Question

The Iandian hockey federation organised friendly hockey match between India & Pakistan on a circular ground the scale procued of this match shall be donate to the orphanage. A rectangular turf on the ground.

1. Find the radius of the stadium.
2. Which social values is depicted.
3. How does donation to charitable organisation help in the development of society.
please solve in a copy step by step​

Answer 1

Answer:

Step-by-step explanation:

Area of the shaded region = Area of semicircle-Area of segment of the sector BAC

Area of the semicircle with BC as diameter =CBSE Class 10 Maths Chapter 12 Areas related to Circles Question 1 Solution Image19.64cm2

Area of segment = Area of quadrant – Area of ΔABC

= 90/360 × 22/7 × 52– ½ × 5 × 5=19.64-12.5=7.14 cm2….. (ii)

Area of the shaded region = (i) – (ii) = 19.64 – 7.14 = 12.5 cm2

The tangents at A and B are parallel, so the radii through A and B should be parallel to each other since the radius is perpendicular to the tangent at the point of contact. Since the radius passes through the center of the circle, the radii through A and B must lie on the same straight line. The line AB will be the diameter of the circle.

Now A and B are on two ends of the diameter of the circle. The distance travelled along the circle from A to B will be half the circumference of the circle.

Now, Circumference

= 2πr=2×π×7/π=14cm

Distance

= 1/2×circumference=1/2×14cm=7cm

A pendulum swings through on angle of 30∘ and describes an arc 8.8 cm in length. Find the length of pendulum in cm.

16.8

17.3

15.1

14.5

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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)