Chapter 11 Perimeter and Area Ex 11.3 Solutions
Question - 1 : - Find the circumference of the circles with the following radius. (Take π =22/7)
(a) 14 cm
(b) 28 mm
(c) 21 cm
Answer - 1 : -
(a) Given: Radius (r) = 14 cm
∴ Circumference = 2πr = 2 ×(22/7) × 14
= 88 cm
(b) Given: Radius (r) = 28 mm
∴ Circumference = 2πr = 2 × (22/7) × 28
= 176 mm
(c) Given: Radius (r) = 21 cm
∴ Circumference = 2πr = 2 × (22/7) × 21
= 132 cm
Question - 2 : - Find the area of the following circles, given that (Take π =22/7)
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm
Answer - 2 : -
(a) Here, r = 14 mm
∴ Area of the circle = πr2
= π × 14 × 14 = (22/7) × 14 × 14
(b) Here, diameter = 49 m 49
(c) Here, radius = 5 cm
Question - 3 : - If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =22/7)
Answer - 3 : -
Given: Circumference = 154 m
∴ 2πr = 154
Question - 4 : - A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds offence. Also find the cost of the rope, if it costs ₹ 4 per metre. (Take π =22/7)
Answer - 4 : -
Diameter of the circular garden = 21 m
∴ Radius = 21/2 m
∴ Circumference = 2πr = 2×(22/7)×(21/2)
= 66 m
Length of rope needed for 2 rounds
= 2 × 66 m = 132 m
Cost of the rope = ₹4 × 132 = ₹ 528
Question - 5 : - From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Answer - 5 : -
Radius of the circular sheet = 4 cm
∴ Area =πr2 = π × 4 × 4 = 16π cm2
Radius of the circle to be removed = 3 cm
∴ Areaof sheet removed = πr2 = 9π cm2
Area of the remaining sheet
= (16π – 9π) cm2 = 7π cm2
= 7 × 3.14 cm2 = 21.98 cm2
Hence, the required area = 21.98 cm2.
Question - 6 : - Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs ₹ 15. (Take π = 3.14)
Answer - 6 : -
Diameterof the table cover = 1.5 m
∴ Radius= 1.5 =0.75 m
∴ Lengthof the lace = 2πr = 2 × 3.14 × 0.75
= 4.710 m
Cost of the lace = ₹ 15 × 4.710 = ₹ 70.65
Question - 7 : - Find the perimeter of the given figure, which is a semicircle including its diameter.
Answer - 7 : -
Given: Diameter = 10 cm
Hence, the required perimeter
= 25.7 cm. (approx.)
Question - 8 : - Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15 m2. (Take π = 3.14)
Answer - 8 : -
Given:
Diameter = 1.6 m
∴ Radius= 1.6 =0.8 m
Area of the table-top = πr2
= 3.14 × 0.8 × 0.8 m2
= 2.0096 m2
∴ Costof polishing = ₹ 15 × 2.0096
= ₹ 30.14 (approx.)
Question - 9 : - Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π =22/7)
Answer - 9 : -
Lengthof the wire to be bent into a circle = 44 cm
2πr = 44
Now, the length of the wire is bent into a square.
Here perimeter of square
= Circumference of line k
Length of each side of the square
= Perimeter/4=44/4=11cm
Area of the square = (Side)2 = (11)2 =121 cm2
Since, 154 cm2 >121 cm2
Thus, the circle encloses more area.
Question - 10 : - From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take π = 22/7)
Answer - 10 : -
Radiusof the circular sheet = 14 cm
∴ Area =πr2 = (22/7) ×14 × 14 cm2
= 616 cm2
Area of 2 small circles = 2 × πr2
= 2 × (22/7) ×3.5 × 3.5 cm2
= 77.0 cm2
Area of the rectangle = l × b
= 3 × 1 cm2 = 3 cm2
Area of the remaining sheet after removing the 2 circles and 1 rectangle
= 616 cm2 – (77 + 3) cm2
= 616 cm2 – 80 cm2 = 536 cm2