radius more

than half of EF, draw two arcs inter-

secting each other at M.

(v) Join PM and produce to L. Then LP

is the required perpendicular at P.

48. Draw a line segment XY. Take a Point A outside it. Draw a line

passing through A and perprpendicular to XY.


(i) Draw a line XY and take a point A outside it

(ii) With centre A and a suitable radius draw an

arc intersecting XY at M and L.

(iii) With centre M and L, radius more than half

of ML, draw two arcs intersecting each other

at N.

(iv) Join AN intersecting XY at P. Then AP is

perpendicular to XY from A.

P FE2.5 cmA B

M

L

4.8 cm

A

N

PX YM L


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49. Construct a triangle PQR such that PQ = QR = RP = 5.1 cm. Name

the triangle.


(i) Draw PQ = 5.1 cm.

(ii) With P as centre and 5.1 cm as radius

draw an arc.

(iii) With Q as centre and 5.1 cm as radius

draw another arc which meets the pre-

vious are at R.

(iv) Join PR and QR. Hence, PQR∆ is the required triangle The

name of the triangle is an equilateral triangle.

50. Draw a ABC∆ in which BC = 4.1 cm, 135C∠ = °and CA = 4 cm.


(i) We draw a line segment BC = 4.1 cm.

(ii) With C as a centre making an angle another 135BCY∠ = °

(iii) With C as centre and radius = 4 cm.

Draw an arc, cutting CY

at A.

(iv) Join AB.

Hence, ABC∆ is the required tri-

angle.

51. Construct a ∆ ABC such that AB = BC = 4.6 cm and B∠ = 75°

Measure A∠ and C∠ .


(i) Draw AB = 4.6 cm

(ii) At B, construct 75ABP∠ = °

(iii) With B as centre and 4.6 cm as radius take

a point C on BP.

5.1 cm

R

P Q

5.1 cm

5.1 cm

YA

B C4.1 cm

4 c

m135°

A B

C

4.6 cm

4.6 cm

P

52.5° 75°

52.5°


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(iv) Join AC. Then ABC is the required triangle.

(v) Measure A∠ and 1

522

C°

∠ =

51. Construct a triangle PQR such that

PQ = 4 cm, QR = 3 cm and 90Q∠ = ° .

Measure PR.


(i) Draw PQ = 4 cm.

(ii) At Q, construct 90PQL∠ = °

(iii) From QL, cut off QR = 3 cm.

(iv) Join PR and measure it. Measuring PR

= 5 cm

52. Construct a triangle PQR given that

QR = 4.9 cm, 45Q∠ = ° and

75P∠ = ° .


(i) We draw QR = 4.9 cm.

(ii) At Q, draw 45RQM∠ = °

(iii) At R, construct 60QRN∠ = °

(iv) Let these angles meet at P.

(v) Then PQR is the required triangle.

53. Construct a triangle ABC given that BC = 5.4 cm, AB = 6 cm and

median CM = 4.6 cm.


(i) We draw BC = 5.4 cm

(ii) With C as centre and radius 4.6 cm draw an

arc.

P Q4 cm

L

R

5 cm

3 c

m

90°

Q R

45° 60°

MP

N

75°

4.9 cm

A

B C

M

3 cm

3 cm

5.4 cm

4.6 cm


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(iii) With A as centre and radius = 1

32

AB = cm, draw an arc to meet

previous arc at M.

(iv) Join BM and produce it to A such that MA = MB.

(v) Join AC.

54. Construct an equilateral triangle PQR

such that the length of its side is 4.8 cm

Measure all the angles of .PQR∆


(i) We draw PQ = 4.8 cm.

(ii) With P and Q as centres and 4.8 cm as

radius, draw two arcs which meet at

R.

Measure, ∠P, ∠Q, ∠R = 60°

55. Construct an equilateral triangle

whose altitude is 4.4 cm.


(i) We draw PQ (any line segment)

(ii) Take a point D on PQ and at D, con-

struct DR⊥ to PQ. From DR, cut

off DA = 4.4 cm.

(iii) At A, construct

160

2DAS DAT∠ = ∠ = × ° = 30°

on either side of AD. Let AS and AT meet PQ at B and C. Then

ABC is the required triangle.

R

P Q

60° 60°

60°

4.8

cm

4.8

cm

4.8 cm

PS

B D C QT

A

R

4.4

cm


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56. Construct an isosceles triangle given that its

base AB = 5 cm and the altitude CM to the

base AB = 4.2 cm.


(i) Draw AB = 5cm

(ii) Draw PQ, the perpendicular bisector of

AB. Let PQ meet AB at M.

(iii) From MP, cut off MC = 4.2 cm.

(iv) Join BC and AC, then ABC is the required

triangle.

57. Construct a triangle ABC such that AB = 4.2 cm, 90A∠ = °and the

hypotenuse BC = 5.8 cm.


(i) We draw AB = 4.2 cm.

(ii) At A, construct 90 .BAP∠ = °

(iii) With B as centre and radius = 5.8 cm, cut

off AP a C.

(iv) Join BC, then ABC is the required triangle.

58. Construct an isosceles right angled triangle ABC such that is

hypotenuse AB = 5.7 cm.



(ii) At A, construct 45 .BAP∠ = °

(iii) At B, construct 45 .ABQ∠ = °

AP and BQ meet at C, then ABC is the re-

quired triangle.

A B

M

5 cm

Q

P

4.2

cm

C

P

A B4.2 cm

P

C

90°

5.8 cm

Q P

A

B C4.5 cm


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59. Construct a triangle ABC given that BC = 6 cm, 60B∠ = °and alti-

tude (height) = 4.6 cm.


(i) We draw BC = 6 cm.

(ii) At B, construct 60CBP∠ = ° . At C

draw .CQ BC⊥

(iv) From CQ, cut off CR = 4.6 cm.

(iv) Through R, draw a line parallel to BC

meet BP at A

(v) Join AC. the ABC is the required tringle.

60. Using ruler and compasses only,construct a triangle ABC having

135 ,C∠ = ° and ∠B = 30°, BC = 5cm, Bisect angles B and C and

measure the distance of A from the point where the bisector meet.

Ans.

B C30°

A

P

5 cm

135°

Steps : (i) Draw BC = 5 cm

(ii) Draw AC such that 135 .C∠ = °

(iii) Draw AB such that 30 .ABC∠ = ° so that both meet at A.Thus

ABC is the required triangle

(iv) Draw bisectors of B∠ and C∠ so that both meet at P.

(v) Join AP and measure it. Hence, AP = 9.4 cm.

B C

P

AR

Q

4.6

cm

6 cm

60°


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61. Construct a triangle ABC given that AB – AD = 1.8 cm,

BC = 6 cm and 60B∠ = ° .


(i) Draw BC = 6 cm.

(ii) At B, draw 60CBP∠ = °

(iii) From BP, cut off BD = 1.8 cm.

(iv) Join CD and draw its perpendicular

bisector to meet BP at A

(v) Join AC, then ABC is the required

triangle.

62. Construct an isosceles triangle PQR with base PQ = 5.7 cm and

altitude RM to the base PQ = 4.3 cm.

Ans.

Steps of construction :

(i) We draw PQ = 5.7 cm.

(ii) Draw perpendicular bisector AB of

PQ. Let AB, meet PQ at M.

(iii) From AB, cut off MR = 4.3 cm.

(iv) Join PR and QR, then PQR is the re-

quired triangle.

63. Construct an isosceles triangle ABC given that base BC = 6 cm and

vertical 120 .A∠ = °

Ans.

Steps of construction:

(i) Draw BC = 6 cm

(ii) At B and C, draw angle of 30°

each.

(iii) Let these angles intersect at

point A. Then ABC is the required isosceles triangle.

PA

B C

60°

D

1.8

cm

6 cm

P Q

M

R

A

B

5.7 cm

4.3

cm

A

B C

30° 30°

120°

6 cm


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64. Draw a quadrilateral ABCD with AB = 6 cm, BC = 4 cm,CD = 4

cm and 90 .∠ = ∠ = °B C

Ans.

Steps of construction :

(i) We draw BC = 4 cm.

(ii) At B and C, construct 90CBP BCQ∠ = ∠ = °

(iii) With B as centre and radius equal to 6 cm

cut off BP at A.

(iv) With C as centre and radius equals to 4 cm

cut off CQ at D.

(v) Join AD. Then ABCD is the required quadrilateral.

65. Construct a quadrilateral ABCD in

which AB = 5 cm, BC =

2.5 cm, CD = 6 cm 90BAD∠ = °and

diagonal AC = 5.5 cm

Ans. Steps ofconstruction :

(i) We draw AB = 5 cm.

(ii) At A, construct 90BAP∠ = °

(iii) With A as centre and 5.5 cm as ra-

dius draw an arc.

(iv) With B as centre and 2.5 cm as ra-

dius draw an arc to meet the previous arc at C.

(v) With C as centre and radius equals to 6 cm draw an arc to meet

AP at D.

(vi) Join CD.

Then, ABCD is the required quadrilateral.

B C

90° 90°

4 cm

4 c

m

6 c

m

D

Q

AP

P

D

A B

C

5 cm

5.5 cm

6 cm

90°

2.5

cm


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66. Construct a quadrilateral ABCD such that AB = 4.5 cm ,BC = 4 cm

CD = 3.9 cm, AD = 3.2 cm and B∠ = 60°.

Ans. Stepsof construction :


(ii) At B, construct 60ABP∠ = ° .

(iii) From BP, cut off BC = 4 cm.

(iv) With C as centre, and 3.9 cm as radiusdraw an arc.

(v) With A as centre and 3.2 cm as radius,draw an arc to meet the previous are atD.

(vi) Join AD and CD.

67. Construct a quadrilateral ABCD in which AB = 3.5 cm, BC

= 5 cm CD = 5.6 cm DA = 4 cm and BD = 5.4 cm

Ans. Step of construction :

(i) We draw AB = 3.5cm.

(ii) With A as centre and radius = 5.4cm draw an arc to meet the previ-ous arc at D. Join AD and BD.

(iii) With B as centre and radius = 5 cmdraw an arc With D as centre andradius = 5.6 cm, draw an are to meetthe previous arcat C.

(iv) Join BC and CD, then ABCD is the required quadrilateral.

68. Construct a parallelogram ABCD such that AB = 5 cm,

BC = 3.2 cm and 120B∠ = ° .


(i) Draw AB = 5 cm

(ii) At B, construct angle = 120°

(iii) With B as centre and 3.2 cm as radius

CD

P

A B4.5 cm

3.9 cm

4 c

m

3.2

cm

60°

A B

5. 6 cm4

cm

5.4 cm

3.5 cm

5 c

m

D C

D C5 cm

5 cmA B

3.2

cm

3.2

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cut off B∠ at C.

(iv) With C as centre and AB as radius draw an arc.

(v) With A as centre and 3.2 cm as as radius draw an arc which

meets the previous are at D.

(vi) Join AD and CD.

Then ABCD is required parallelogram.

69. Construct a rectangle such that one diagonals is 6.6 cm and angle

between the two is 120°.


(i) Draw AO 1 1

6.62 2

AC

= = ×

cm

= 3.3 cm and produce AO to such that

OC = OA = 3.3 cm

(ii) At O, construct 120COP∠ = °

(iii) From OP, cut off OD

= 1

3.32

AC = cm.

(iv) Produce DO to B such that OB = OD

= 3.3 cm.

(v) Join AB, BC, CD and DA. Then ABCD is the required rectangle.

70. Construct a rectangle whose one diagonal is 7 cm and an angle

between two diagonals is 45°.


(i) We draw AO = 1 1

72 2

AC

= ×

cm = 3.5cm and produce AO to

C such that OC = OA = 3.5 cm.

(ii) At O, construct 45 .COP∠ = °

(iv) From OP, cut off OD

A

D

C

B

P

3.3 cm

3.3 cm

120°

O

A

B C

D

P

3.5 cm

3.5 cm

O45°


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= 1

2AC = 3.5cm produce DO to B such that OB = OD = 3.5 cm.

(v) Join AB, BC, CD, DA, Then ABCD is the required rectangle.

70. Construct a rhombus whose one side is 5cm and one angle is 45°


(i) We draw AB = 5

(ii) At A, construct

45BAP∠ = °.

(iii) From AP, cut off AD = 5

cm.

(iv) With B as centre and radius

= 5 cm, draw an arc.

(v) With D as centre and radius = 5 cm, draw an arc meet the previ-

ous arc at C.

(iv) Join BC and CD. Then ABCD is the required rhombus.

72. Using ruler and compasses only, draw a ABC∆ such that AB = 4.5

cm AC = 5.4 cm and 90 .A∠ = ° Draw the circumcircle of the

triangle and measure its radius.

Ans.

Steps :

(i) We draw AB = 4.5 cm

(ii) Draw 90 .CAB∠ = °

(iii) Draw perpendicular bisectors of

AB and AC which meet at O.

(iv) With O as centre, and radius OA,

draw a circle which passes through

A, B and C. It is the required cir-

cumcircle.

(v) Measure the radius OA, OB or OC. Radius = 3.5 cm.

A B

P

DC

5 cm

45°

C

OP

Q

A B

5.4

cm

4.5 cm


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73. Construct a triangle ABC in which base BC = 6.0 cm, 60B∠ = °and

altitude (height) is 4.5 cm. In the same figure, construct a circle

which passes throught A, B and C. Name the circle drawn and

measure its radius.


(i) We draw BC = 6 cm

(ii) Draw PQ | | BC at a distance

of 4.5 cm.

(iii) Draw 60ABC∠ = ° which

meets PQ at A.

(iv) Join AC. Thus ABC is the

required triangle.

(v) Draw ⊥ bisectors of BC and AC to intersect at O.

(vi) With O as centre and radius OC, draw circle to pass through A B

and C.

(vii) Measure the radius OC.

Radius = 3.3 cm.

74. Draw an equilateral triangle of side 5 cm. Draw its circumcircle.

Ans. Steps of Constructions :

(i) We draw BC = 5 cm

(ii) With B and C as centre draw arcs of5 cm each which intersect at A.

(iii) Join AB and AC. Thus ABC is the re-quired triangle.

(iv) Draw perpendicular bisectors of AC

and BC to meet at O.

(v) With centre O as centre and radiusOA, draw the circumcircle whichpasses through A and C.

A

4.5 cm

B C

S

6 cm

P Q

M

60°

O

R

L

A

P

B CG5 cm

R

H

Q

S


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75. Construct a triangle ABC such that : BC = 6 cm, 60B∠ = ° . In thesame figure, find a point which is equidistant from the vertices ofthe triangle. Name this point. Draw circumcircle of the triangle.


(i) Draw BC = 6 cm and draw

60PBC∠ = °

(ii) Draw 45BCA∠ = ° to meet BP at A.

Thus ABC is the required triangle.

(iii) Draw perpendicular bisectors of BC

and AC which intersect at O.

Thus O is the point equidistant fromvertices A, B and C.

(iv) With O as centre and radius OA, drawa circumcircle to pass through A, Band C.

76. Draw an equilateral triangle of side 5 cm and construct its circum-circle.


(i) Draw a triangle ABC with BC = 5cm, AC = 5 cm and AB =

5 cm.

(ii) Draw perpendicular bisectors of

any two sides, say BC and AC. Let

these perpendicular bisectors meet

at O.

(iii) With O as centre and radius equal

to OA, draw a circle which passes

through A, B and C.

P

A

B C

Y

X

60° 45°Q

O

R

6 cm

B C

A

5 cm

5 cm

O

5 cm


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77. Draw an isosceles triangle whose one of equal sides is 5cm and

vertical angle is 75°. Construct its circumcircle.


(i) Draw a triangle ABC with given data.

(ii) Draw perpendicular bisectors of any

two sides of triangle say BC and AC.

Let these perpendicular bisectors

meet at O.

(iii) With O as centre, and OA as radius,

draw a circle which passes through A,

B and C.

78. Draw a triangle with sides 6 cm, 5 cm

and 4 cm. Construct its incircle.


(i) Construct a ABC∆ .

(ii) Draw the bisectors of B∠ and

C∠ . Let these bisectors meet atthe point I.

(iii) From I, draw. IN perpendicular

to BC.

(iv) With I as centre and radius equal

to IN, draw a circle which

touches all the sides of the triangle ABC and is the required in

circle.

B C

5 c

m

O

5 cm

A

75°

I

B CN6 cm

5 cm

4 cm

A

Documents

Lines, angles and circles7. Two complementary angles are in the ratio 7 : 1 1. Find the angles. Ans. Let first angle be x. Then its complementary angle is 90 – x According to the