In the figure given below, $\Delta \ ABC$ is right $\Delta $, $\angle B = 90 ^{\circ}$, $AB = 28 \ cm$ and $BC = 21 \ cm$. With $AC$ as diameter, a semicircle(though it may seem like one but it is) is drawn and with $BC$ as radius, a quarter circle is drawn. Find the area of shaded region(in red colour).
I named different parts of the figure as I, II, III and IV for the sake of convenience. such as in the following figure:
So now I did the following
ar(semicircle) = II + III, ar(triangle) = I + IV, ar(quadrant) = I + II
We need to find the area of III + IV
III + IV = (II + III) + (I + IV) - (I + II)
Thus ar(shaded region) = ar(semicircle) + ar(traingle) - ar(quadrant), which in my case turns out to be $428.75 cm^2$ while the answer to the question given in the book is $688.625 cm^2$.
Is my approach to the question wrong or is the answer given in the book wrong?
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$\begingroup$$$ \begin{align} \text{red area} &= \text{area of }\Delta ABC + \text{area of semi-circle} - \text{area of quadrant} \\ &= \tfrac12 \times 21 \times 28 + \tfrac12 \pi \left(\tfrac{35}{2}\right)^2 - \tfrac14 \pi (21)^2 \\ &= 294 + 481.0563 - 346.3606 \\ &= 428.6957 \end{align} $$ So it looks like you are right and your book is wrong.
$\endgroup$ $\begingroup$$S=S_{ABC}+S_{semicircle}-S_{quatercircle}=28.21/2+(7\sqrt{7})^2\pi/2-21^2\pi/4=6.49+\frac{7}{2}\pi.49-49\pi$
The semi-circle with $AC$ has area$= \frac{\pi}{8} AC^2 = \frac{\pi(28^2+21^2)}{8}=\frac{1225\pi}{8}$
Area of $\triangle ABC =\frac{1}{2}AB\times BC= 294$
Area of Quadrant$= \frac{\pi}{4} BC^2=\frac{441\pi}{4}$
Required area$= \frac{1225\pi}{8}+294- \frac{441\pi}{4}=294-\frac{784\pi}{8}$ sq. unit.
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