An equilateral triangle of side \(4\sqrt[4]{3}\) cm formed out of a sheet is converted into a rectangle such that there is no loss of the area of the triangle. Then the least perimeter of the rectangle (in cm) will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(8\sqrt{3}\)
The area of an equilateral triangle of side a is \(\frac{\sqrt{3}}{4}a^2\). Here \(a^2 = 16\sqrt{3}\):
\( \displaystyle \text{Area} = \frac{\sqrt{3}}{4}\times 16\sqrt{3} = 12 \text{ cm}^2 \)
Let the rectangle have length l and breadth b, with lb = 12. Its perimeter is
\( \displaystyle P(l) = 2\left(l + \frac{12}{l}\right) \)
\( \displaystyle P'(l) = 2\left(1 - \frac{12}{l^2}\right) \)
P’(l) = 0 gives \(l^2 = 12\), so \(l = 2\sqrt{3}\). Then
\( \displaystyle P''(l) = \frac{48}{l^3} \gt 0 \)
so this is a minimum. Then \(b = 12/l = 2\sqrt{3}\) too: the rectangle is a square.
\(P = 2\left(2\sqrt{3} + 2\sqrt{3}\right) = 8\sqrt{3}\) cm
Tip: Among rectangles of a fixed area, the square has the least perimeter; so the answer is \(4\sqrt{\text{area}}\) straight away.