[5 marks sum]

Question 15 Marks

The angles A, B, C and D of a quadrilateral are in the ratio 2:3: 2 : 3. Show this quadrilateral is a parallelogram.

Answer

Given, Angles of a quadrilateral are in the ratio 2: 3: 2: 3
i.e. A : B : C : D are in the ratio
2: 3: 2: 3
To prove: Quadrilateral ABCD is a parallelogram
Proof: Let us take ∠A = 2x, ∠B = 3x, ∠C = 2x and ∠D = 3x
We know, that the sum of interior angles of a quadrilateral = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
⇒ 2x + 3x + 2x + 3x = 360°
⇒ 10x = 360°
$\Rightarrow x=\frac{360^{\circ}}{10}=36^{\circ}$
∴ ∠A = ∠C = 2x = 2 × 360° = 72°
∠B = ∠D = 3x = 3 × 36° = 360°
Now, A quadrilateral ABCD is considered as a parallelogram.
(i) When opposite angles are equal,
i.e. ∠A = ∠C = 72° and ∠B = ∠D = 108°
(ii) When adjacent angles are supplementary
i.e. ∠A + ∠B = 180°
and ∠C = ∠D = 180°
⇒ 72° + 108° and 72° + 108° = 180°
⇒ 180° = 180° and 180° = 180°
Since quadrilateral ABCD fulfills the conditions.
∴ Quadrilateral ABCD is a parallelogram.

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Question 25 Marks

Two opposite angles of a parallelogram are 100° each. Find each of the other two opposite angles.

Answer

Given: Two opposite angles of a parallelogram are 100° each.

∵ Adjacent angles of a parallelogram are supplementary,
∴ ∠A + ∠B = 180°
⇒ 100° + ∠B = 180°
⇒ ∠B = 180° – 100°
⇒ ∠B = 80°
Also, opposite angles of a parallelogram are equal
∴∠D = ∠B = 80°
∴∠B = ∠D = 80°

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Question 35 Marks

The angles A, B, C and D of a trapezium ABCD are in the ratio 3: 4: 5: 6. Le. ∠A : ∠B : ∠C : ∠D = 3:4: 5 : 6. Find all the angles of the trapezium. Also, name the two sides of this trapezium which are parallel to each other. Give reason for your answer.

Answer

As the trapezium ABCD is a quadrilateral,
∴ Sum of its interior angles = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
⇒ 3x + 4x + 5x + 6x = 360°
⇒ 18x = 360°
$\Rightarrow x=\frac{360^{\circ}}{18}=20^{\circ}$
∴ ∠A = 3x = 3 × 20° = 60°
and ∠B = 4x = 4 × 20° = 80°
and ∠C = 5x = 5 × 20° = 100°
and ∠D = 6x = 6 × 20° = 120°
AB is parallel to DC.
∵ ∠A + ∠D = 180°
∠A and ∠D are co-interior angles whose sum = 180°

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Question 45 Marks

In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x° and ∠D = (3x – 20)°; find the value of x.

Answer

∵ AB || DC and BC is transversal
∴ ∠A and ∠B are Co-interior angles with their sum = 180°
i.e. ∠A + ∠D = 180°

⇒ x° + (3x – 20)° = 180°
⇒ x° + 3x° – 20° = 180°
⇒ 4x° = 180° + 20°
$x^{\circ}=\frac{200}{4}=50^{\circ}$
∴ Value of x = 50°

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Question 55 Marks

From the following figure find ;
(i) x
(ii) ∠ABC
(iii) ∠ACD

Answer

(i) In Quadrilateral ABCD,

x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x = 312
$x=\frac{312}{12}=26^{\circ}$
(ii) ∠ABC = 4x4 × 26 = 104°
(iii) ∠ACD = 180° - 4x - 48°
= 180° - 4 × 26° - 48°
= 180° - 104° - 48°
= 180° - 152° = 28°

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Question 65 Marks

In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD

Answer

∵ ∠A : ∠D = 1 : 2
Let ∠A = x and ∠B = 2x
∵ ∠C : ∠B = 4 : 5
Let ∠C = 4y and ∠B = 5y
∵ AB || DC
∴ ∠A + ∠D = 180°
x + 2x = 180°
3x = 180°
x = 60°
∴ A = 60°
∠D = 2x = 2 × 60 = 120°
Again ∠B + ∠C = 180°
5y + 4y = 180°
9y = 180°
y = 20°
∴ ∠B = 5y = 5 20 = 100°
∠C = 4y = 4 20 = 80°
Hence ∠A = 60°; ∠B = 100°; ∠C = 80°
and ∠D = 120°

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Question 75 Marks

Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C

Answer

∵ ∠A = 90° (Given)
∠B = (2x + 4°)
∠C = (3x - 5°)
∠D = (8x - 15°)
∠A + ∠B + ∠C + ∠D = 360°
90° + (2x + 4°) + (3x - 5°) + (8x - 15°) = 360°
90° + 2x + 4° + 3x - 5° + 8x - 15° = 360°
⇒ 74° + 13x = 360°
⇒ 13x = 360° - 74°
⇒ 13x = 286°
⇒ x = 22°
∵ ∠B = 2x 4 = 2 × 22° + 4 = 48°
∠C = 3x - 5 = 3 × 22° - 5 = 61°
Hence (i) 22° (ii) ∠B = 48°, ∠C = 61°

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Question 85 Marks

Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°. Find :
(i) the value of x.
(ii) each angle of the quadrilateral.

Answer

Angles of quadrilateral are,
(4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°
∴ 4x + 5(x + 2) + (7x - 20) + 6(x + 3) = 360°
4x + 5x + 10 + 7x - 20 + 6x + 18 = 360°
22x + 8 = 360°
22x = 360° - 8°
22x = 352°
x = 16°
Hence angles are,
(4x)° = (4 × 16)° = 64°
5(x + 2)° = 5(16 + 2)° = 90°
(7x - 20)° = (7 × 16 - 20)° = 92°
6(x + 3)° = 6(16 + 3) = 114°

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Question 95 Marks

In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other
(i) Is PS also parallel to QR?
(ii) Assign a special name to quadrilateral PQRS.

Answer

∵ ∠P : ∠Q : ∠R : ∠S = 3: 4: 6: 7
Let ∠P = 3x
∠Q = 4x
∠R = 6x
∠S = 7x
∴ ∠P + ∠Q + ∠R + ∠S = 360°
3x + 4x + 6x + 7x = 360°
20x = 360°
x = 18°
∴ ∠P = 3x = 3 × 18 = 54°
∠Q = 4x = 4 × 18 = 72°
∠R = 6x = 6 × 18 = 108°
∠S = 7x = 7 × 18 = 126°
∠Q + ∠R = 72° + 108° = 180°
or ∠P + ∠S = 54° + 126° = 180°
Hence PQ || SR
As ∠P + ∠Q = 72° + 54° = 126°
Which is ≠ 180°
∴ PS and QR are not parallel.
PQRS is a Trapezium as its one pair of opposite side is parallel.

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