$AB = AC$, $BC = 2$ and $∠BAC = 90$°. Then find the value of $\dfrac{\text{$a$}}{\text{$b$}}$.

Question

In the figure, $AB = AC$, $BC = 2$ and $∠BAC = 90$°. $BD = CD = 1$. $DE$ and $DF$ are the arcs whose centers are at $B$ and $C$ respectively. The area of the shaded region is $x$. If $x = a - b\pi$ where $a$ and $b$ are rational numbers, then find the value of $\dfrac{\text{$a$}}{\text{$b$}}$.

Answer 1

Because $AB = AC$, we know that $\measuredangle B = \measuredangle C$ by base angles theorem and since $\measuredangle A = 90^{\circ}$, $\measuredangle B = \measuredangle C = 45^{\circ}$.

This gives us a $45^{\circ} 45^{\circ} 90^{\circ}$ triangle meaning $AB = AC = \frac{BC}{\sqrt{2}} = \sqrt{2}$

Since $\Delta ABC$ is a right triangle, we can calculate its area as $\frac{AB\cdot AC}{2} =1$.

Since $\measuredangle B = 45^{\circ}$, the area of sector $BED$ is $\frac{45^{\circ}}{360^{\circ}} = \frac{1}{8}$ of the area of a circle with radius $BD$. The area of that circle can be found with $\pi r^2 \rightarrow \pi \cdot 1^2 \rightarrow \pi$. We then take the fraction: $\frac{1}{8} \cdot \pi = \frac{\pi}{8}$ so the area of the sector $BED$ is $\frac{\pi}{8}$.

Same thing for sector $FCD$ so that area is also $\frac{\pi}{8}$

X is just the whole triangle minus those two sectors so $1 - \frac{\pi}{8} - \frac{\pi}{8} = 1-\frac{\pi}{4}$ so $4$

Answer 2

Hint: The shaded area is the difference between the area of the right (isosceles) triangle $ABC$ and the sum of the areas of the circular sectors $BDE$ and $CDF$.