Math, asked by kirtidasari437, 2 months ago

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Answered by amitnrw
0

Given : (complete Question is )

The areas allotted for a hockey court and a shooting range of NCC TRAINING CAMP

The shapes of the hockey court and the shooting range are square and triangle respectively. Both of the courts have a common vertex that coincide with the centre of circular camp. The construction of the shooting range is such that the angle to centre is 90°. The radius of the circular camp is 180 metres.  

To Find :

1) What is the area allotted to shooting  range ?

(a) 12, 600 m²  (b) 22, 000 m²  (c) 16,200 m²  (d) 16, 880 m²

2) What is the area allotted to hockey court ?

(a) 12, 600 m²  (b) 22, 000 m²  (c) 20, 000 m²   (d) 16, 200 m²

3) What is the area of training camp.

(a) 101736 m²  (b) 32400 m²  (c) 16200 m²   (d) 20000 m²

4) If the management of NCC TRAINING  Camp likes to allot space for some more  sports, how much area is available to  them?

(a) 85, 536 m² (b) 95, 800 m²  (c) 60, 040 m² (d) 69,336 m²

Solution  :

OA = OB = Radius = 180 m

angle = 90°

Area of shooting  range = (1/2) * 180 * 180 =  16200 m²

OD = Radius = 180 m

OE = 180/√2  = 90√2  m  

Area of Hockey court = (90√2)² = 16200 m²

area of training camp. = π(180)² = 3.14(180)² =    1,01,736 m²

area is available =  1,01,736 -  16200 -  16200  = 69,336 m²

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Answered by Aarkes
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Answer:

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