Area of a region bound by three circular arcs, why doesn't this approach work? Three circular arcs of radius $5$ units bound the region shown. Arcs AB and AD are quarter-circles, and arc BCD is a semicircle.
I tried to find the answer by calculating the total area of the circle if it was a full one,and then I subtracted the area of the two quarter circles:
$$
\pi 5^{2} - 1/4\pi 5^{2} - 1/4 \pi r 5^{2} = 39.25
$$
but the answer should be $50$, why doesn't this approach work
${\large ?}$.
To answer the question as to why it doesn't work: draw the entire circle. Your cuts are not actually removing the entirety of two quarter circles, but only a section thereof.
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In order to evaluate the area $\ds{\mathcal{A}}$ of the $\ds{\color{blue}{\mbox{blue region}}}$, the below evaluation subtracts $\ds{\color{red}{twice}}$ the $\ds{\color{maroon}{maroon}}$ area from the total area $\ds{25\pi}$ of the &quot;big circle&quot;. Namely,
\begin{align}
\mathcal{A} &amp; \equiv
\bbox[5px,#ffd]{25\pi - \color{red}{2}\,\,\,
\overbrace{\iint_{\large\pars{0,5}^{2}}
\bracks{x^{2} + y^{2} &lt; 25}\bracks{\pars{x - 5}^{2} + \pars{y - 5}^{2} &lt; 25}
\dd x\,\dd y}^{\ds{\color{maroon}{Maroon}\ \mbox{Area Value}}}}
\\[5mm] &amp; =
25\pi
\\[2mm] &amp; - 2\iint_{\large\pars{0,5}^{2}}
\bracks{0 &lt; r &lt; 5}
\bracks{r^2 - 10r\cos\pars{\phi} -10r\sin\pars{\phi} + 25 &lt; 0}
r\,\dd r\,\dd\phi
\\[5mm] &amp; =
25\pi
\\[2mm] &amp; - 2\iint_{\large\pars{0,5}^{2}}\bracks{0 &lt; r &lt; 5}
\bracks{r^{2} - 10\root{2}r\cos\pars{\phi - {\pi \over 4}} + 25 &lt; 0}
r\,\dd r\,\dd\phi
\\[5mm] &amp; =
25\pi
\\[2mm] &amp; - 2\iint_{\large\pars{0,5}^{2}}\bracks{0 &lt; r &lt; 5}
\bracks{r &gt; 5\root{2}\cos\pars{\phi - {\pi \over 4}} -
5\root{\sin\pars{2\phi}}}\ \times
\\[2mm] &amp; \phantom{-2\iint_{\large\pars{0,5}^{2}}\,\,\,}
\bracks{r &lt; 5\root{2}\cos\pars{\phi - {\pi \over 4}} +
5\root{\sin\pars{2\phi}}}r\,\dd r\,\dd\phi
\end{align}

$$
\mbox{However,}\quad 5\root{2}\cos\pars{\phi - {\pi \over 4}}
\color{red}{\large +}
5\root{\sin\pars{2\phi}} \color{red}{&gt; 5}\quad
\mbox{when}\quad \phi \in \pars{0,{\pi \over 2}}
$$

Then,
\begin{align}
\mathcal{A} &amp; \equiv
\bbox[5px,#ffd]{25\pi -\color{red}{2}\iint_{\large\pars{0,5}^{2}}
\bracks{x^{2} + y^{2} &lt; 25}\bracks{\pars{x - 5}^{2} + y^{2} &lt; 25}
\dd x\,\dd y}
\\[5mm] &amp; =
25\pi - 2\iint_{\large\pars{0,5}^{2}}
\bracks{ 5\root{2}\cos\pars{\phi - {\pi \over 4}} -
5\root{\sin\pars{2\phi}} &lt; r &lt; 5}r\,\dd r\,\dd\phi
\\[5mm] &amp;  =
25\pi - 4\int_{0}^{\pi/4}
\int_{5\root{2}\cos\pars{\phi} - 5\root{\cos\pars{2\phi}}}^{5}
r\,\dd r\,\dd\phi
\\[5mm] &amp;  =
25\pi - 2\int_{0}^{\pi/4}
\braces{25 - \bracks{5\root{2}\cos\pars{\phi} - 5\root{\cos\pars{2\phi}}}^{2}}\,\dd\phi
\\[5mm] &amp;  =
{25\pi \over 2} + 50\ \underbrace{\int_{0}^{\pi/4}
\bracks{\root{2}\cos\pars{\phi} - \root{\cos\pars{2\phi}}}^{2}\,\dd\phi}
_{\ds{1 - {\pi \over 4}}}\ =\
\bbx{\large 50} \\ &amp;
\end{align}