ABCD is a square. If the shaded area is A, the side of the s

Say the length of the side of the square is x, then:

The area of the square is x^2;
The diameter of the circle is also x and hence the area of the square is \(\pi{(\frac{x}{2})^2}=\pi{\frac{x^2}{4}}\).

Now, the area of the shaded region is a quorter of the difference between the areas of the square and the circle: \(A=\frac{1}{4}(x^2-\pi{\frac{x^2}{4}})\) --> \(x^2=\frac{16A}{4-\pi}\) --> \(x=\sqrt{\frac{16A}{4-\pi}}\).

Answer: D.

GEOMETRY: Shaded Region Problems!

Hope it helps.
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Refer diagram below:



Consider the red marked small square (Its 1/4 th of the big square)

Let the side = x

Area of square \(x^2 = a + \frac{\pi x^2}{4}\)

\(x = \sqrt{\frac{4a}{4-\pi}}\)

\(2x = 2 * \sqrt{\frac{4a}{4-\pi}} = \sqrt{\frac{16a}{4-\pi}}\)

Answer = D

tough one.
Area of the shaded region is 1/4 area of square - 1/4 area of circle
suppose that length of the square is x, tthat means that radius of circle is 0.5x. Area of square is x^2 while area of circle is 0.25x^2*pi
now we can set the formula:
A=(1/4)x^2 - (1/4)*0.25x^2*pi - multiply everything by 16 to get rid of fractions
16A=4x^2 - x^2*pi
factor x^2
16A=x^2(4-pi)
x^2 = 16A/(4-pi)
x=sqr[16A/(4-pi)]
answer choice D.

hi debbiem,
refer your message..

In these Qs, the first thing is you have to move towards finding the area A..
what is this A? - It is the 1/4th of the area of square - 1/4th area of the circle..
( remember this always when ever you see such a figure and you can see this concept in different forms)
let side of square = b, so radius = b/2..
your Q is solved..
\(\frac{1}{4} *( b^2 - pi*\frac{b}{2}^2)=A\)..
\(A= \frac{1}{4} *( b^2 - pi*\frac{b}{2}^2)= \frac{1}{4} *b^2( \frac{4 - pi}{4})\)..
=\(b^2 = A*\frac{16}{4-pi}\)..
\(b= \sqrt{\frac{16A}{4-pi}}\)
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The moment I see a variable in the options, my instinct is to assume a value.
For ease, I say the side of the square is 2 which means the diameter of the circle is 2. So the radius of the circle is 1.
A quarter of the square will have area \(1^2 = 1\) and a quarter of the circle will have area \(\frac{\pi*r^2}{4} = \frac{\pi}{4}\).
\(A = 1 - \frac{\pi}{4} = \frac{(4 - \pi)}{4}\)

Now look for the option in which if you put this value of A, you will get 2. It might seem complicated but if you observe properly, you will get the answer in a moment. You should get 2 so the pi should get cancelled. You see there are two options, (A) and (D), in which the term \((4 - \pi)\) is in the denominator and A is in the numerator. You see that when you put this value of A in (D), you get 2.

Answer (D)
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