# Find the area of shaded region

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?

• Could you find an equation for the circle with radius AB and with radius BC? In that case, integrate the former and subtract the integral of the latter. Feb 24 '14 at 14:09
• radius AB or AC? Feb 24 '14 at 14:12
• Sorry, diameter AC. Feb 24 '14 at 14:13
• i did area of semicircle with radius AC is $1/2 * \pi * 35^2$ and area of quarter with BC as radius is $1/4 * \pi * 21^2$ Feb 24 '14 at 14:14

\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.

$S=S_{ABC}+S_{semicircle}-S_{quatercircle}=28.21/2+(7\sqrt{7})^2\pi/2-21^2\pi/4=6.49+\frac{7}{2}\pi.49-49\pi$

• I did it can you please verify the answer. Feb 24 '14 at 14:17

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.

• Your semi-circle area is wrong, I think. Feb 24 '14 at 14:46