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Tamil Nadu Board of Secondary EducationHSC Science Class 11

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

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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

Sum
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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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Properties of Triangle
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Chapter 3: Trigonometry - Exercise 3.9 [Page 143]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 3 Trigonometry
Exercise 3.9 | Q 10 | Page 143
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