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Bunuel
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Bunuel
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?



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 1/4th of the area of the circle minus the area of the square.

The area of the circle is \(\pi{r^2}=6\pi\);

Since the diagonal of the square equals to the diameter of the circle then \(\text{diagonal}=2\sqrt{6}\). The area of a square equals \(\frac{\text{diagonal}^2}{2}=12\);

The area of the shaded region equals \(\frac{1}{4}(6\pi - 12)=3(\frac{\pi}{2} - 1)\).


Answer: D

How do we get the area of the square to equal : \(\frac{\text{diagonal}^2}{2}=12\)
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Bunuel
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?



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 1/4th of the area of the circle minus the area of the square.

The area of the circle is \(\pi{r^2}=6\pi\);

Since the diagonal of the square equals to the diameter of the circle then \(\text{diagonal}=2\sqrt{6}\). The area of a square equals \(\frac{\text{diagonal}^2}{2}=12\);

The area of the shaded region equals \(\frac{1}{4}(6\pi - 12)=3(\frac{\pi}{2} - 1)\).


Answer: D

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

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


aqeelnaqvi10
I would like to know if we can use the 45-45-90 triangle properties here
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what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.
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RigelKent
what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.

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.
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A different way of solving-
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The figure below shows a square inscribed in a circle with a radius of \(\sqrt{6}\). What is the area of the shaded region?




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. \(3 \frac{\pi}{4} - \frac{3}{2}\)
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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