The sides of a triangular field are 55m, 300 m and 300 m. Its area is equal to
15
- 22
What is the perimeter (in m) of the circular park (correct to one decimal place)? (Take pi=22/7)
Answers
Answer:
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 A with sides a, b, c and semi- perimeter s is v[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 = V[327.5* (327.5 - 55) * (327.5 - 300) * (327.5 - 300)] = V[327.5 * 272.5 * 27.5 27.5] = 27.5/(327.5 * 272.5) = 27.5 * 298 = 8195 m?
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,
+ Tr? = 17560.7
+ (22/7) * p2 = 17560.7
r2 = (17560.7 * 7)/ 22
+r = 74.74 m.
Hence,
Perimeter of the circular Park = 2nr = 2
* (22/7) * 74.74 = (3288.56)/7 = 469.8m.
(Ans.)
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