3420.61 – Two Triangles and a Square

In the diagram below two triangles and a square are drawn, but not to scale. The triangles $I$ and $III$ are equilateral of area $32\sqrt{3}$ and $8\sqrt{3}$ square inches, respectively. The square $II$ has area $32$ square inches.

If the segment $AD$ is decreased by $12.5\%$ while lengths $AB$ and $CD$ are unchanged, what is the percent change in the area of the square?

Solution

The area of an equilateral triangle of side $s$ is $\dfrac{s^2\sqrt{3}}{4}$ . Referring to the lengths in the diagram above, this implies that

$\begin{aligned}\text{For I: }&& \dfrac{x^2\sqrt{3}}{4} &= 32\sqrt{3} \\[1em]&& \dfrac{x^2}{4} &= 32 \\[1em]&& x^2 &= 128 \\[0.5em]&& x &= 8\sqrt{2}\\ \\\text{For II: }&& y^2 &= 32 \\[0.5em]&& y =& 4\sqrt{2} \\ \\\text{For III: }&& \dfrac{z^2\sqrt{3}}{4} &= 8\sqrt{3} \\[1em]&& z^2 &= 32 \\[0.5em]&& z &= 4\sqrt{2}\end{aligned}$

So $AD = 8\sqrt{2} + 4\sqrt{2} + 4\sqrt{2} = 16\sqrt{2}$ . If $AD$ decreases by $12.5\%$ , that is by one-eighth, then it becomes $14\sqrt{2}$ . Lengths $AB$ and $BD$ are unchanged so $BC$ absorbs the whole decrease of $2\sqrt{2}$ and shrinks from $4\sqrt{2}$ to $2\sqrt{2}$ .

The area of the square is now $\left( 2\sqrt{2} \right)^2 = 8$ , down from $32$ , a loss of $75\%$ .