Practicing S Chand Class 9 Maths Solutions ICSE Chapter 16 Area of Plane Figures Ex 16(A) is the ultimate need for students who intend to score good marks in examinations.

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

Question 1.

Find the perimeter and area of the rectangle whose dimensions are

(i) 8 cm by 5 cm

(ii) 12 m by 9 m

(iii) 3.5 cm by 2 m

Solution:

(i) Length of rectangle (l) = 8 cm

Breadth (b) = 5 cm

∴ Perimeter = 2 (l + b) = 2 (8 + 5) cm

= 2 × 13 = 26 cm

and area = l × b = 8 × 5 = 40 cm^{2}

(ii) Length of rectangle (l) = 12 m

and breadth (b) = 9 m

∴ Perimeter = 2 (l + b) = 2 (12 + 9) m

= 2 × 21 = 42 m

and area = l × b = 12 × 9 = 108 m^{2}

(iii) Length of the rectangle (l) = 3.5 m

and breadth (b) = 2 m

∴ Perimeter = 2 (l + b) = 2(3.5 + 2) m

= 2 × 5.5 = 11 m

and area = l × b = 3.5 × 2 m^{2}

= 7 m^{2}

Question 2.

Find the perimeter and area of the square whose each side is

(i) 3 cm

(ii) 5 m

(iii) 1.3 m

(iv) 2.4 m

Solution:

(i) Side of the square (a) = 3 cm

∴ Perimeter = 4a = 4 × 3 = 12 cm

and area = a^{2} = (3)^{2} = 9 cm^{2}

(ii) Side of the square (a) = 5 m

∴ Perimeter = 4a = 4 × 5 = 20 m

and area = a^{2} = (5)^{2}

= 25 m^{2}

(iii) Side of square (a) = 1.3 m

∴ Perimeter = 4a = 4 × 1.3 m

= 5.2 m

and area = a^{2} = (1.3)^{2} m^{2}

= 1.69 m^{2}

(iv) Side of the square (a) = 2.4 m

∴ Perimeter = 4a = 4 × 2.4 = 9.6 m

and area = a^{2} = (2.4)^{2} m^{2}

= 5.76 m^{2}

Question 3.

Find the length of one side and perimeter of a square whose area is :

(a) 25 m^{2}

(b) 144 m^{2}

(c) 256 cm^{2}

(d) 961 dm^{2}

Solution:

(i) Area of the square = 25 m^{2}

∴ Side of square (d) = \(\sqrt{\text { Area }}\) = \(\sqrt{25}\) = 5 m

∴ Perimeter = 4a = 4 × 5 m = 20 m

(ii) Area of the square = 144 m^{2}

∴ Side (a) = \(\sqrt{\text { Area }}\) = \(\sqrt{144}\) m = 12 m

and perimeter = 4a = 4 × 12 = 48 m

(iii) Area of the square = 256 cm^{2}

∴ Side (a) = \(\sqrt{\text { Area }}\) = \(\sqrt{256}\) = 16 cm

and perimeter = 4 a = 4 × 16 = 64 cm

(iv) Area of the square = 961 dm^{2}

∴ Side (a) = \(\sqrt{\text { Area }}\) = \(\sqrt{961}\) = 31 cm

and perimeter = 4a = 4 × 31 = 124 cm

Question 4.

The perimeter of a rectangle is 30 cm, and its breadth is 6 cm, find its length and its area.

Solution:

Perimeter of a rectangle = 30 cm

and breadth (b) = 6 cm

∴ l + b = \(\frac{\text { Perimeter }}{2}\) = \(\frac { 30 }{ 5 }\) = 15 cm

∴ Length = 15 – 6 = 9 cm

and area = l × b = 9 × 6 = 54 cm^{2}

Question 5.

The area of a rectangular courtyard, 30 m long, is 450 m^{2}. Find its breadth and perimeter.

Solution:

Area of rectangular compound = 450 m^{2} Length (l) = 30 m

∴Breadth = \(\frac{\text { Area }}{\text { Length }}\) = \(\frac { 450 }{ 30 }\) = 15 cm

and perimeter = 2 (l + b)

= 2 (30 + 15) m

= 2 × 45 = 90 m

Question 6.

The perimeter of a rectangular field is \(\frac { 3 }{ 5 }\) km. If the length of the field is twice its width, find its area in sq. metres.

Solution:

Perimeter of a rectangular field = \(\frac { 3 }{ 5 }\) km

= \(\frac { 3 }{ 5 }\) × 1000 = 600 m

Let width of the field = x m

Then length = 2 x m

Now l + b = \(\frac{\text { Perimeter }}{2}\) = \(\frac { 600 }{ 2 }\) = 300 m

∴ 2x + x = 300 ⇒ 3x = 300

⇒ x = \(\frac { 300 }{ 3 }\) = 100

∴ Length (l) = 2x = 2 × 100 = 200 m

and breadth (b) x = 100 m

and area = l × b = 200 × 100 = 20000 m^{2}

Question 7.

If the side of a square is doubled, find the ratio of the area of the resulting square to that of the given square.

Solution:

Let side of square = a m

Then area = (side)^{2} = a^{2} m^{2}

If the side is doubled, then

Side = 2a

and area = (side)^{2}

= (2a)^{2} = 4a^{2}

∴ Ratio in the two area = 4 a^{2} : a^{2} = 4 : 1

Question 8.

The side of a square field is 89 m. By how many square metres does its area fall short of a hectare ?

Solution:

Side of a square field (a) = 89 m

Area = a^{2} = (89)^{2} m^{2} = 7921 m^{2}

∴ Difference between 1 hectare and 7921 m^{2}

= 10000 – 7921 (1 hectare = 10000 m^{2})

= 2079 m^{2}

Question 9.

The side of a square is 25 cm. Find the length of its diagonal correct to 2 decimal places.

Solution:

Side of a square (a) = 25 cm

∴ Length of its diagonal = √2 a

= 1.414 × 25 cm

= 35.350 = 3535 cm

Question 10.

If the diagonal of a rectangle is 10 cm and its width is 6 cm, find its area.

Solution:

Diagonal of rectangle = 10 cm

and breadth (b) = 6 cm

∴ (Diagonal)^{2} = l^{2}+ b^{2}

⇒ (10)^{2} = (6)^{2} + l^{2}

⇒ 100 = 36 + l^{2} ⇒ l^{2} = 100 – 36 = 64

⇒ l^{2} = 64 = (8)^{2}

∴ l = 8

∴ Length = 8 cm

Now area = l × b = 8 × 6 = 48 cm^{2}

Question 11.

The length of a rectangle is twice its width. If the length of its diagonal is 16√5 cm, find its area.

Solution:

Length of a diagonal of a rectangle = 16√5 cm

Let breadth (b) = x cm

Then length (l) = 2x cm

∴ (Diagonal)^{2} = l^{2} + b^{2}

\((16 \sqrt{5})^2\) = (2x)^{2} + x^{2}

256 × 5 = 4x^{2} + x^{2} ⇒ 5x^{2} = 256 × 5

⇒ x^{2} = \(\frac{256 \times 5}{5}\) = 256 = (16)^{2}

∴ x = 16

∴ Length (l) = 2x = 2 × 16 = 32 cm

and breadth = x = 16 cm

and area = l × b

= 32 × 16 cm^{2} = 512 cm^{2}

Question 12.

The sides of two squares are in the ratio 2 : 3 . Find the ratio of their perimeters and areas.

Solution:

Ratio in the sides of two squares =2 : 3

Let side of the first square (a) = 2x

and of second (A) = 3x

Now perimeter of the first square = 4a

= 4 × 2x = 8x

and that of the second square = 4 × 3x = 12x

∴ Ratio in their perimeter = 8x : 12x

=2 : 3

(ii) Area of the first square (a)^{2} = (2x)^{2} = 4x^{2}

and area of the second square (A)^{2} = (3x)^{2} = 9x^{2}

and ratio between their areas = 4x^{2} : 9x^{2} = 4 : 9

Question 13.

How many carpets, 3 m by 2 m, are required to cover the floor of a hall 30 m by 12 m ?

Solution:

Length of the floor (L) = 30 m and breadth (B) = 12 m

∴ Area of floor = L × B = 30 × 12 = 360 m^{2}

Now length of carpet (l) = 3 m

and breadth (b) = 2m

∴ Area of one carpet = l × b = 3 × 2 = 6 m^{2}

∴ No. of carpets = \(\frac{\text { Area of floor }}{\text { Area of one carpet }}\)

= \(\frac { 360 }{ 6 }\) = 60 carpets

Question 14.

A room is 5 m by 4 m. Find the cost of cementing its floor at Rs. 7.50 per m^{2}.

Solution:

Length of a room (l) = 5 m and breadth (b) = 4 m

∴ Area = l × b = 5 × 4 = 20 m^{2}

Cost of cementing per m^{2} = Rs. 7.50

Total cost = Rs.7.50 × 20 = Rs.150.00

= Rs. 150

Question 15.

A room is 12 m by 10 m by 10 m. Find the cost of covering its floor with bricks 20 cm by 6 cm, if the cost of one thousand bricks is Rs. 300.

Solution:

Length of a room (l) = 12 m

Breadth (b) = 10 m

and height (h) = 10 m

∴ Area of floor = l × b = 12 × 10 = 120 m^{2}

Length of one brick = 20 cm

and breadth = 6 cm

∴ Area of one brick = 20 × 6 = 120 cm^{2}

= \(\frac{120}{100 \times 100}\) = \(\frac{12}{1000}\) cm^{2}

No. of bricks required = \(\frac{\text { Area of floor }}{\text { Area of one brick }}\)

= \(\frac{120}{12}\) × 1000 = 10000 bricks

Question 16.

In exchange for a square plot of land, one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as the square plot. Determine the width of the rectangular plot.

Solution:

Side of the square field (a) = 84 m

∴ Area = a^{2} = (84)^{2} m^{2}

= 84 × 84 = 7056 m^{2}

∴ Area of rectangular field = 7056 m^{2}

Length of the field (l) = 144 m

∴ Breadth = \(\frac{\text { Area }}{\text { Length }}\) = \(\frac{7056}{144}\) m = 49 m

Question 17.

A plot 110 metres long and 80 metres broad is to be covered with grass leaving 5 metres all around. Find the area to be laid with grass.

Solution:

Length of plot (l) = 110 m and breadth (b) = 80 m

∴ Area of the plot = l × b

= 110 × 80 = 8800 m^{2}

Leaving 5 m all around, then

Length of inner part = 110 – 2 × 5

= 110 – 10 = 100 m

and inner breadth = 80 – 2 × 5 = 80 – 10 = 70 m

∴ Inner area where grass is laid

=100 × 70 = 7000 m^{2}

Question 18.

The dimensions of a rectangular field are in the ratio 6 : 5. Find the cost of constructing a fence around the field at the rate of Rs. 1.25 per metre, given that the area of the field is 27000 m^{2}.

Solution:

Ratio in the dimensions of the field = 6 : 5

Let length (l) = 6x m

and breadth (b) = 5x m

∴ Area = l × b = 6x × 5x = 30 x^{2} m^{2}

Now area of the field = 27000 m^{2}

∴ 30x^{2} = 27000

⇒ x^{2} = \(\frac{27000}{30}\) = 900 = (30)^{2}

∴ x = 30

∴Length = 6x = 6 × 30 = 180 m

and breadth = 5x = 5 × 30 = 150 cm

and perimeter = 2(l + b) = 2(180 + 150) m

=2 × 330 = 660 m

Cost of fence = Rs. 1.25 per m

∴ Total cost = Rs. 1.25 × 660 = Rs. 825

Question 19.

The cost of enclosing a rectangular garden with a fence all round at the rate of 75 paise per metre is Rs. 300. If the length of the garden is 120 metres, find the area of the garden in square metres.

Solution:

Total cost of fencing = Rs. 300

Perimeter of the garden = Rs.\(\frac{300}{\frac{3}{4}}\) m

= \(\frac{300 \times 4}{3}\) = 400 m

∴ l + b = \(\frac{\text { Perimeter }}{2}\)

= \(\frac { 400 }{ 2 }\) = 200 m

Length (l) = 120 m

∴ Breadth (b) = 200 – 120 = 80 m

Now area of the garden = l × b

= 120 × 80 m^{2} = 9600

Question 20.

Find the area and perimeter of a square plot of land, the length of whose diagonal is 14 metres. Give your answer correct to 2 places of decimals.

Solution:

Diagonal of a square plot = 14 m

∴ Side (a) = \(\frac{\text { Diagonal }}{\sqrt{2}}\) = \(\frac{14}{\sqrt{2}}\)

∴ Area = \(\frac{(\text { Diagonal })^2}{2}\) = \(\frac{(14)^2}{2}\) m^{2}

= \(\frac { 196 }{ 2 }\) = 98 m^{2}

and perimeter = 4a = \(\frac{4 \times 14}{\sqrt{2}}\) m

= \(\frac{4 \times 14 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}\) = \(\frac{4 \times 14 \times \sqrt{2}}{2}\) m

= 28√2 = 28(1.4142) m

= 39.5976 = 39.60 m

Question 21.

A lawn 10 m by 8 m is surrounded by a path 1 m wide. Find the area of the path.

Solution:

Length of lawn (l) = 10 m

and breadth (b) = 8 cm

Width of path = 1 m

∴ Outer length (L) = 10 + 1 × 2 = 10 + 2 = 12 m

and outer width (B) = 8 + 1 × 2

= 8 + 2 = 10 m

Now area of path = Outer area – Inner area

= L × B – l × b

= 12 × 10 -10 × 8

= 120 – 80 = 40 m^{2}

Question 22.

The diagram given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. Find the area of the shaded portion.

Solution:

Length of the rectangular field (l) = 50 m and breadth (b) = 35 m

Width of each path = 5 m

∴ Area of two paths = 50 × 5 + 35 × 5 – (5)^{2}

= 250 + 175 – 25

= 425 – 25 = 400 m^{2}

∴ Area of shaded portion = 400 m^{2}

Question 23.

Find the area of the four walls of a room, 7 m long, 5 m broad, and 3 m high. Find the cost of distempering the walls at the rate of Rs. 3.15 per m^{2}.

Solution:

Length of room (l) = 7 m

Breadth (b) = 5 m

and height (h) = 3 m

∴ Area of four walls = 2 (l + b) h

= 2 (7 + 5) × 3 = 2 × 12 × 3 m^{2}

= 72 m^{2}

Rate of distempering = Rs. 3.15 per m^{2}

∴ Total cost = Rs. 3.15 × 72 = Rs. 226.80

Question 24.

Find the cost of the paper for the walls of the room if each piece of paper is 40 cm wide and 60 cm long : the room is 6 m long, 4 m wide, 3 m high and the cost of paper is Rs. 1.40 per piece; allow 11 m^{2} for doors, etc., and assume that a whole number of pieces has to be bought.

Solution:

Length of room (l) = 6 m = 600 cm

Breadth (b) = 4 m = 400 cm

and height (h) = 3 m = 300 cm

∴ Area of 4-walls = 2 (l + b) × h = 2 (600 + 400) × 300 cm^{2} = 2 × 1000 × 300 = 600000 cm^{2}

Length of paper = 60 cm

and breadth = 40 cm

∴ Area of 1 paper = 40 × 60 = 2400 cm^{2}

Area of doors etc. = 11 m^{2}

= 11 × 100 × 100 cm^{2} = 110000 cm^{2}

∴ Remaining area = 600000 – 110000 = 490000 cm^{2}

∴ No. of papers required = \(\frac { 490000 }{ 2400 }\)

= \(\frac { 4900 }{ 24 }\) = 204.166 = 205 sheets

Cost of one sheet = Rs. 1.40

∴ Total cost = 205 × Rs.1.40 = Rs. 287

Question 25.

A cistern 6 m long, 4 m wide, contains water to a depth of 1 m 25 cm : find the area of the wet surface.

Solution:

Length of cistern (l) = 6 m

Breadth (b) = 4 m

Level of water (h) = 1.25 m

Area of wet surface = area of 4 – walls + area of base

=2(6 + 4) × 1.25 + 6 × 4 m^{2}

= 2 × 10 × 1.25 + 24 = 25 + 24 m^{2} = 49 m^{2}

Question 26.

Find the area and perimeter of the shape illustrated in figure given below. The corners are all right angles.

Solution:

In the figure by joining the upper portion, we get the rectangles

Length of outer rectangle (l) = 6 + 2 + 2 = 10 cm

and breadth (B) = 5 cm

Length of inner rectangle (l) = 6 cm

and breadth (b) = 3 cm

(i) Area of the given figure = L × B – l × h

= 10 × 5 – 6 × 3

= 50 – 18 = 32 cm^{2}

(ii) Perimeter = (5 + 10 + 5 + 2 + 3 + 6 + 3 + 2) cm = 36 cm

Question 27.

Find the area of the figure, in which all corners are right-angles and the dimensions are shown in centimetres.

Solution:

In the figure all the dimensions are in cm By joining as shown in the figure we have 3 rectangles

Area of rectangle I = 4\(\frac { 1 }{ 4 }\) × 4 = 17 cm^{2}

Area of rectangle II = 7 × 3\(\frac { 1 }{ 2 }\) = \(\frac{7 \times 7}{2}\)

= \(\frac { 49 }{ 2 }\) = 24.5 cm^{2}

Area of rectangle III = 3 × 4\(\frac { 1 }{ 2 }\) = 3 × \(\frac { 9 }{ 2 }\)

= \(\frac { 27 }{ 2 }\) = 13.5 cm^{2}

Total area of the figure = 17 + 24.5 + 13.5 cm^{2} = 55 cm^{2}

Question 28.

How many tiles, each 12 cm by 6 cm are needed for a floor of a room 162 cm long, 144 cm wide

Solution:

Length of the floor of a room = 162 cm

and breadth = 144 cm

∴ Area of floor = length × breadth = 162 × 144 cm^{2}

Now length of one tile =12 cm

Breadth = 6 cm

∴ Area of one tile = 12 × 6 = 72 cm^{2}

∴ No. of tiles required = \(\frac{\text { Area of floor }}{\text { Area of one tile }}\)

= \(\frac{162 \times 144}{72}\) = 324 tiles

Question 29.

How many dusters, each 16 cm square can be cut from material 84 cm long, 36 cm wide? What areas remains over?

Solution:

Length (l) = 84 cm

Breadth (b) = 36 cm

Area = l × b = 84 × 36 cm^{2}

Side of square duster = 16 cm

∴ Area of one duster = 16 × 16 cm^{2}

∴ No. of total duster = \(\frac{\text { Area of total cloth }}{\text { Area of one duster }}\)

= \(\frac{84 \times 36}{16 \times 16}\) = 11.8125

∴ Complete duster to be cut = 11 and remaining cloth = 84 × 36 = 11 × 256 = (3024 – 2816) cm^{2} = 208 cm

Question 30.

A floor which measure 15 m × 8 m is to be laid with tiles measuring 50 cm × 25 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m eixst between its edges and the edges of the floor, what fraction of the floor is uncovered?

Solution:

Length of floor (l) = 15 = 1500 cm

Breadth (b) = 8 m = 800 cm

∴ Area of floor = l × b

=1500 × 800 cm^{2} = 1200000 cm^{2}

Area of one tile = 50 cm × 25 cm = 1250 cm^{2}

∴ No. of tiles = \(\frac{\text { Area of floor }}{\text { Area of one tile }}\)

= \(\frac { 1200000 }{ 1250 }\) = 960

Space left for carpeting = 1 m

∴ Inner length = 15 – 1 × 2 = 15 – 2 = 13 m

and inner breadth = 8 – 1 × 2 = 8 – 2 = 6 m

Area of the space left uncovered

= outer area – inner area

= 15 × 8 – 13 × 6 = 120 – 78 m^{2} = 42 m^{2}

∴ Fraction of the floor which is uncovered = \(\frac { 42 }{ 120 }\) = \(\frac { 7 }{ 20 }\) part