**S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16A**

__Exercise – 16A__

__Exercise – 16A__**(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 m by 2 m**

** Solution:**

** (i) **Perimeter = 2 (8 + 5) cm

= 26 cm

Area = (8 x 5) cm^{2}

= 40 cm^{2}

(ii) Perimeter = 2 (12 + 9)

= 42 m

Area = (12 x 9) m^{2}

= 108 m^{2}

(iii) Perimeter = 2 (3.5 + 2) m

= 11 m

Area = (3.5 x 2) m^{2}

= 7 m^{2}

**(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) Perimeter = ( 4 x 3) = 12 cm

Area = 3^{2} = 9 cm^{2}

(ii) Perimeter = (4 x 5) = 20 cm

Area = 5^{2} = 25 cm^{2}

(iii) Perimeter = (4 x 1.3) m

= 5.2 m

Area = (1.3)^{2} m^{2}

= 1.69 m^{2}

(iv) Perimeter = (4 x 2.4 m)

= 9.6 m

Area = (2.4)^{2} m^{2}

= 5.76 m^{2}

**(3) Find the length of one side and perimeter of a square whose area is :**

**(i) 25 m ^{2}**

**(ii) 144 m ^{2}**

**(iii) 256 cm ^{2}**

**(iv) 961 dm ^{2}**

**Solution: **

(i) Let, the length of one side = a

∴ a^{2} = 25

a = 5,

∴ Perimeter = (4 x 5)

= 20 m

(ii) a^{2} = 144

a = 12,

Perimeter = (4 x 12)

= 48 m.

(iii) a^{2} = 256

a = 16

Perimeter = 4 x 16 = 64 cm.

(iv) a^{2} = 961

a = 19,

Perimeter = 4 x 19 = 126 dm

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

**Solution:** Let, length of the rectangle is a,

2 (a + 6) = 30

Or, a + 6 = 15

Or, a = 9

∴ length = 9 cm

Let, the breadth is = b

∴ 30 x b = 450

b = 15 m

**(6)** **The perimeter of a rectangular field is 3/5 km. If the length of the field is twice its width , fins its area in sq metres.**

**Solution:** Let, the width is = b km

∴ length = 2b km

2 (2b + b) = 3/5

Or, 2 x 3b = 3/5

Or, b = 1/10

∴ length = 2/10 = 1/5 km.

∴ Area = 1/10 x 1/5 km^{2}

= (100 x 200) m^{2}

= 20000 m^{2}

**(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 the side of the square is = a

∴ No, the side is = 2a

∴

**(8) The side of a square field is 89 m. By how many square metres does its area fall short of a hectre.**

**Solution:** Area of the square field is = (89)^{2} m^{2}

= 7921 m^{2}

1 hectre = 10,000 m^{2}

∴ Area short of (10,000 – 7921) m^{2}

= 2079 m^{2}

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

**Solution: **

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

**Solution:** By the phythogorus theorem,

(10)^{2} = (6)^{2} + Length^{2}

∴ (length)^{2} = 100 – 36 = 64

Length = 8

∴ Area = (8 x 6) cm^{2}

= 48 cm^{2}

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

**Solution:** Let the width is b cm

∴ Length = 2b cm

∴ (2b)^{2} + b^{2} = (16√5)^{2}

5b^{2} = 256 x 5

b^{2 }= 256

b = 16

∴ Area = (16 x 32) cm^{2}

= 512 cm^{2}

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

**Solution: **

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

**Solution:** Area of the floor = (30 x 12) m^{2}

= 360 m^{2}

Area of the carpet = (3 x 2) m^{2}

= 6 m^{2}

∴ Number of carpet = 360 / 6 = 60

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

**Solution:** Area of the room = (5 x 4) m^{2}

= 20 m^{2}

Cost of cementing = (20 x 7.50) Rs.

= 150 Rs.

**(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:** Area of the floor = (12 x 10) m^{2}

= 120 m^{2}

Area of one bricks = (20 x 6) cm^{2}

= 120 m^{2}

∴ Total number of bricks required = 12 x 10000 / 120

= 10000

∴ Total bricks required = 10000,

∴ Cost of 1000 bricks = Rs. 300

∴ Cost of 10,000 bricks = Rs. 3000

(**16) In exchange for a square plot of land, one of whose sides is 84 m, a man want 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: **

**(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:** Area of to be laid with gram = (110 – 10) (80 – 10) m^{2}

= 100 x 70 m^{2}

= 7000 m^{2}

**(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. 1.25per metre, given that the area of the field is 27000 m ^{2}.**

**Solution:** let the common ratio be x

∴ 6x x 5x = 27000

Or, x^{2} = 27000 / 6 x 5

x= 30

∴ Perimeter = 2 (180 + 150)

= 2 x 330

= 660 m

Cost = (660 x 1.25) Rs.

= 825 Rs.

**(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: **

**(20) Find the area and perimeter of a square plot of land, the length of whose diagonal is14 metres. Give your answer correct to 2 places of decimals.**

**Solution:**** Let, the length be ‘a’ meter**

∴ a^{2} + a^{2} = 196

a^{2} = 98

∴ Area = 98 m^{2}

Perimeter = 4√98 m

= 39.60 m

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

**Solution:** Area of the lawn = (10 x 8) m^{2}

= 80 m^{2}

Total area of lawn & Path is = 12 x 10 = 120 m^{2}

∴ Path = (120 – 80) m^{2}

= 40 m^{2}

**(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: **Area of the shaded portion = Area of the rectangle – (Area of the portion without path)

= (50 x 35) – (50 – 5) (35 – 5)

= 50 x 35 – 45 x 30

= (1750 – 1350)

= 400 m^{2}

**(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:** Area of the four wall = 2 (7 x 3 + 5 x 3) m^{2}

= 2 (21 + 15)

= 2 x 36 m^{2}

= 72 m^{2}

Cost of distempering = (72 x3.15) Rs.

= 226.8 Rs.

**(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:** Area of each paper = (40 x 60) cm^{2}

= 2400 cm^{2}

Area of walls = 2(6 x 3 + 4 x 3) m^{2}

= 2 (18 + 12) m^{2}

= 2(30) m^{2}

= 60 m^{2}

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

**Solution: **

**(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:** Area of one tiles = (12 x 6) cm^{2}

= 72 cm^{2}

Area of room = (162 x 144) cm^{2}

∴ Number of tiles = 162 x 144 / 72

= 324 tiles.

**(29) How many dusters, each 16 cm square can be cut from material 84 cm long, 36 cm wide? What area remains over?**

** ****Solution: **

Number of dusters = 11

Area remain over = 208 cm^{2}

**(30) A floor which measures 15 m x 8 m is to be laid with tiles measuring 50 cm x 25 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges of the floor, what fraction of the floor is uncovered.**

**Solution: **

Thank You netexplations for giving answers. But can you also provide me the important questions for this chapter for my exams. Thank you……