11 triangles edulabz - testlabz · © edulabz interna tional 1. one angle of ... ⇒ x = 30 ° ......

Math Class VIII 1 Question Bank

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1. One angle of a triangle is 78° and the other two angles are in the ratio

7 :10. Calculate the two unknown angles of the triangle.

Ans. Let 78A∠ = ° and : 7 :10B C∠ ∠ =

Let 180A B C∠ + ∠ + ∠ = °

⇒ 78° +7x + 10x = 180°

⇒ 17x = 180° – 78° ⇒ 17x = 102°

⇒ 102

17x = ⇒ x = 6°

Therefore, 7 7 6 42B x∠ = = × = °

And, 10 10 6 60C x∠ = = × = °

Hence, unknown angles are 42° and 60°.

2. In a triangle ABC; 2 3 .A B C∠ = ∠ = ∠ Find each angle of the

triangle.

Ans.

∴ 2 3A B C∠ = ∠ = ∠ (Given)

Let ∠ =A x

∴ 2

xB∠ = and

3

xC∠ =

∵ 180A B C∠ + ∠ + ∠ = ° ⇒ 1802 3

x xx + + = °

⇒ 6x + 3x + 2x = 180° × 6 ⇒ 11x = 1080°

⇒ 1080

11

°=x

11TRIANGLES


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So, 2

9811

x

°

=

Therefore, 2

9811

A

°

∠ =

1080 540

2 2 11 11

° °∠ = = =

×

xB

149

11

°

=

1080

3 11 3

xC∠ = =

×

360

11=

832

11

°

=

Hence, each angles of triangle are 1

98 ;11

°

1

4911

°

and 8

3211

°

.

3. Use the informations given in the figure below to calculate the val-

ues of x and y.

A

B C

35°25°

y

72°

xx

Ans. In ,ABC∆ 180A B C∠ + ∠ + ∠ = °

⇒ 72 (35 25 ) ( ) 180x x° + ° + ° + + = °

⇒ 132 2 180x° + = ° ⇒ 2x = 180° – 132°

⇒ 2x = 48

∴ x = 24° ...(1)

Again x + y + 25° = 180° ⇒ 24° + y + 25° = 180° [form (i)]

⇒ y = 180° – 49° ⇒ y = 131°

Hence, x = 24° and y = 131°.


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4. In ∆ ABC, (2 15) ,A x∠ = + ° B x∠ = ° and (3 15) .C x∠ = − ° Find

each angle of the triangle and hence show that it is an isosceles tri-

angle.

Ans. In ABC∆ , 180A B C∠ + ∠ + ∠ = ° (by property)

⇒ 2x + 15 + x + 3x – 15 = 180°

⇒ 6x = 180° ⇒ 180

6x

°=

⇒ x = 30°

∴ (2 15) (2 30 15)A x∠ = + ° = × + °

= (60 + 15)° = 75°

30B x∠ = ° = °

and (3 15) (3 30 15)C x∠ = − ° = × − ° = (90 – 15)° = 75°

we see that 75A C∠ = ∠ = °

⇒ AB = AC

Hence ∆ ABC is an isosceles triangle.

5. In a ∆ ABC, ( 18) ,A x∠ = + ° 2( 12)B x∠ = − ° and

33 .

2

xC

∠ = − °

Find each angle of the triangle and hence show

that it is an equilateral triangle.

Ans. In ∆ ABC, 180A B C∠ + ∠ + ∠ = ° (by property)

⇒3

18 2( 12) 3 1802

+ ° + − ° + − ° = °x

x x

⇒ 3

18 2 24 3 1802

+ ° + − ° + − ° = °x

x x

⇒ 3

3 18 27 1802

+ + ° − ° = °x

x ⇒ 3

3 9 1802

+ − ° = °x

x

⇒ 3

3 180 92

+ = ° + °x

x ⇒ 6 3

1892

+= °

x x

B C

A


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⇒9

1892

= °x

⇒ 189 2

9

°×=x ⇒ x = 42°

∴ ( 18) (42 18) 60A x∠ = + ° = + ° = °

2( 12) 2(42 12) 2 30 60∠ = − ° = − ° = × ° = °B x

3 3 423 3 (63 – 3) 60

2 2

xC

° °×

∠ = − = − = ° = °

Thus, 60A B C∠ = ∠ = ∠ = °

⇒ AB = BC = CA

Hence∆ ABC is an equilateral triangle.

6. In a ,ABC∆ 3 6 ,A B C∠ = ∠ = ∠ Find each angle of the triangle.

Ans. Let 3A∠ = 6B∠ = C x∠ = °

⇒ A x∠ = ° , 3

xB

° ∠ =

and

6

xC

° ∠ =

In ABC∆ ,

180A B C∠ + ∠ + ∠ = ° (by property)

⇒ 1803 6

+ + = °x x

x ⇒ 6 2

1806

+ += °

x x x

⇒ 9180

6= °

x ⇒ 180 6

9

°×=x ⇒ x = 120

Hence 120A x∠ = ° = °

12040

3 3

xB

° °

∠ = = = °

and120

20 .6 6

xC

° °

∠ = = = °


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7. In the adjoining figure, it is being given that

∠ ABC = 40°, ∠ ACD = 70° ∠ ACB = x°,

and ∠ EAC = y°, find the values of x and y.

BC

D

A E

40° 70°x°

y°

Ans. 180ACB ACD∠ + ∠ = ° (linear pair)

⇒ x + 70° = 180° ⇒ x = 180° – 70° = 110°

In ,ABC∆

CAE B ACB∠ = ∠ + ∠ (property of exte-

rior angle)

⇒ y = 40° + x = 40° + 110° = 150°

Hence, x = 110° and y = 150°

8. In the adjoining figure, it is being given that

50 ,A∠ = ° 70B∠ = ° and BO and CO are the bisectors of B∠ and

C∠ respectively. Find the measure of .BOC∠

Ans. In ABC∆ , 180A B C∠ + ∠ + ∠ = ° [by property]

⇒ 50 70 180C° + ° + ∠ = ° ⇒ 120 180C° + ∠ = °

⇒ 180 120 60C∠ = ° − ° = °

CO is the bisector of C∠

Therefore, 60

302

OCB

°

∠ = = °

In BOC∆ , 180BOC OBC OCB∠ + ∠ + ∠ = ° [by property]

⇒ 35 30 180∠ + ° + ° = °BOC

BC

D

A E

40° 70°x°

y°

B C

A

O

35°35°

50°


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⇒ 65 180∠ + ° = °BOC ⇒ 180 65BOC∠ = ° − °

Hence, 115 .BOC∠ = °

9. Using the information given in the adjoining figure, find the values

of x and y.

B C

A

64°

y°36°

24°

x°

x°

O

Ans. In ABC∆ , (24 36) 60B∠ = + ° = °, And ( ) 2C x x x∠ = + ° = °

and 180A B C∠ + ∠ + ∠ = ° [by property]

⇒ 64° + 60° + 2x = 180° ⇒ 124° + 2x = 180°

⇒ 2x = 180° – 124° ⇒ 2x = 56°

⇒56

2x

°= ⇒ x = 28°

In BOC∆ , 180BOC OCB CBO∠ + ∠ + ∠ = ° [by property]

⇒ y + x + 36° = 180° ⇒ y + 28° + 36° = 180°

⇒ y + 64° = 180° ⇒ y = 180° – 64° ⇒ y = 116°

Hence, x = 28° and y = 116°.

10. In the adjoining figure DBCE is a straight line, 135ABD∠ = ° and

125 .ACE∠ = ° Find .BAC∠

B C ED

A

135° 125°


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Ans. In ABC∆ ,

180ABC ABD∠ + ∠ = ° [linear pair of angles]

⇒ 135 180ABC∠ + ° = ° ⇒ 180 135ABC∠ = ° − °

⇒ 45ABC∠ = ° and 180ACB ACE∠ + ∠ = °

[linear pair of angles]

⇒ 125 180ACB∠ + ° = °

⇒ 180 125ACB∠ = ° − ° ⇒ 55ACB∠ = °

In ABC∆ ,

180BAC ABC ACB∠ + ∠ + ∠ = ° [By property]

⇒ 45 55 180BAC∠ + ° + ° = ° ⇒ 180 100BAC∠ = ° − °

Hence 80 .BAC∠ = °

11. In ,ABC∆ side AC has been produced to D. If 125BCD∠ = ° and

: 2 : 3.A B∠ ∠ = Find the measures of A∠ and .B∠

A

B C

D125°

Ans. Let 2A x∠ = and 3B x∠ =

Thus, In ABC∆ , BCD A B∠ = ∠ + ∠ (by property)

⇒ 125° = 2x + 3x ⇒ 125° = 5x ⇒ 5x = 125°

Thus 125

5

°=x ⇒ x = 25°

Therefore, 2A x∠ = ° = 2 × 25° = 50°

And, 3B x∠ = = 3 × 25° = 75°


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12. Calculate the size of each lettered angles in the following figures :

54°

xy

80°

z

z44°

48° yx

50°

B

DA

E

C

(i) (ii)

Ans. (i) AOD BOC∠ = ∠ (Vertically opposite angles)

In ,AOD∆ 180AOD OAD ODA∠ + ∠ + ∠ = °

(Sum of all angles in a triangle is 180°)

⇒ 80° + x + 54° = 180° ⇒ 134° + x = 180°

⇒ x = 180° – 134° ⇒ x = 46°

Now, AB = AD ⇒ 54ABD ADB∠ = ∠ = ° ⇒ 54ABD∠ = °

(i) In ,ABD∆

180ABD ADB BAD∠ + ∠ + ∠ = °

(sum of all angles in a triangle is 180°)

⇒ 54° + 54° + (x + y) = 180° (∵ BAD∠ = x + y)

⇒ 108° + 46° + y = 180° (∵ x∠ = 46°)

⇒ 154° + y = 180°

⇒ y = 180° – 154° = 26°

Also, z = x + y = 46° + 26° = 72°

Hence, x = 46°, y = 26° and z = 72°.

(ii)z 44°

48° yx

50°

B C

ED

A

54°

xy

80°

z

BA

CO

D


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In ,ABD∆

∵ BD = AD (given)

∴ 44ABD BAD∠ = ∠ = °

∴ z = 180° – ( )ABD BAD∠ + ∠

⇒ z = 180° – (44° + 44°) ⇒ z = 180° – 88° ⇒ z = 92°

In ,AEC∆

∵ AE = EC ∴ 1

2x = (180 )AEC° − ∠

⇒1

2x = (180° – 50°) ⇒

1130

2x = × ° ⇒ x = 65°

∵ AB || EC ⇒ 48° + (x + y) = 180°

(∵ Sum of co interior angles is 180°) (∵ ECB∠ = (x + y))

⇒ 48° + 65° + y = 180° ⇒ y = 180° – 113°

y = 67°

Hence, x = 65°, y = 67° and z = 92°.

Q.13 From the following figures, find the values of x and y:

(y+7)°(4x–5)°

(y–

10)°

5x°

A

BC

D

E

(y+15)°

(x–15)°

(2y–

15

)°

x°

B C

A

(i) (ii)

Ans. (i)

(y+15)°

(4x–5)°

(y–10)°

5x°

B CD

E

A


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In ,ABC∆ 180A B ACB∠ + ∠ + ∠ = °

(∵ sum of all angles in a triangle is 180°)

⇒ (y – 10°) +(y + 7°) + (4x – 5°) = 180°

⇒ 4x + 2y – 8° = 180° ⇒ 4x + 2y = 180° + 8°

⇒ 4x + 2y = 188° ⇒ 2 (2x + y) = 188°

⇒ 2x + y = 94° ...(i)

Also AB || EC

∴ ABC ECA∠ = ∠ (corresponding angles)

∴ ( 7)ECD y∠ = + °

Now (4x – 5)° + 5x° + (y + 7)° = 180° (linear pair)

⇒ 9x + y + 2° = 180° ⇒ 9x + y = 180° – 2°

⇒ 9x + y = 178° ...(ii)

Subtracting (ii) from (i), we get

2x + y = 94°

9x + y = 178°

– – –

– 7x = – 84°

⇒84

7x

−=

−⇒ x = 12°

Substituting the value of x = 12° in (i), we get

2 × 12° + y = 94° ⇒ 24 + y + 94°

⇒ y = 94° – 24° ⇒ y = 70°

Hence, x = 12°, y = 70°.

(ii) In ,ABC∆ 180BAC B C∠ + ∠ + ∠ = °

⇒ (2y – 15)° + (y + 15)° + (x – 15)° = 180°

⇒ 3y + x – 15° = 180°

⇒ x + 3y = 180° + 15°

(y+15)°

(x–15)°

(2y–

15)°

x°

B C

A


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⇒ x + 3y = 195° ...(i)

Also, x = y + 15° (corresponding angles)

⇒ x – y = 15 ...(ii)

Subtract (ii) from (i), we get

x + 3y = 195°

x – y = 15°

– + –

4y = 180°

⇒ 180

454

y°

= = °

Putting the value of y = 45° in (ii), we get

x – 45° = 15° ⇒ x = 15° + 45° ⇒ x = 60°

Hence, the value of x = 60°, y = 45°.

14. Calculate the size of each lettered angles in the following figures :

(iii)

30°24°

y xx

56°

62°

27°

(i) (ii)

23°

x

y 130°30° x°

64°

A

D

B CB

C

E

A

B

C

D

A

D

Ans. (i) In ABC∆ , 180ABC BAC ACB∠ + ∠ + ∠ = °

(Sum of all angles in triangle is 180°)

⇒ (24° + 30°) + 56° + (x + x) = 180°

⇒ 54° + 56° + 2x = 180°

⇒ 110° + 2x = 180°

⇒ 2x = 180° – 110° ⇒ 2x = 70°

⇒ 70

2x

°= ⇒ x = 35°

30°24°

y xx

56°

(i)

A

B C

D


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In BCD∆

180DBC BCD BDC∠ + ∠ + ∠ = °

⇒ 30° + x + y = 180°

30° + 35° + y = 180° (∵ x = 35°)

⇒ 65° + y = 180°

⇒ y = 180° – 65° ⇒ y = 115°

Hence, x = 35° and y = 115°.

(ii) In ,ABC∆ BC || DE (given)

⇒ 23BCD∠ = ° (Alternate angles)

Now, 180A B ACB∠ + ∠ + ∠ = °


⇒ y + 62° + (27° + 23°) = 180° (∵ BCD∠ = 23°)

⇒ y + 62° + 50° = 180°

⇒ y + 112° = 180° ⇒ y = 180° – 112°

⇒ y = 68°

Also, x EDC DCE= ∠ + ∠

(∵ exterior angle is equal to the sum the interior opposite angles)

⇒ x = 23° + 27° ⇒ x = 50°

Hence, x = 50° and y = 68°.

(iii) ∵ ADC EAD AED∠ = ∠ + ∠

(∵ exterior angle is equal to the sum of the opposite interior

angles)

⇒ 130° = 30° + AED∠

⇒ 130 – 30AED∠ = ° °

⇒ 100AED∠ = °

∵ 180AED DEB∠ + ∠ = ° (linear pair)

⇒ 100 180DEB° + ∠ = °

62°

27°

(ii)

23°

x

y

E

A

D

B C

A

B

C30° 130°

64°

E

x


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⇒ 180 100DEB∠ = ° − ° ⇒ 80DEB∠ = °

In ,BEC∆

180B ECB BEC∠ + ∠ + ∠ = °

(Sum of all the angles in a triangle) is 180°)

⇒ 64° + x° + 80° = 180° (∵ BEC DEB∠ = ∠ )

⇒ 144° + x° = 180° ⇒ x° = 180° – 144°

⇒ x° = 36°

Hence, x = 36°.

15. Find the values of x and y each in the following figures :

x

y

66°

117°

(i)

64° y

(ii)

x

E A

B C D B Cm

Al

Ans. (i) x

y

66°

117°

(i)B C D

E A

∵ In fig. (i), ACD ABC BAC∠ = ∠ + ∠

⇒ 117° = ABC∠ + 66°

⇒ 117 66ABC∠ = ° − ° ⇒ 51ABC∠ = °

Thus, x = 51° (∵ AE || BC so these are alternate angles)

In triangle ABE, 180ABE AEB BAE∠ + ∠ + ∠ = °

(∵ sum of all angles of a triangle is 180°)


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⇒ y + 90° + x = 180°

⇒ y + 90° + 51° = 180° ⇒ y + 141° = 180°

⇒ y = 180° – 141° ⇒ y = 39°

Hence, x = 51° and y = 39°.

(ii) In fig. (ii) 64ABC x∠ = = °

(∵ l || m, these are alternate angles)

and ACB y∠ = (alternate angles)

In ,ABC∆

180ABC ACB BAC∠ + ∠ + ∠ = °


⇒ x + y + BAC∠ = 180°

⇒ x + y + x = 180° (∵ BC = AC ∴ BAC x∠ = )

⇒ 2x + y = 180°

⇒ 2 × 64° + y = 180° (∵ x = 64°)

⇒ 128° + y = 180°

⇒ y = 180° – 128° ⇒ y = 52°

Hence, the value of x = 64° and y = 52°.

16. In an isosceles triangle, the vertical angle is 15° more than each of its

base angles. Find the base angles of the triangle.

Ans. Let the base angles of the triangle ABC be x°.

∴ Vertical angle of triangle ABC = (x + 15°)

∴ 180A B C∠ + ∠ + ∠ = °

⇒ x + x + 15° + x = 180°

⇒ 3x + 15° = 180°

⇒ 3x = 180° – 15°

⇒ 3x = 165° ⇒ 165

3x

°= ⇒ x = 55°

∴ x + 15° = 55° + 15° = 70°

Hence, base angles of triangle are 55° and 55° and vertical angle 70°

64° y

(ii)

x

Al

mB C

B C

A

x° x°

(x+15)°


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17. The ratio between the vertical angle and a base angle of an isosceles

triangle is 4 : 3. Find all the angles of the triangle.

Ans. Let the vertical angle and base angle of an isosceles triangle is

4x and 3x respectively. Then other base angle will also be 3x.

Now, we know that, Sum of all angles in a triangle is 180°

⇒ 4x + 3x + 3x = 180° ⇒ 10x = 180° ⇒ 180

10

°=x ⇒ x = 18°

Hence, all the angles in triangle ABC are 4 × 18°, 3 × 18° and 3 ×

18° i.e. 72°, 54° and 54°.

18. In the adjoining figure AB = AC and DE || BC. Find

(i) x

(ii) y

(iii) BAC∠

Ans. In ,ABC∆ given that AB = AC and DE || BC

∴ .B C∠ = ∠

and ADE ABC∠ = ∠ (corresponding angles)

⇒ x + y – 36° = 2x

⇒ x + y – 2x = 36°

⇒ – x + y = 36° ...(i)

Also, 2x = y – 2° (∵ B C∠ = ∠ )

⇒ 2x – y = – 2° ...(ii)

Adding (i) and (ii), we get

– x + y = 36

2x – y = – 2°

x = 34°

Substituting the value of x = 34 in (i), we get

– 34° + y = 36°

A

B C

2x°(y-2)°

D E(x+y

–36)

°

A

B C

2x°(y-2)°

D E(x+y

–36)

°


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⇒ y = 36° + 34° ⇒ y = 70°

Also, 180 (2 2)BAC x y∠ = ° − + − °

180 (2 34 70 2)= ° − × ° + ° − °

= 180° – 68° – 68° = 112° – 68° = 44°

Hence, (i) x = 34°, (ii) y = 70° and (iii) 44 .BAC∠ = °

19. In the adjoining figure, it is being given that BC || DE, 70 ,CED∠ = °

84CBA∠ = °and .BAC x∠ = ° Find the value of x.

E

D BA

C

x°84°

70°

Ans.BC || DE and AE is the transversal

∴ ACB CED∠ = ∠ (corresponding angles)

In ∆ ABC, 180∠ + ∠ + ∠ = °ABC ACB CAB [by property]

⇒ 84° + 70° + x = 180°

⇒ 154° + x = 180°

⇒ x = 180° – 154° ⇒ x = 26°.

20. In the adjoining figure, it is being given that DE || BC,

25 ,EDC∠ = ° 20 ,ECD∠ = ° 70 ,ABC∠ = ° BAC x∠ = ° and

.DEA y∠ = ° Find the values of x and y.

A

B C

D E25°

70°

20°

y°

x°


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Ans. DE || BC, and DC is the transversal

∴ BCD CDE∠ = ∠ ⇒ 25BCD∠ = ° (alternate angles)

∴ 20 25 45ACB∠ = ° + ° = °

In ABC∆ , 180A B C∠ + ∠ + ∠ = ° [by property]

⇒ x + 70° + 45° = 180°

⇒ x + 115° = 180°

⇒ x = 180° – 115° ⇒ x = 65°

DE || BC and AC is the transversal

AED ECB∠ = ∠ (corresponding angles)

∴ y = 45°

Hence, x = 65° and y = 45°.

21. In the adjoining figure, AE || BC,

,DAE x∠ = ° ( 15)∠ = − °ACB x

2

xBAC y

°

∠ = +

and ( 15)∠ = + °ABC y

Find the values of x and y.

Ans. AE || BC and AC is the transversal

CAE ACB∠ = ∠ (alternate angles)

⇒ ( 15)CAE x∠ = − °

180DAE CAE CAB∠ + ∠ + ∠ = °

[Angles at a point on a line]

⇒ 15 1802

+ − ° + + = °x

x x y ⇒ 2 180 152

+ + = ° + °x

x y

⇒4 2

1952

+ += °

x x y ⇒ 5x + 2y = 195 × 2°

⇒ 5x + 2y = 390° ...(i)

Now in ABC∆

B C

A E

D

(y+15)° (x–15)°

x°

x2—+y

°


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180BAC ABC BCA∠ + ∠ + ∠ = ° (by property)

⇒ 15 15 1802

+ + + ° + − ° = °x

y y x ⇒ 2 1802

xx y+ + =

⇒2 4

1802

+ += °

x x y ⇒ 3x + 4y = 180° × 2

⇒ 3x + 4y = 360° ...(ii)

Multiply (i) by 2, we have

10x + 4y = 780° ...(iii)

Subtracting (iii) from (ii), we have

⇒ – 7x = – 420° ⇒ 7x = 420°

⇒420

7

°=x ⇒ x = 60°

Putting the value of x in (i), we have

5 × 60° + 2y = 390°

⇒ 300° + 2y = 390° ⇒ 2y = 390° – 300°

⇒ 2y = 90° ⇒ 90

2

°=y ⇒ y = 45°

Hence, x = 60° and y = 45°.

22. In the adjoining diagram BAC BDC∠ = ∠ and .ACB DBC∠ = ∠

Prove that AC = BD.

B C

A D

Ans. In the adjoining diagram,

Given : BAC BDC∠ = ∠


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and ACB DBC∠ = ∠

To prove : AC = BD

Proof: In ∆ ABC and ∆ BCD

BAC BDC∠ = ∠ (given)

ACB DBC∠ = ∠ (given)

BC = BC (common)

∴ ABC BCD∆ ≅ ∆ (AAS axiom of congruency)

⇒ AC = BD (c.p.c.t.).

23. In the adjoining figure, AB = BC and AD = CD. Prove that .A C∠ = ∠

D

AB

C


Given : AB = BC and AD = CD

To prove : A C∠ = ∠

Proof : In ∆ ABD and ∆ BCD

AB = BC (given)

AD = CD (given)

BD = BD (common).

∴ ABD BCD∆ ≅ ∆

(SSS axiom of congruency)

⇒ A C∠ = ∠ (c.p.c.t.)

B C

A D

D

A B

C


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24. In the adjoining figure, ,ACB EDF∠ = ∠ BA || EF and AC = DE.

Prove that

(i) AB = EF (ii) BD = CF

B

A

C

D

E

F


Given : ,ACB EDF∠ = ∠ BA || EF and AC = DE

To prove: (i) AB = EF, (ii) BD = CF

B

A

C

D

E

F

Proof: (i) ∆ ABC and ∆ DEF

ACB EDF∠ = ∠ (given)

AC = DE (given)

ABC DFE∠ = ∠ (Alternate angles)

∴ ∆ ABC ≅ ∆ DEF (AAS axiom of congruency)

⇒ AB = EF (c.p.c.t.)

(ii) BC = DF (c.p.c.t.)

⇒ BD + DC = DC + CF ⇒ BD = CF


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25. In the adjoining figure, line segments AB and CD bisect each other at

O. Prove that

(i) AC = DB (ii) AD = CB

C B

A D

O

Ans. In the adjoining figure

C B

A D

O

Given : AO = BO

CO = DO (∵ AB and CD bisect each other at O)

To prove: (i) AC = DB (ii) AD = CB

Proof : (i) In ∆ AOC and ∆ BOD

AO = BO (given)

CO = DO (given)

AOC BOD∠ = ∠ (Vertically opposite angles)

∴ AOC BOD∆ ≅ ∆ (By SAS Axiom of congruency)

⇒ AC = DB (c.p.c.t.)

AOD COB∆ ≅ ∆

⇒ AD = CB (c.p.c.t.)


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26. In the adjoining figure, ABC is an equilateral triangle and BCDE is a

square.A

B

E

C

D

Prove that: (i) 15BAE∠ = ° (ii) AE = AD

Ans. In the adjoining figure

A

B

E

C

D

Given : ABC is an equilateral triangle and BCDE is a square.

To prove : (i) 15BAE∠ = ° (ii) AE = AD.

Proof :

(i) ∵ ABC is an equilateral triangle

∴ AB = BC = AC ...(i)

Also BCDE is an square

∴ BC = BE = ED = DC ...(ii)

From (i) and (ii), we have

AB = BE


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In ∆ ABE, ABE ABC EBC∠ = ∠ + ∠ = 60° + 90° = 150°

∵ AB = BE

⇒ BAE BEA∠ = ∠

∴ 1

(180 )2

BAE ABE∠ = ° − ∠ 1

(180 150 )2

= ° − °

1

30 152

= × ° = °

(ii) ∵ ACD ACB BCD∠ = ∠ + ∠ = 60° + 90° = 150°

Now in ACD∆

AB = AC (By equation (i))

BE = CD (By equation (ii))

ABE ACD∠ = ∠ (each 150°)

∴ ABE ACD∆ ≅ ∆ (By SAS axiom of congruency)

Thus, AE = AD (c.p.c.t.)

27. In ,ABC∆ it is being given that AB = AC and AD is the bisector of

,A∠ meeting BC at D. Prove that:

An (i) ABD ACD∆ ≅ ∆ (ii) .AD BC⊥

Ans. (i) In ABD∆ and ACD∆

A

B D C

AB = AC (given)

AD = AD (common)

BAD CAD∠ = ∠ (given)


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ABD ACD∆ ≅ ∆ [SAS]

(ii) ADB ADC x∠ = ∠ = (say) (c.p.c.t.)

But 180ADB ADC∠ = ∠ = ° (Linear pair of angles)

⇒ x + x = 180° ⇒ 2x = 180°

⇒ 180

2x

°= ⇒ x = 90°

90ADB ADC∠ = ∠ = °

Hence, .AD BC⊥

28. In the adjoining figure AB || GF, AC || DE and BF = CE. Prove that :

BDE FGC∆ ≅ ∆

Ans. BF = CE (given)

Adding FE both sides

BF + FE = CE + FE

⇒ BE = FC

BF E

C

GD

A

In BDE∆ and FGC∆

B F∠ = ∠ (corresponding angles)

E C∠ = ∠ (corresponding angles)

BE = CF (Proved above)

.BDE FGC∆ ≅ ∆


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29. In the adjoining figure, ABCD is a square and CEB is an isosceles

triangle in which EC = EB.

D

A

C

B

E

Show that .∆ ≅ ∆DCE ABE

Ans. Given : ABCD is a square and CEB is an isosceles traiangle such

that EC = EB.

To prove : DCE ABE∆ ≅ ∆

Proof : In ,BEC∆ BE = CE (given)

∴ 90DCE DCB BCE BCE∠ = ∠ + ∠ = ° + ∠

and 90ABE ABC CBE CBE∠ = ∠ + ∠ = ° + ∠

∴ DCE ABE∠ = ∠

Now in DCE∆ and ABE∆

DC = AD (sides of a square)

CE = BE (equal sides of an isosceles triangle)

DCE ABE∠ = ∠ (proved)

Hence, DCE ABE∆ ≅ ∆ (SAS axiom)

30. In the adjoining figure, ABCD is a square and P, Q, R are points on

the sides AB, BC and CD respectively such that AP = BQ = CR and

90PQR∠ = °

D

A

C

B

R

P

QL


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Prove that : (i) PB = QC (ii) PQ = QR (iii) 45QPR∠ = °

Ans. Given : In square ABCD,

P, Q, R are points on sides AB, BC and CD respectively such that

AP = BQ = CR and 90PQR∠ = °

PQ, QR and RP are joined

To prove : (i) PB = QC (ii) PQ = QR (iii) 45QPR∠ = °

Proof : ∴ AB = BC ...(i) (Sides of a square)

and AP = BQ ...(ii) (given)

Subtracting (ii) from(i), we have

AB – AP = BC – BQ

⇒ PB = QC

In ,CQR∆ RQB QRC C∠ = ∠ + ∠

⇒ RQP PQB QRC C∠ + ∠ = ∠ + ∠

⇒ 90 90PQB QRC° + ∠ = ∠ + ° ⇒ PQB QRC∠ = ∠

In PBQ∆ and ,QCR∆

PB = QC (proved)

BQ = CR (given)

90B C∠ = ∠ = ° (Proved)

∴ PBQ QCR∆ ≅ ∆ (SAS axiom)

∴ PQ = QR (c.p.c.t.)

(iii) In ,PQR∆

PQ = QR (Proved)

∴ QPR QRP∠ = ∠ (angles opposite to equal sides)

But 90PQR∠ = ° (given)

∴ 90QPR QRP∠ + ∠ = °

Hence, 45QPR∠ = °


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31. In the adjoining figure ABCD is a square, EF || AC and R is the mid

point of EF

D

A

C

B

F

E

R

Prove that: (i) AE = CF (ii) DE = DF (iii) DR bisects EDF∠

Ans. Given: ABCD is a square in which EF || AC and R is the mid-point of

EF then DR, DE and DF are joined.

To Prove: (i) AE = CF

(ii) DE = DF

(iii) DR bisects EDF∠

Proof : ∵ AC || EF ⇒ ∠BEF = ∠BAC and ∠BFE = ∠BCA

∴ EFB FEB∠ = ∠ (correspondign angles)

But ∠ACB = ∠BAC (AB = BC) ⇒ ∠EFB = ∠FEB

∴ EB = BF (sides opposite to equal angles)

But AB = BC (sides of a squares)

Subtracting

AB – EB = BC – BF

AE = CF

Now in ADE∆ and ,CDF∆

AD = DC (sides of a square)

AE = CF (proved)

A C∠ = ∠ (each 90°)

∴ ADE CDF∆ ≅ ∆ (SAS axiom)

∴ DE = DF (c.p.c.t.)

Now in ,EDF∆

DE = DF (Proved)


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R is mid-point of EF

Hence, DR is the bisector of .EDF∠

32. The supporting wire to a top of a vertical pole is 17 m long and it is

fastened to the ground 8 m from the foot of the pole. How high is the

pole?

Ans. Let AB = h m be the pole

Thus AC = 17 m

BC = 8 m

∴ In right-angled triangle ABC

AC2 = BC2 + AB2

⇒ AB2 = AC2 – BC2

⇒ 2 2h AB AC BC= = −

⇒ 2 2(17) (8)h = −

⇒ 289 64h = − , 225h = , h = 15

Hence, height of pole AB = 15 m.

33. A 15 m long ladder rests against a vertical wall and reaches to a

window. If the window is 12 m above the ground, how far is the

foot of the ladder from the wall ?

Ans. Let AC be ladder and AB be the height of window.

AB = 12 m

AC = 15 m

In right-angled ,ABC∆ using

Pythagoras theorem, we have

AC2 = AB2 + BC2

BC2 = AC2 – AB2

2 2BC AC AB= −

A

B C

17 m

8 m

A

BC

12 m

15 m


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2 2(15) (12)= −

225 144= −

81= , = 9

BC = 9 m.

34. The length of the diagonal of a rectangle is 40 cm. If the length of its

shorter side is 24 cm ; find :

(i) the length of its longer side.

(ii) the area of the rectangle.

(iii) the perimeter of the rectangle.

Ans. Let ABCD be the rectangle in which diagonal AC = 40 cm and BC =

24 cm.

(i) In right-angled triangle ABC,

AC2 = AB2 + BC2

AB2 = AC2 – BC2

2 2AB AC BC= −

2 2(40) (24)= − 1600 576= −

1024= = 32

∴ Longer side = 32 cm

(ii) Area of rectangle ABCD = AB × BC = 32 × 24 = 768 cm2

(iii) Perimeter of rectangle ABCD = 2 (AB + BC)

= 2 (32 + 24) = 112 cm

35. Use the diagram given below to prove that :

AC2 + CD2 = BD2 + AB2

D

A

C

B

24 cm40 cm


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A

B C

D

Ans. From right-angled triangle ABC, AC2 = AB2 + BC2 ...(i)

From right angled triangle DBC BD2 = BC2 + CD2 ...(ii)

Subtracting (ii) from (i), we have

AC2 – BD2 = AB2 + BC2 – BC2 – CD2

⇒ AC2 – BD2 = AB2 – CD2

Hence AC2 + CD2 = BD2 + AB2

36. A man goes 1.2 km due North and then 0.5 km due West. Find the

least distance between his initial and final positions.

Ans.

W

N

B

A E

C 0.5 km

1.2

km

Let AB = 1.2 km, BC = 0.5 km

Since 90ABC∠ = °

[∵ AB in the North direction and BC is in the west direction]

∴ 2 2AC AB BC= + 2 2(1.2) (0.5) 1.44 0.25= + = +


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1.69= = 1.3 km

Hence, the least distance between the initial and final positions is

1.3 km.

37. The base BC of an isosceles triangle is 28 cm long and AB = AC =

50 cm. Let .AD BC⊥ Find.

(i) the length of AD and (ii) the area of .ABC∆

Ans.

B D C

A

50 cm

28 cm

(i) ABC∆ is an isosceles triangle and AD BC⊥

⇒ BD = DC = 28

2 cm = 14 cm

In right angled ADC∆ using Pythagoras theorem

(AD)2 = (AC)2 – (DC)2

= (50)2 – (14)2

= 2500 – 196 = 2304

∴ 2304AD =

⇒ AD = 48 cm.

(ii) Area of 1

2ABC BC AD∆ = × ×

21

28 48 cm2

= × × = 672 cm2.


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38. A diagonal of a rhombus is 16 cm long and each of its sides measure

10 cm. Find the length of the other diagonal of the rhombus.

D C

BA 10 cm

O

8 cm

8 cm

Ans. In rhombus ABCD, Each side = 10 cm, and one diagonal AC =

16 cm

∴ Diagonals bisects each other at right angles.

∴ In right triangle AOB,

AB = 10 cm, 1

2AO = =

116 cm

2= ×AC = 8 cm

∴ AB2 = AO2 + OB2 (Pythagoras theorem)

(10)2 = (8)2 + OB2 ⇒ 100 = 64 + OB2

⇒ OB2 = 100 – 64 = 36 = (6)2. Thus, OB = 6 cm

∴ BD = 2 OB = 2 × 6 = 12 cm.

39. The supporting wire to the top of a vertical pole is 13 m long and

it is fastened to the ground at a stake 5 m away from the foot of the

pole. How high is the pole?

Ans. Here, AB is the length of the pole and AC = 13 m is the length of

the wire. In right angle, ∆ABC by Pythagoras Theorem

(AB)2 = (AC)2 – (BC)2 = (13)2 – (5)2

= 169 – 25 = 144

∴ 144AB =

⇒ AB = 12 m

Hence, length of the pole is 12 m.

A

BC

13 m

5 m


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40. In the adjoining figure, ABC is a triangle in which 90 .B∠ = ° If D is

the mid-point of BC, prove that AC2 = AD2 + 3CD2

Ans.

B D C

A

Given : In ,ABC∆ 90B∠ = °. D is mid point of BC. AD is joined

To prove : AC2 = AD2 + 3 CD2

Proof : In right triangle ABD, we have

AD2 = AB2 + BD2 ⇒ AB2 = AD2 – BD2

and in right triangle ABC, we have

AC2 = AB2 + BC2 = AD2 – BD2 + (2CD)2 (∵ 2CD = BC)

= AD2 – CD2 + 4CD2 (∵ BD = DC)

= AD2 + 3CD2

41. In the adjoining figure, it is being given that AB = 27cm, CD = 12

cm, AC = 36 cm, 90BAC DCA∠ = ∠ = ° and AM = CM. Find

(i) BM 2 (ii) MD2 (iii) BD2

B

E

A M C

D15 cm

12 cm

18 cm 18 cm

Ans.In the figure, AB = 27 cm, CD = 12 cm, AC = 36 cm

90BAC DCA∠ = ∠ = ° and AM = MC


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(i) In right triangle ABM, we have

BM 2 = AB2 + AM 2 = (27)2 + (18)2

1 1

( 36 18 cm)2 2

AM AC= = × = = 729 + 324 = 1053

(ii) In triangle MDC, we have MD2 = DC2 + MC2

= (12)2 + (18)2 = 144 + 324 = 468 cm

(iii) BE = AB – AE = AB – CD = 27 – 12 = 15 cm

ED = AC = 36 cm

Now in right triangle BED, we have

BD2 = BE2 + ED2 = (15)2 + (36)2

= 225 + 1296 = 1521 cm2.

42. In the adjoining figure, 90PQR QRS∠ = ∠ = °

Prove that PR2 – PQ2 = QS2 – SR2

P

Q R

S

Ans. Given : In the figure, 90PQR QRS∠ = ∠ = °

To prove : PR2 – PQ2 = QS2 – SR2

Proof : In ,PQR∆ 90PQR∠ = °

PR2 = PQ2 + QR2

⇒ QR2 = PR2 – PQ2 ...(1)

Similarly in ,QSR∆ 90QRS∠ = °

∴ QS2 = QR2 + SR2, QR2 = QS2 – SR2 ...(2)


PR2 – PQ2 = QS2 – SR2


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43. In the adjoining figure, 90ABC∠ = ° . Prove that AC2 + PQ2

= AQ2 + PC2

A

P

B QC

Ans. Given : In figure, 90ABC∠ = °

P, Q are any two points on the sides AB and BC of ABC∆

AQ, CP are joined

To prove : AC2 + PQ2 = AQ2 + PC2

Construction : Join PQ

Proof : In ,ABC∆ 90B∠ = °

∴ AC2 = AB2 + BC2 ...(i)

(Pythagoras theorem)

Similarly, in ,PBQ∆ we have, PQ2 = PB2 + BQ2 ...(ii)

In ,ABQ∆ we have, AQ2 = AB2 + BQ2

In ,PBC∆ we have, PC2 = PB2 + BC2

Now, AC2 + PQ2 = AB2 + BC2 + PB2 + BQ2

= AB2 + BQ2 + PC2 + PB2

= AQ2 + PC2

44. In a ,ABC∆ if D and E are mid points of AB and AC respectively

and 90 .BAC∠ = ° Prove that, BE2 + CD2 = 5DE2


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A

CB

D E

Ans. In right triangle ABE BE2 = AB2 + AE2...(i) (Pythagoras theorem)

and in right triangle ACD, we have CD2 = AC2 + AD2 ...(ii)

Adding (i) and (ii) we have

BE2 + CD2 = AB2 + AE2 + AC2 + AD2

= AB2 + AC2 + AE2 + AD2

= (BC)2 + DE2

= (2DE)2 + DE2 (∵ BC = 2 ED)

= 4DE2 + DE2 = 5DE2

45. In quadrilateral ABCD, 90B D∠ = ∠ = °.

Prove that, AB2 – AD2 = CD2 – CB2

A B

C

D

Ans. Given : In quadrilateral ABCD, 90B D∠ = ∠ = °

To prove : AB2 – AD2 = CD2 – CB2

Construction : Join AC

Proof : In right ,ABC∆ 90B∠ = °

AC2 = AB2 + BC2 ...(i) (Pythagoras theroem)

Similarly, in right triangle ADC, 90D∠ = °

∴ AC2 = AD2 + CD2 ... (ii)


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AB2 + BC2 = AD2 + CD2

AB2 – AD2 = CD2 – CB2

46. In the adjoining figure, all measurements are in the centimeres. Find

the length AD.

B

A C

123

D

4

Ans. In right angled ∆ABC, by Pythagoras theorem, we have.

AC2 = AB2 + BC2 = (3)2 + (12)2 = 9 + 144 = 153

In right angled triangle ACD, we have,

By Pythagoras theorem,

AD2 = AC2 + CD2

= 153 + (4)2 = 153 + 16 = 169

⇒ 169AD = = 13

Hence, the length of AD is 13 cm.

47. In the adjoining figure, AB = 6 cm, BD = 3 cm and AD = BC, if

90ABC∠ = °. Find the length of AC.

Ans. We have,

AB = 6 cm, BD = 3 cm (given)

and AD = BC

In right-angled triangle ABD, we have


AD2 = AB2 + BD2 = (6)2 + (3)2

= 36 + 9 = 45

A C

12

4

B

3

D

A

B D CA

B D C

6 cm

3 cm


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⇒ 45AD =

∴ 45BC = (∵ AD = BC)

Now, in right triangle ABC, we have

2 2 2AC AB BC= + ( )

22(6) 45= + = 36 + 45 = 81

⇒ 81AC = = 9

Hence, the length of AC is 9 cm.

48. In the adjoining figure, all measurements

are in centimetres. Find (i) ED (ii) AE.

Hence, prove that AE2 + ED2 = AD2.

State the size of .AED∠

Ans. (i) In right angle triangle ECD,

By Pythagoras theorem, we have

ED2 = CD2 + EC2 = (1.8)2 +

(2.4)2

∴ = 3.24 + 5.76 = 9

⇒ 9ED = = 3

∴ Length of ED = 3 cm

(i) In right angled triangle AEB, we have


AE2 = AB2 + BE2

= (AF + BF)2 + BE2

= (AF + CD)2 + BE2

( )22(1.2 1.8) 7= + +

( )22(3) 7= + = 9 + 7 = 16

⇒ 16AE = = 4

A

D

CB

F

5

1.8

1.2

E7 2.4


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Thus, length of AE = 4 cm

Now, AE2 + ED2 = (4)2 + (3)2 = 16 + 9 = 25

And, AD2 = (5)2 = 25

∴ AE2 + ED2 = AD2

Then, by converse of Pythagoras theorem

if AE2 + ED2 = AD2 then AED∆ is a right angle triangle thus

90 .AED∠ = °

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