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Question 66
A copper wire is bent in the form of an equilateral triangle of area $$4 \sqrt{3} cm^2$$.If the same wire is bent into the form of a square, the area of the square will be:
Solution
Area of a equilateral triangle=$$\sqrt{3}a^{2}/4$$
$$\sqrt{3}a^{2}/4$$=$$4\sqrt{3}$$
a=4 cm
Total length =3*4
=12 cm
Side of each square=12/4
=3 cm
Area=9 sqcm
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