Question

The area of a square is $64x^2$. An equilateral triangle is constructed on one of its sides. What is the altitude of this triangle?

Answer 1

We know that the area of square with side $a$ is $A=a^2$.

Here, it is given that the area of the square is $64x^2$, therefore,

$64x^2=a^2\Rightarrow a=\sqrt{64x^2}=8x$

Therefore, the side of the square is $8x$.

It is given that an equilateral triangle is constructed on one of its sides, therefore, the base of the triangle is $8x$.

Let the altitude of the equilateral triangle be $h$.

Now the area of equilateral triangle is:

$\frac{1}{2}\times 8x\times h=\frac{\sqrt{3}}{4}(8x{)}^2\Rightarrow 4xh=\frac{\sqrt{3}}{4}\times 64x^2\Rightarrow 4xh=16\sqrt{3}{x}^2\Rightarrow h=\frac{16\sqrt{3}{x}^2}{4x}=4\sqrt{3}x$

Hence, the altitude of the triangle is $4\sqrt{3}x$.