Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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). enter image description here

I named different parts of the figure as I, II, III and IV for the sake of convenience. such as in the following figure: enter image description here

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?

share|improve this question
    
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. –  naslundx Feb 24 at 14:09
    
radius AB or AC? –  kool_kartikey Feb 24 at 14:12
    
Sorry, diameter AC. –  naslundx Feb 24 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 $1/4 * \pi * 21^2$ –  kool_kartikey Feb 24 at 14:14

3 Answers 3

up vote 2 down vote accepted

$$ \begin{align} \text{red area} &= \text{area of }\Delta ABC + \text{area of semi-circle} - \text{area of quadrant} \\ &= \tfrac12 \times 21 \times 28 + \tfrac12 \pi \left(\tfrac{35}{2}\right)^2 - \tfrac14 \pi (21)^2 \\ &= 294 + 481.0563 - 346.3606 \\ &= 428.6957 \end{align} $$ So it looks like you are right and your book is wrong.

share|improve this answer

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

enter image description here

share|improve this answer
    
I did it can you please verify the answer. –  kool_kartikey Feb 24 at 14:17

the semi-circle with AC has area= $\pi AC^2/8 = \pi(28^2+21^2)/8=\frac{1225\pi}{8}$

Area of ABC =$\frac{1}{2}AB*BC= 294$

Area of Quadrant=$ \pi BC^2/4=\frac{441\pi}{4}$

reqd. area= $\frac{1225\pi}{8}+294- \frac{441\pi}{4}=294-\frac{784\pi}{8}$ sq. unit

share|improve this answer
1  
Your semi-circle area is wrong, I think. –  bubba Feb 24 at 14:46
    
thanks for notice –  mahavir Feb 24 at 15:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.