Students can track their progress and improvement through regular use of S Chand Class 9 Maths Solutions ICSE Chapter 16 Area of Plane Figures Ex 16(C).

## S Chand Class 9 ICSE Maths Solutions Chapter 16 Area of Plane Figures Ex 16(C)

Question 1.

Find the area of a parallelogram whose base and height are as given below.

Base | 8 cm | 2.8 cm | 12 cm | 6.5 cm |

Height | 3 cm | 5 cm | 8.7 cm | 4.8 cm |

Solution:

(i) Base (b) = 8 cm

Height (h) = 3 cm

∴ Area = bh = 8 × 3 = 24 cm^{2}

(ii) Base (b) = 2.8 cm

Height (h) = 5 cm

∴ Area = bh = 2.8 × 5 = 14 cm^{2}

(iii) Base (b) = 12 mm

Height (h) = 8.7 mm

Area = bh = 12 × 8.7 = 104.4 mm^{2}

(iv) Base (b) = 6.5 m

Height (h) = 4.8 m

Area = bh = 6.5 × 4.8 = 31.20 m^{2}

Question 2.

The area of a parallelogram is 1\(\frac { 1 }{ 2 }\) ares. Its base is 20 m. Find its height (1 are = 100 m^{2}).

Solution:

Area of a parallelogram = 1\(\frac { 1 }{ 2 }\) ares

= 1.5 × 100 = 150 m^{2}

Base (b) = 20 m

Height (h) = \(\frac{\text { Area }}{\text { Base }}\) = \(\frac { 150 }{ 20 }\) = 7.5 m

Question 3.

In a parallelogram ABCD, AB = 8 cm, BC = 5 cm, perp. from A to DC = 3 cm. Find the length of the perp. drawn from B to AD.

Solution:

In ||gm ABCD,

AB = 8 cm, BC = 5 cm

and perp. AL on DC = 3 cm

Let perp. BM on DA = x

Now area of the ||gm ABCD

= Base × Height

= DC × AL = 8 × 3 = 24 cm^{2}

If base is AD = BC = 5 cm

Then height BM = \(\frac{\text { Area }}{\text { Base }}\)

Question 4.

A parallelogram has sides 34 cm and 20 cm. One of its diagonals is 42 cm. Calculate its area.

Solution:

In ||gm ABCD,

AB = 34 cm, AD = 20 cm

and diagonal BD = 42 cm

∵ Each diagonals of a ||gm bisect it into two triangles of equal area

∴ Area of ||gm ABCD = 2 area △ABD Now in △ABD,

Question 5.

ABCD is a parallelogram with side AB = 12 cm. Its diagonal AC and BD are of lengths 20 cm and 16 cm respectively. Find the area of ||gm ABCD.

Solution:

Side AB of ||gm ABCD = 12 cm

Diagonal AC = 20 cm

and diagonal BD = 16 cm

∵ Diagonals of a ||gm bisect each other

∴ AO = OC = \(\frac { 20 }{ 2 }\) = 10 cm

and BO = OD = \(\frac { 16 }{ 2 }\) = 8 cm

Now area of △AOB with sides 12 cm, 10 cm, 8 cm

∴ Diagonals of a ||gm form 4 equal triangles

∴ Area of ||gm ABCD = 4 × area △AOB

= 4 × 15 √7 = 60√7 cm^{2}

Question 6.

What is the area of a rhombus which has diagonals of 8 cm and 10 cm.

Solution:

Diagonals of a rhombus ABCD are AC = 8 cm, BD= 10 cm

∴ Area = \(\frac{\text { Product of diagonals }}{2}\) = \(\frac{\mathrm{AC} \times \mathrm{BD}}{2}\)

= \(\frac{8 \times 10}{2}\) = 40 cm^{2}

Question 7.

The area of a rhombus is 98 cm^{2}. If one of its diagonals is 14 cm, what is the length of the other diagonal ?

Solution:

Area of rhombus = 98 cm^{2}

and length of one diagonal = 14 cm

Second diagonal = \(\frac{\text { Area } \times 2}{1 \text { st diagonal }}\) = \(\frac{98 \times 2}{14}\) = 14 cm^{2}

Question 8.

PQRS is a rhombus.

(i) If it is given that PQ = 3 cm, calculate the perimeter of PQRS.

(ii) If the height of the rhombus is 2.5 cm, calculate its area.

(iii) The diagonals of a rhombus are 8 cm and 6 cm respectively. Find its perimeter.

Solution:

(i) In rhombus PQRS,

Side PQ = 3 cm

∴ Perimeter = 4 × side = 4 × 3 = 12 cm

(ii) Height of rhombus PQRS = 2.5 cm

∴ Area = base + height = 3 × 2.5 = 7.5 cm^{2}

(iii) Diagonals are 8 cm and 6 cm of rhombus PQRS

∵ Diagonals of a rhombus bisect each other at right angles

∴ In right △POQ,

PO = 4 cm, QO = 3 cm

∴ PQ^{2} = PO^{2} + QO^{2} (Pythagoras Theorem)

(4)^{2} + (3)^{2} = 16 + 9 = 25

= (5)^{2}

∴ Side PQ = 5 cm

and perimeter = 4 × side = 4 × 5 = 20 cm

Question 9.

The sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate,

(i) The length of the other diagonal and

(ii) The area of the rhombus.

Solution:

Each side of rhombus = 5 cm

and one diagonal = 8 cm

∵ Diagonals of a rhombus bisect each other at right angle

(i) ∴ In right △AOB,

AO = \(\frac { 1 }{ 2 }\) AC = \(\frac { 8 }{ 2 }\) = 4 cm and BO = \(\frac { 1 }{ 2 }\) BD

= AB^{2} = AO^{2} + BO^{2}

(5)^{2} = (4)^{2} + BO^{2}

⇒ 25 = 16 + BO^{2}

⇒ BO^{2} = 25 – 16 = 9 = (3)^{2}

∴ BO = 3 cm

∴ BD = 2 × BO = 2 × 3 = 6 cm

(ii) Area = \(\frac{\text { Product of diagonals }}{2}\)

= \(\frac{8 \times 6}{2}\) = 24 cm^{2}

Question 10.

In the figure, ABCX is a rhombus of side 5 cm. Angles BAD and ADC are right angles. If DC = 8 cm, calculate the area of ABCX.

Solution:

In the figure, ABCX is a rhombus with each side = 5 cm.

∠BAD and and ∠ADC are 90° each DC = 8 cm

∴ DX = DC – XC

= 8 – 5 = 3 cm

Now in right △ADX

AX^{2} = AD^{2} + DX^{2} (Pythagoras Theorem)

(5)^{2} = AD^{2} + (3)^{2}

25 = AD^{2} + 9

⇒ AD^{2} = 25 – 9 = 16 = (4)^{2}

∴ AD = 4 cm

Now area of rhombus ABCX = base × height

= AB × AD

= 5 × 4 = 20 cm^{2}