The sides of triangular field are 55m, 300m and 300m. It's area is equal to (7/15)th area of circular park. What is perimeter (in m) of the circular park (corrected to one deciml place)? (Take pi = 22/7)
Answers
Given :- The sides of triangular field are 55m, 300m and 300m. It's area is equal to (7/15)th area of circular park. What is perimeter (in m) of the circular park (corrected to one deciml place) ? (Take pi = 22/7)
Solution :-
we know that,
- Area of a ∆ with sides a, b , c and semi- perimeter s is √[s * (s - a) * (s - b) * (s - c)] .
- semi - perimeter = (a + b + c)/2 .
given sides of triangular field are 55m, 300m and 300m..
So,
→ s = (55 + 300 + 300) / 2 = 327.5m.
Than,
→ Area of triangular field = √[327.5 * (327.5 - 55) * (327.5 - 300) * (327.5 - 300)] = √[327.5 * 272.5 * 27.5 * 27.5] = 27.5√(327.5 * 272.5) = 27.5 * 298 = 8195 m².
Now given that,
→ Area of triangular field = (7/15)th area of circular park.
So,
→ 8195 = (7/15) * Area of circular park.
→ (8195 * 15)/7 = Area of circular park.
→ Area of circular park = 17560.7 m².
Therefore,
→ πr² = 17560.7
→ (22/7) * r² = 17560.7
→ r² = (17560.7 * 7) / 22
→ r = 74.74 m.
Hence,
→ Perimeter of the circular Park = 2πr = 2 * (22/7) * 74.74 = (3288.56)/7 = 469.8m. (Ans.)
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Answer:
471.4
Step-by-step explanation:
As, the base BASE<<SIDE of a triangle we can assume the height to be approx the side
Therefore, Height = 300
Area of Δ = 300*55/2 = 8250
Area of Δ = (7/15)*(Area of Circle)
8250 = (7/15)* π * R*R
R = 75
Perimeter of a Circle = 2*π * R
Therefore Ans is 471.4