A composite geometrical shape is made up of a square, equilateral and right triangles. Calculate the area of hatched triangle?

Apr 26, 2016

$156.25$

Explanation:

As $\triangle A B D$ is a right triangle, we have, using the Pythagorean theorem, $\overline{B D} = \sqrt{{25}^{2} + {25}^{2}} = 25 \sqrt{2}$. Because $\triangle B D E$ is equilateral, this implies that $\overline{E D} = 25 \sqrt{2}$.

Next, we can note that $\triangle A B D$ is a $45 - 45 - 90$ right triangle, meaning the angle $\angle A D B = {45}^{\circ}$. Additionally, as $\triangle B D E$ is equilateral, we have $\angle B D E = {60}^{\circ}$. Using those facts, along with $\angle A D F = {180}^{\circ}$, we have:

$\angle E D F = \angle A D F - \angle B D E - \angle A B D$

$= {180}^{\circ} - {60}^{\circ} - {45}^{\circ}$

$= {75}^{\circ}$

As $\triangle E D F$ is a right triangle, we can use the sine and cosine functions to say

$\frac{\overline{E F}}{\overline{E D}} = \sin \left({75}^{\circ}\right)$

$\implies \overline{E F} = \overline{E D} \sin \left({75}^{\circ}\right) = 25 \sqrt{2} \sin \left({75}^{\circ}\right)$

and

$\frac{\overline{D F}}{\overline{E D}} = \cos \left({75}^{\circ}\right)$

$\implies \overline{D F} = \overline{E D} \cos \left({75}^{\circ}\right) = 25 \sqrt{2} \cos \left({75}^{\circ}\right)$

As $\overline{D F}$ and $\overline{E F}$ are the base and height, respectively, of the hatched triangle, we can use the formula $A = \frac{1}{2} b h$ to obtain the area of the hatched triangle as

$A = \frac{1}{2} \overline{D F} \cdot \overline{E F}$

$= \frac{1}{2} \left(25 \sqrt{2} \cos \left({75}^{\circ}\right)\right) \left(25 \sqrt{2} \sin \left({75}^{\circ}\right)\right)$

$= 625 \cos \left({75}^{\circ}\right) \sin \left({75}^{\circ}\right)$

$= 156.25$