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Question
An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park
Solution
Let ∆ABC be the triangular-shaped park.
a, b, c be the length of the sides.
Given perimeter of the park = 120 m
2s = a + b + c = 120 m ......(1)
For a fixed perimeter 2s.
The area of a triangle is maximum when a = b = c.
(1) = a + a + a = 120
3a = 120
⇒ a = 40 m
Length of the sides 40 m, 40 m, 40 rn.
Maximum area ∆ = `8^2/(3sqrt(3))`
= `60^2/(3sqrt(3))`
= `(60 xx 60)/(3sqrt(3))`
= `(20 xx 20 xx 3)/sqrt(3)`
= `40sqrt(3)` sq.m.
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