The availability of S Chand Class 9 Maths Solutions ICSE Chapter 16 Area of Plane Figures Chapter Test encourages students to tackle difficult exercises.

## S Chand Class 9 ICSE Maths Solutions Chapter 16 Area of Plane Figures Chapter Test

Question 1.

What is the area of an equilateral triangle having altitude equal to 2√3 cm?

(a) √3cm^{2}

(b) 2√3cm^{2}

(c) 3√3cm^{2}

(d) 4√3cm^{2}

Solution:

Altitude = \(\frac{\sqrt{3}}{2}\) sides ⇒ \(\frac{\sqrt{3}}{2}\) = 2√3 cm

∴ Side = \(\frac{2 \sqrt{3} \times 2}{\sqrt{3}}\) = 4 cm

Area = \(\frac{\sqrt{3}}{4}\) side^{2} = \(\frac{\sqrt{3}}{4}\) × 4 × 4

= 4√3cm^{2} (d)

Question 2.

The altitude of an equilateral triangle is √3 cm. What is its perimeter?

(a) 3 cm

(b) 3√3 cm

(c) 6 cm

(d) 6√3 cm

Solution:

Altitude of an equilateral triangle = √3

∴ \(\frac{\sqrt{3}}{2}\) side = √3 ⇒ side = 2 cm

and perimeter = 3 × Side

=3 × 2 = 6 cm

Question 3.

The sides of a triangle are in the ratio 2 : 3 : 4. The perimeter of the triangle is 18 cm. The area (in cm^{2}) of the triangle is

(a) 9

(b) 36

(c) √42

(d) 6√15

Solution:

Sides of a triangle are in the ratio = 2 : 3 : 4

and perieter = 18 cm

Sum of ratios = 2 + 3 + 4 = 9

∴ First side = \(\frac { 18 }{ 9 }\) × 2 = 4 cm

Second side = \(\frac { 18 }{ 9 }\) × 3 = 6 cm

Third side = \(\frac { 18 }{ 9 }\) × 4 = 8 cm

∴ s = \(\frac{a+b+c}{2}\)

= \(\frac { 18 }{ 9 }\) = 9

Question 4.

A parallelogram has sides 15 cm and 7 cm long. The length of one of the diagonals is 20 cm. The area of the parallelogram is

(a) 42 cm^{2}

(b) 60 cm^{2}

(c) 84 cm^{2}

(d) 96 cm^{2}

Solution:

Sides of a parallelogram = 15 cm and 7 cm

and length of one diagonal = 20 cm

Question 5.

The perimeter of two squares are 68 cm and 60 cm. The perimeter of a third square whose area is equal to the difference of the areas of the two squares.

(a) 64 cm

(b) 60 cm

(c) 32 cm

(d) 8 cm

Solution:

Perimeter of one square = 68 cm

∴ Side = \(\frac { 68 }{ 4 }\) = 17 cm

Perimeter of second square = 60 cm

∴ Side = \(\frac { 60 }{ 4 }\) = 15 cm

Area of first square = (Side)^{2}

= (17)^{2} = 289 cm^{2}

Area of first square = (Side)^{2}

=(17)^{2} = 289 cm^{2}

Area second square = (15) = 225 cm^{2}

∴ Area of third square = 289 – 225 = 64 cm^{2}

∴ Side = \(\sqrt{\text { Area }}\) = \(\sqrt{64}\) = 8 cm

and perimeter =4 × side = 4 × 8 = 32 cm (c)

Question 6.

A square is of area 200 m^{2}. A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. Then the area of the new square formed is

(a) 200√2 sq m

(b) 400√2 sq m

(c) 400 sq m

(d) 800 sq m

Solution:

Area of a square = 200 m^{2}

∴ Its diagonal = \(\sqrt{\text { Area } \times 2}\)

= \(\sqrt{200 \times 2}\) = \(\sqrt{400}\) = 20 m

∴ Diagonal of second square = √2 × 20 m

Area of second square = \(\frac{(\text { diagonal })^2}{2}\)

= \(\frac{(\sqrt{2} \times 20)^2}{2}\) = \(\frac{2 \times 400}{2}\) = 400 m^{2} (c)

Question 7.

The sides of a parallelogram are in the ratio 5 : 4. Its area is 1000 sq. units. Altitude on the greater side is 20 units. Altitude on the smaller side is

(a) 30 units

(b) 25 units

(c) 10 units

(d) 15 units

Solution:

Ratio in the sides of a parallelogram = 5 : 4

Area = 1000 sq. units

Altitude on the greater side = 20 units

Let greater side = 5x

and smaller side = 4x

∴ Area = Greater side × Altitude on it

1000 = 5x × 20 ⇒ x = \(\frac{1000}{5 \times 20}\) = 10

∴ Greater side = 5 × 10 = 50 units

and altitude on smaller side = \(\frac{\text { Area }}{\text { Smaller side }}\) = \(\frac{1000}{4 \times 10}\) = 25 units

Question 8.

One diagonal and perimeter of a rhombus are 24 cm and 52 cm respectively. The other diagonal is

(a) 13 cm

(b) 12 cm

(c) 10 cm

(d) 15 cm

Solution:

Perimeter of a rhombus = 52 cm

∴ Side = \(\frac{\text { Perimeter }}{4}\) = \(\frac{52}{4}\) = 13 cm

One diagonal = 24 cm

By joining the diagonals

They bisect each other at O at right angle

∴ AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm and BO = OD

In right △AOB

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

⇒ (13)^{2} = 12^{2} + B^{2}

⇒ 169 = 144 + B^{2}

⇒ BO^{2} = 169 – 144 = 25 = (5)^{2}

BO = 5 cm

∴ Diagonal BD = 2 × BO

= 2 × 5 = 10 cm

Question 9.

The area of a field in the shape of a trapezium measures 1440 m^{2}. The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel sides is:

(a) 45 m

(b) 60 m

(c) 75 cm

(d) 120 m

Solution:

Area of a trapezium shaped field = 1440 m^{2}

Perpendicular distance = 24 m

∴ Sum of parallel sides = \(\frac{\text { Area } \times 2}{\text { Height }}\) = \(\frac{1440 \times 2}{24}\) = 120 cm

Ratio in side = 5 : 3

Sum of ratio = 5 + 3 = 8

∴ First longer side = \(\frac { 120 }{ 8 }\) × 5 = 75 m

Question 10.

A lawn 30 m long and 16 m wide is surrounded by a path 2 m wide. What is the area of the path?

(a) 200 m^{2}

(b) 280 m^{2}

(c) 300 m^{2}

(d) 320 m^{2}

Solution:

Length of a lawn = 30 m

Breadth of a lawn = 16 m

Width of path around it = 2 m

∴ Outer length = 30 + 2 × 2

= 30 + 4 = 34 m

and outer breadth = 16 + 2 × 2

=16 + 4 = 20 m

∴ Area of path = Outer Area – Inner area

= 34 × 20 – 30 × 16 m^{2}

& = 680 – 480 = 200 m^{2} (a)