A Quick Problem!

Geometry Level 3

In Right $\triangle{ABC}, \:\ \overline{AB} = b + 1, \overline{AC} = b$ , and $M$ is a midpoint of $\overline{BC}$ and $C_{2}$ is a semicircle with diameter $\overline{BC}$ and $C_{1}$ is a quarter of a circle with radius $\overline{AC}$ .

Let $A$ be the red shaded region.

Find the value of $b$ for which $A = \dfrac{2\pi + 3}{8}$ .

The answer is 0.5.

1 solution

Rocco Dalto
Apr 2, 2021

$R_{2} = \dfrac{\sqrt{(b + 1)^2 + b^2}}{2}$ and $R_{1} = b$

$\implies$

$A = A_{\triangle{ABC}} + A_{C_{2}} - A_{C_{1}} = \dfrac{b^2 + b}{2} + \dfrac{2b + 1}{8}\pi =$ $\dfrac{4b^2 + 2(2 + \pi)b + \pi}{8} = \dfrac{2\pi + 3}{8}$

$\implies 4b^2 + 2(2 + \pi)b - (\pi + 3) = 0 \implies b = \dfrac{-(2 + \pi) \pm (\pi + 4)}{4}$

dropping the negative root $\implies b = \dfrac{1}{2} = \boxed{0.5}$ .

A Quick Problem!

The answer is 0.5.

1 solution

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