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Question
A rope of length 42 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed
Solution
Let a, b, c be the lengths of the sides of the triangle.
Given the perimeter of the triangle
2s = a + b + c = 42 m ......(1)
For a fixed perimeter 2s, the area of a triangle is maximum
when a = b = c.
(1) ⇒ a + a + a = 42
3a = 42 ⇒ a = `42/3`
a = 14 m
∴ a = b = c = 14 m
The maximum area of the triangle ∆ = `8^2/(3sqrt(3))`
= `21^2/(3sqrt(3))`
= `(21 xx 21)/(3sqrt(3))`
= `(7 xx 3 xx 7)/sqrt(3)`
= `49sqrt(3)` sq.m.
∴ The dimensions of the triangle are 14 m, 14 m, 14 m.
Maximum area = `49sqrt(3)` sq.m.
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