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## S Chand Class 9 ICSE Maths Solutions Chapter 11 Rectilinear Figures Chapter Test

Question 1.

Which of the following terms best describes the figure below?

(a) Rhombus

(b) Trapzoid

(c) Quadrilateral

(d) Square

(e) Parallelogram

Solution:

(b) Trapzoid

The given figure is a trapezium.

∵ A pair of two opposite sides are parallel.

Question 2.

Give the quadrilateral below, what value of x would allow you to conclude that the figure is a parallelogram?

(a) – 2

(b) 0

(c) 1

(d) 2

(e) 3

Solution:

(e) 3

∵ The given figure is of a parallelogram.

∴ Their opposite sides are equal and parallel

∴ x – 1 = 2x – 4 ⇒ – 1 + 4 = 2x – x ⇒ x = 3

2x + 1 = 3x – 2 ⇒ 3x – 2x = 1 + 2 ⇒ x = 3

∴ x = 3

Question 3.

In the figure, if ABCD is a rectangle, what type of triangle must ∆ABD be?

(a) Equilateral

(b) Right

(c) Equiangular

(d) Isosceles

(e) Scalene

Solution:

(b) Right

In figure, ABCD is a rectangle whose diagonals bisect each other at O.

∆ABD is a right triangle as ∠A = 90°.

Question 4.

Find the measures of the lattered angles in each figure.

(i) ABCD is a parallelogram.

(ii) rectangle PQRS

(iii) rhombus MNPQ

Solution:

(i) In figure, ABCD is a parallelogram

∴ Their opposite sides are equal and parallel and opposite angles are equal

∴ x + 6 = 5x – 8 ⇒ 5x – x = 6 + 8

⇒ 4x = 14 ⇒ x = \(\frac { 14 }{ 4 }\) = 3.5 4

⇒ 14y + 6y = 180 (co-interior angles)

20y = 180° ⇒ y = \(\frac { 180° }{ 20° }\) = 9

6y = a ⇒ 6 x 9 = a ⇒ a = 54°

b = 14y = 14 x 9 = 126°

∴ a = 54°, b = 126°

(ii) PQRS is a rectangle

∴ Opposite sides are equal and parallel. Diagonals bisect each other

c = 33° (Alternate angles)

Similarly, a = e (Alternate angles)

e = ∠1 (∵ OQ = OR)

and ∠1 + 33° = 90°

⇒ ∠1 = 90° – 33° = 57°

∴ e = ∠1 = 57° and a = e = 57°

∵ OP = OQ

∴ c = ∠2 = 33°

d = 180° – c – ∠2 = 180° – 33° – 33°

= 180 – 66° = 114°

and b + d = 180°

⇒ b = 180° – d = 180°- 114° = 66°

Hence,

a = 57°, b = 66°, c = 33°, d = 114°, e = 57°

(iii) MNPQ is a rhombus in which all sides are equal and diagonals bisect each other at 90° and diagonals bisect opposite angles

b = 53° (alternate angles)

a = d (alternate angles)

e = 53°

but e + d = 90° (in ∆PON)

53° + d = 90° ⇒ d = 90° – 53° = 37° a = d = 31°

∠c = 90°

(∵ Diagonals bisect each other at 90°)

a = 37°, b = 53°, ∠c = 90°, d = 37°, e = 53°

Question 5.

In kite ABCD, ∠OBC = 58°, and ∠DAB = 50°, and the measures of

(i) ∠BCD

(ii) ∠DAO

(iii) ∠ODA

(iv) ∠ADC

Solution:

In the figure, ABCD is a kite in which BC = CD, AB = AD

and diagonals AC and BD intersect at right angles at O and AC bisects the oppoiste angles.

∠OBC = 58°, ∠DAB = 50°

(i) In ABCD, BC = CD

∴ ∠OBC = ∠ODC = 58°

∴ ∠BCD = 180° – ∠OBC – ∠ODC

= 180° – 58° – 58°

= 180°- 116° = 64°

(ii) ∠DAO = \(\frac { 1 }{ 2 }\) ∠BAD = \(\frac { 1 }{ 2 }\) x 50° = 25°

(iii) ∠ODA = 90° – ∠DAO = 90° – 25° = 65°

(iv) ∠ADC = ∠ODA + ∠ODC = 65° + 58° = 123°

Question 6.

Given: PQ and QR are mid segment of ∆ABC and AB = BC

Prove: BPQR is a rhombus.

Solution:

In ABC, PQ and QR are mid segments AB = BC

∴ PQ || BC and PQ = \(\frac { 1 }{ 2 }\) BC = BR … (i)

(R is mid point of BC)

But AB = BC and P, R their mid points

∴ PB = BR … (ii)

and QR || AB and QR = \(\frac { 1 }{ 2 }\) AB = PB

From (i), (ii) and (iii),

PB = BR = QR = PQ

∴ BPQR is a rhombus or a square

But ∠B ≠ 90°

∴ BPQR is a rhombus.