M06-32

Question

The figure below shows a square inscribed in a circle with a radius of  \sqrt{6} . What is the area of the shaded region?
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m06-32.png

A.  \pi  
B.  6\pi - \sqrt{12}  
C.  6\pi - 2\sqrt{6}  
D.  3(\frac{\pi}{2} - 1)  
E.  3 \frac{\pi}{4} - \frac{3}{2}

Bunuel · Accepted Answer

Official Solution:

The figure below shows a square inscribed in a circle with a radius of  \sqrt{6} . What is the area of the shaded region?
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m06-32.png

A.  \pi  
B.  6\pi - \sqrt{12}  
C.  6\pi - 2\sqrt{6}  
D.  3(\frac{\pi}{2} - 1)  
E.  3 \frac{\pi}{4} - \frac{3}{2}

The area of the shaded region is equal to one-fourth of the difference between the area of the circle and the area of the square.
 
The area of the circle is given by  \pi{r^2}=6\pi .
 
Given that the diagonal of the square is equal to the diameter of the circle, the diagonal is  2\sqrt{6} . The area of the square can be expressed as  \frac{	ext{diagonal}^2}{2}=12 .
 
Thus, the area of the shaded region is  \frac{1}{4}(6\pi - 12) = 3(\frac{\pi}{2} - 1) .

Answer: D

aqeelnaqvi10 · Answer

I would like to know if we can use the 45-45-90 triangle properties here

Prostar · Answer

How do we get the area of the square to equal :  \frac{	ext{diagonal}^2}{2}=12

Bunuel · Answer

You should brush up fundamentals: the area of a square is given by  side^2  as well as  \frac{	ext{diagonal}^2}{2}

susheelh · Answer

HI  aqeelnaqvi10 ,

I think you can use 45-45-90 triangle here. In fact this formula -  Area = \frac{(diagonal)^2}{2}  is got after using the phythagoras theorem.

Consider a square of side 'a'. The area of the square =  a^2 .

In a 45-45-90 triangle, formed by the diagonal and the sides of the square we get -

a^2+a^2 = (diagonal)^2 
 2a^2 = (diagonal)^2 
 a^2 = \frac{(diagonal)^2}{2} 
Area of square  = a^2 = \frac{(diagonal)^2}{2}

Bunuel has done this directly. I think this formula is a definite time saver!

RigelKent · Answer

what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.

Bunuel · Answer

Factor out 12 from  (6\pi - 12) :

\frac{1}{4}(6\pi - 12)=\frac{1}{4}*12(\frac{\pi}{2} - 1)=3(\frac{\pi}{2} - 1) .

Hope it's clear.

sankarsh · Answer

A different way of solving-

hiranmay · Answer

The figure below shows a square inscribed in a circle with a radius of  \sqrt{6} . What is the area of the shaded region?
 
 https://gmatclub.com/tests-beta/resources/import/img/m06-32.png

A.  \pi  
B.  6\pi - \sqrt{12}  
C.  6\pi - 2\sqrt{6}  
D.  3(\frac{\pi}{2} - 1)  -->   correct : let's say radius of the circle =r =  \sqrt{6}  & side of the square = a, so  a\sqrt{2} = 2r  =>   a = r\sqrt{2} , so the area of the shaded region =  \frac{1}{4}(\pi r^2 - a^2) = \frac{1}{4}(\pi r^2 - (r\sqrt{2})^2) = \frac{1}{4} r^2(\pi- 2) = \frac{1}{4} (\sqrt{6})^2(\pi- 2) = \frac{3}{2} (\pi- 2) = 3(\frac{\pi}{2} - 1)    
E.  3 \frac{\pi}{4} - \frac{3}{2}

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

M06-32

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