Question

The shapes of the hockey court and the shooting range are square
and triangle respectively. Both of the courts have a common edge
that touches the centre of stadium. The construction of the shooting
range is such that the angle to centre is 90°. The radius of the
stadium is 180 metres.

i. What is the area allotted to shooting range ?
a) 12,600 m2
b) 22,000 m2
c) 20,000m2
d) 16,880 m2

ii. What is the area allotted to hockey court ?
a) 12,600 m2
b) 20.000 m2
c) 22,000 m2
d) 16,880 m2

iii. If the team of the curators managing the stadium, likes to allot space for some more sports, how much area is available to them?
a) 85,600 m2
b) 95,800m2
c) 60,040 m2
d) 76, 980 m2

iv. If the boundaries of the hockey court and shooting range are to be fenced, then what is the required length of the fence ?
a) 200(2 + 5√3) m
b) 200(2 + 3√2) m
c) 200(2 + 5√2) m
d) 200(2 + 3√3) m

v. If the cost of fencing is Rs 6 per metre, what is the total cost of fencing ?
a) Rs. 2400( 2 + 3√2)
b) Rs. 1200(2 + 5√2)
c) Rs. 1200 (2 + 3√2)
d) Rs. 2400(2 + 3√2)​

Answer 1

Answer:

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Answer 2

Correct Radius = 200 m .

Solution :-

→ Area alloted to shooting range = (1/2) * OA * OB * sin (AOB) = (1/2) * 200 * 200 * sin 90° = 100 * 200 = 20000 m² (Ans.i)

now,

→ Diagonal of hockey court = radius = 200 m .

→ √2 * side of hockey court = 200 m .

→ side of hockey court = (200/√2) m .

then,

→ Area of hockey court = (200/√2)² = 100 * 200 = 20000 m²

(Ans.ii)

now,

→ Area available for other teams = πr² - 2(20000) = 3.14*200*200 - 40000 = 125600 - 40000 = 85600 m². (Ans.iii) .

now,

→ perimeter of hockey court and shooting range = (200 + 200 + 200√2) + 4 * (200/√2) = 400 + 200√2 + 400*√2 = (400 + 600√2) = 200(2 + 3√2) m (Ans.iv)

then,

→ Cost of fencing = 6 * 200(2 + 3√2) = Rs.1200(2 + 3√2) (Ans.v) .

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