[4 marks sum]

Question 14 Marks

Angle A of an isosceles trapezium ABCD is 115°; find the angles B, C and D.

Answer

Since the base angles of an isosceles trapezium are equal,
∴ ∠A = ∠B = 115°

Also, ∠A and ∠D are co-interior angles and their sum = 180°
∴ ∠A + ∠D = 180°
⇒115° + ∠D = 180°
⇒ ∠D = 180° - 115°
⇒ ∠D = 65°
Also, ∠D = ∠C = 65°
∴ ∠B = 115°, ∠C = 65°, ∠D = 65°

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Question 24 Marks

Two diagonals of an isosceles trapezium are x cm and (3x – 8) cm. Find the value of x.

Answer

∵ The diagonals of an isosceles trapezium are of equal length

∴ 3x – 8 = x
⇒ 3x – x = 8 cm
⇒ 2x = 8 cm
⇒ x = 4 cm
∴ The value of x is 4 cm

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Question 34 Marks

Each angle of a quadrilateral is x + 5°. Find:
(i) the value of x
(ii) each angle of the quadrilateral.
Give the special name of the quadrilateral taken.

Answer

(i) We have,
∠A + ∠B + ∠C + ∠D = 360°
∵ We know that the sum of interior angles of a quadrilateral is 360°
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ x + 5° + x + 5° + x + 5° + x + 5° = 360°
⇒ 4x + 20° = 360°
⇒ 4x = 360° - 20°
$\Rightarrow x =\frac{340^{\circ}}{4}=85^{\circ}$
(ii) Each angle of the quadrilateral ABCD = x + 5°
= 85° + 5°
= 90°
The name of the given quadrilateral is a rectangle.

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Question 44 Marks

In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78° and ∠C = 120. find angles B and D.

Answer

∵ AB || DC and BC is transversal
∴∠B and ∠C, ∠A and ∠D are Cointerior angles with their sum = 180°
i.e. ∠B + ∠C = 180°
⇒ ∠B + 120° = 180°
⇒ ∠B = 180° – 120°
⇒ ∠B = 60°
Also ∠A + ∠D = 180°
⇒ 78° + ∠D = 180°
⇒ ∠D = 180° – 78°
∠D = 102°

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Question 54 Marks

Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.

Answer

∵ ∠C = 64° (Given)
∴ ∠D = ∠C – 8° = 64°- 8° = 56°
∠A = 5(a+2)°
∠B = 2(2a+7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a+2)° + 2(2a+7)° + 64° + 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360° – 144°
9a = 216°
a = 24°
∴ ∠A = 5 (a + 2) = 5(24+2) = 130°

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Question 64 Marks

The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.

Answer

Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [AB||DC]
4x + 5x = 180°
⇒ 9x = 180°
⇒ x = 20°
∠A = 4 (20) = 80°,
∠D = 5 (20) = 100°
Again ∠B + ∠C = 180° [ AB||DC]
3x – 15° + 4x + 20° = 180°
7x = 180° – 5°
⇒ 7x = 175°
⇒ x = 25°
∠B = 75° – 15° = 60°
and ∠C = 4 (25) + 20 = 100°+ 20°= 120°

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MATHS [4 marks sum]

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