Answer:
Step-by-step explanation:
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.)
Hope this answer helps you :)
Have a great day
Mark brainliest
