Jawaharlal Nehru Stadium in New Delhi is conducting the annual sports competition soon. The curator of the stadium is tasked with preparing the grounds for various sports as per the technical requirements of sports invigilator. The engineer assigned to assist the curator is tasked with figuring out the dimensions for carving out some areas allotted for a 'HOCKEY COURT' and a 'JAVELIN RANGE', as shown in the figure below. The shapes of the 'hockey court' and the 'javelin range' are square and triangle respectively. Both of the courts have a common edge that touches the Centre of stadium. The construction of the javelin range is such that the angle to Centre is 90°. The radius of the stadium is 200 meters.
On the basis of the above information, answer the following questions.
FIGURE :- CASE BASED QUESTION - 1
(A) The area ( in sq. m) allotted to javelin range' is:
1) 11400
2) 20000
3) 31400
4) 40000
(B) If the team of the curators managing the stadium, likes to allot space for some more sports, how much area (in sq. m) is available to them?
1) 125600
2) 85600
3) 20000
4) 57600
(C) If the boundaries of the hockey court and javelin range are to be fenced, then the total length ( in m) of the fence required is:
1) 624.26
2) 1248.52
3) 907.10
4) 1814.21
(D) Whose area is larger?
1) Hockey court
2) Both are equal
3) Javelin Range
4) None of these
Answers
Given : areas allotted for a 'HOCKEY COURT' and a 'JAVELIN RANGE', as shown in the figure
To Find : (A) The area ( in sq. m) allotted to javelin range' is:
(B) If the team of the curators managing the stadium, likes to allot space for some more sports, how much area (in sq. m) is available to them?
(C) If the boundaries of the hockey court and javelin range are to be fenced, then the total length ( in m) of the fence required is:
(D) Whose area is larger?
Solution:
Two perpendicular sides of triangle are radius
Hence Area of triangle = (1/2) * 200 x 200
= 20000 m²
The area ( in sq. m) allotted to javelin range' is: 20000
Diagonal of Square = Radius
Area of Square = (1/2)* 200² = 20000 m² ( area of square = (1/2) diagnoal²)
Area of Hockey ground = 20000 m²
Both Areas are Equal
Area of circle = πR² = 3.14 x 200² = 125600 m²
Remaining Area = 125600 - 20000 - 20000
= 85600 m²
Hypotenuse of Triangle = 200√2
Side of square = 200/√2
Fence required = 200 + 200 + 200√2 + 4(200/√2)
= 400 + 600√2
= 1248.52 m
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