Question

. The Indian Hockey Federation organized a friendly hockey match between
India and Pakistan on a circular ground. The sale proceeds of this match
shall be denoted to an orphanage. A rectangular turf is spread on the
ground as shown in the figure.
1. the radius of the staduim is?

2. find the area of rectangular field

3. fine the perimeter of circular ground ( take circumference as 22/7)

4. find the area pf circular ground.

5. find the area pf shaded portion ​

Answer 1

Answer:

1. 80*60

I don't know about the others lol

Answer 2

Given :-

• Length of Rectangular turf = 80 m.
• Breadth = 60 m .

To Find :-

1. the radius of the staduim is?

2. find the area of rectangular field

3. fine the perimeter of circular ground ( take circumference as 22/7)

4. find the area of circular ground.

5. find the area of shaded portion

Solution :-

from image we can see that,

→ ABCD = Rectangle .

→ AB = DC = 60 m .

→ CB = 80 m.

→ ∠DCB = 90° . (Each angle of a rectangle is equal to 90°.)

then, in Right ∆DCB, we have,

→ DC² + CB² = BD²

→ 60² + 80² = BD²

→ BD² = 3600 + 6400

→ BD² = 10000

→ BD = √(10000)

→ BD = 100 m

so,

→ Radius of the stadium = BD/2 = 100/2 = 50 m (Ans.1)

and,

→ Area of rectangular field = Length * Breadth = 60 * 80 = 4800 m². (Ans.2)

also,

→ The perimeter of circular ground = 2πr = 2 * (22/7) * 50 = (2200/7) = 314.28 m . (Ans.3)

and,

→ The area of circular ground = π(r)² = (22/7) * 50 * 50 = (55000/7) = 7857.14m² (Ans.4)

therefore,

→ The area of shaded region = The area of circular ground - Area of rectangular field = 7857.14 - 4800 = 3057.14m² (Ans.5)