A wire when bent in the form of an equilateral triangle encloses an area of 64 -3 cm2- Then- find the area enclosed by the same wire when bent to form a square-

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Standard VIII

Mathematics

Triangle as Part of Rectangle

A wire when b...

Question

A wire when bent in the form of an equilateral triangle encloses an area of $64 \sqrt{3} c m^{2} .$ Then, find the area enclosed by the same wire when bent to form a square.

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Solution

Let the side of the equilateral $Δ l e$ 'a' and the side of the square be 'x' we know, area of equilateral $Δ = \frac{\sqrt{3}}{4} a^{2}$

$64 \sqrt{3} = \frac{\sqrt{3}}{4} a^{2}$

$∴ a^{2} = 8^{2} \times 2^{2}$

$a = 16 c m$

Perimeter of equilateral $Δ = 3 (16)$

= 48 cm

Hence, perimeter of square = 48 cm

i.e., $4 \times = 48 \Rightarrow x = 12 c m$

Area of square $= x^{2}$

$= (12)^{2}$

$= 144$

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A wire when bent in the form of an equilateral triangle encloses an area of 64√3 cm2. Then, find the area enclosed by the same wire when bent to form a square.

Let the side of the equilateral Δle 'a' and the side of the square be 'x' we know, area of equilateral Δ=√34a2 64√3=√34a2 ∴a2=82×22 a=16 cm Perimeter of equilateral Δ=3(16) = 48 cm Hence, perimeter of square = 48 cm i.e., 4×=48⇒x=12 cm Area of square =x2 =(12)2 =144

A wire when bent in the form of an equilateral triangle encloses an area of $64 \sqrt{3} c m^{2} .$ Then, find the area enclosed by the same wire when bent to form a square.