Practice Set 15.6 8th Std Maths Answers Chapter 15 Area
Solution & Step-by-Step Answer:
i. Radius of the circle (r) = 28 cm … [Given] Area of the circle = πr² = x (28)² = x 28 x 28 = 22 x 4 x 28 = 2464 sq. cm
ii. Radius of the circle (r) = 10.5 cm … [Given]
Area of the circle = πr²
= x (10.5)²
= x 10.5 x 10.5
= 22 x 1.5 x 10.5
= 346.5 sq. cm
iii. Radius of the circle (r) = 17.5 cm … [Given]
Area of the circle = πr²
= x(17.5)²
= x 17.5 x 17.5
= 22 x 2.5 x 17.5
= 962.5 sq. cm
Solution & Step-by-Step Answer:
i. Area of the circle =176 sq. cm...[Given] Area of the circle = πr² ∴ 176 = x r² ∴ r² = 176 x ∴ r² = 56 ∴ r = √56 … [Taking square root of both sides] Diameter = 2r = 2√56 CM
ii. Area of the circle = 394.24 sq. cm … [Given]
Area of the circle = πr²
∴ Diameter = 2r = 2 x 11.2 = 22.4 cm
iii. Area of the circle = 12474 sq. cm …[Given]
Area of the circle = πr²
∴ 12474 = x r²
∴ r² = 12474 x
∴ r² = 567 x 7
∴ r² = 3969
∴ r = 63 …[Taking square root of both sides]
∴ Diameter = 2r = 2 x 63 = 126cm
Solution & Step-by-Step Answer:
Diameter of the circular garden is 42 m. … [Given] ∴ Radius of the circular garden (r) = = 21 m Width of the road = 3.5 m …[Given] Radius of the outer circle (R) = radius (r) + width of the road = 21 + 3.5 = 24.5 m Area of the road = area of outer circle – area of circular garden = πR² – πr² = π (R² – r²) = [(24.5)² – (21)²] = (24.5 + 21) (24.5 – 21) …..[∵ a²-b² = (a+b)(a-b)] = x 45.5 x 3.5 = 22 x 45.5 x 0.5 = 500.50 sq. m ∴ The area of the road is 500.50 sq. m.
Solution & Step-by-Step Answer:
Circumference of the circle = 88 cm …[Given] Circumference of the circle = 2πr ∴ 88 = 2 x x r ∴ ∴ r = 14cm Area of the circle = πr² = x (14)² = x 14 x 14 = 22 x 2 x 14 = 616 sq. cm ∴ The area of circle is 616 Sq cm
Maharashtra Board Class 8 Maths Chapter 15 Area Practice Set 15.6 Intext Questions and Activities
Solution & Step-by-Step Answer:
(Students should do this activity on their own.)