Page 177 - Ganit Kaushal

Solved Examples

Ex 1. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet are needed to cover the
floor of the room?
Sol. Length of the room (L) = 4 m; Breadth of the room (B) = 3 m 50 cm = 3.50 m
Area of the floor of the room: A = L × B = 4 × 3.50 = 14 m 2
2
Hence, 14 m of carpet is needed to cover the floor of the room.
Ex 2. A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the floor. Find the area of the
floor that is not carpeted. (NCERT)
2
Sol. Area of the floor = L × B = 5 × 4 = 20 m ; Side of the square carpet = 3 m
2
2
Area of the square carpet ‘A’ = (Side) = (3 m) = 9 m 2
2
Hence, area of the floor which is not carpeted = 20 – 9 = 11 m .
Ex 3. The perimeter of a rectangle is 40 cm and its breadth is 6 cm. Find its length.
1 1
Sol. Length, L = P – B = × 40 – 6 = 20 – 6 = 14 cm.
2 2 (15 – 12) = 3 ft
Ex 4. Below is the house plan of Chaman. It is in a 15 ft Utility (___ ft × ___ ft)
rectangular plot. Look at the plan. Some of the Area = ___
measurements are given and some are missing. Master Bedroom Toilet (5 ft × 10 ft)
(15 ft × 15 ft) Kitchen
(a) Find the missing measurements. 15 ft Area = 225 sq ft 12 ft Area = 180 sq ft
(15 ft × 12 ft)
(b) Find out the area of his house. 5 ft
(c) What is the perimeter of Chaman’s house? 15 ft
(NCERT) 30 ft
Sol. (a) Utility: (15 ft × 3 ft); Area = 45 sq ft Small Bedroom
Hall
Small Bedroom: (15 ft × 12 ft); Area = 180 sq ft 12 ft (15 ft × _____ ft) Area = _____
Area = 180 sq ft
Hall: Area = (20 ft × 12 ft) + (5 ft × 5 ft) = 265 sq ft
Garden: (20 ft × 3 ft); Area = 60 sq ft. Garden (___ ft × ___ ft) Parking (_____ ft × _____ ft)
Parking: (15 ft ×3 ft); Area = 45 sq ft. 3 ft Area = ___ Area = _____
(b) Area of the house = (225 + 50 + 45 +180 + 180 + 265 + 60 + 45) sq ft.
= 1050 sq ft (or 30 ft × 35 ft = 1050 sq ft.)
(c) Perimeter of Chaman’s house = 2 × (30 ft + 35ft) = 130 ft.
Ex 5. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden? (NCERT)
Sol. Width/Breadth of the garden, B = (A ÷ L) = 300 ÷ 25 = 12 m [ Given: A = 300 sq m, L = 25 m]
Ex 6. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of `8 per
hundred sq m? (NCERT)
Sol. Area of the plot = 500 m × 200 m = 1,00,000 sq m.
Cost of tiling of 100 sq m = `8 (Given)
\ Cost of tiling of 1 sq m = `(8 ÷ 100)
Hence, the cost of tiling 1,00,000 sq m of land = `(8 ÷ 100) × 1,00,000 = `8000.
Ex 7. Give the dimensions of a rectangle whose area is the sum of the areas of two rectangles having
measurements: 5 m × 10 m and 2 m × 7 m. (NCERT)
Sol. Sum of the areas of rectangles = 5 m × 10 m + 2 m × 7 m = 64 sq m.
Possible dimensions of rectangle whose area is 64 sq m are 1 m × 64 m, 2 m × 32 m, 4 m × 16 m,
8 m × 8 m

Perimeter and Area 175