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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
Math Class VIII 2 Question Bank
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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°.
Math Class VIII 3 Question Bank
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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
Math Class VIII 4 Question Bank
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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
° °
∠ = = = °
Math Class VIII 5 Question Bank
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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°
Math Class VIII 6 Question Bank
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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°
Math Class VIII 7 Question Bank
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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°
Math Class VIII 8 Question Bank
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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
Math Class VIII 9 Question Bank
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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
Math Class VIII 10 Question Bank
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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
Math Class VIII 11 Question Bank
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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
Math Class VIII 12 Question Bank
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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∠ + ∠ + ∠ = °
(Sum of all angles in a triangle is 180°)
⇒ 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
Math Class VIII 13 Question Bank
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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°)
Math Class VIII 14 Question Bank
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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∠ + ∠ + ∠ = °
(Sum of all angles in a triangle is 180°)
⇒ 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)°
Math Class VIII 15 Question Bank
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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)
°
Math Class VIII 16 Question Bank
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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°
Math Class VIII 17 Question Bank
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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
°
Math Class VIII 18 Question Bank
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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∠ = ∠
Math Class VIII 19 Question Bank
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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
Ans. In the adjoining diagram,
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
Math Class VIII 20 Question Bank
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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
Ans. In the adjoining diagram,
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
Math Class VIII 21 Question Bank
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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.)
Math Class VIII 22 Question Bank
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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
Math Class VIII 23 Question Bank
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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)
Math Class VIII 24 Question Bank
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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∆ ≅ ∆
Math Class VIII 25 Question Bank
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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
Math Class VIII 26 Question Bank
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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∠ = °
Math Class VIII 27 Question Bank
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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)
Math Class VIII 28 Question Bank
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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
Math Class VIII 29 Question Bank
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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
Math Class VIII 30 Question Bank
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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= + = +
Math Class VIII 31 Question Bank
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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.
Math Class VIII 32 Question Bank
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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
Math Class VIII 33 Question Bank
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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
Math Class VIII 34 Question Bank
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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)
From (i) and (ii), we have
PR2 – PQ2 = QS2 – SR2
Math Class VIII 35 Question Bank
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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
Math Class VIII 36 Question Bank
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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)
Math Class VIII 37 Question Bank
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From (i) and (ii), we have
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
By Pythagoras theorem,
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
Math Class VIII 38 Question Bank
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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
By Pythagoras theorem,
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
Math Class VIII 39 Question Bank
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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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