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M06-32

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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?

Image


A. \(\pi\)
B. \(6\pi - \sqrt{12}\)
C. \(6\pi - 2\sqrt{6}\)
D. \(3(\frac{\pi}{2} - 1)\)
E. \(3 \frac{\pi}{4} - \frac{3}{2}\)
[Reveal] Spoiler: OA

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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?

Image


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
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M06-32 [#permalink]

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New post 26 Apr 2016, 22:38
Bunuel wrote:
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?

Image


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


Jumping steps there got me confused.

So we get the square with diagonal \(\sqrt{6} * \sqrt{2} = \sqrt{12}\).
The area of the square is \(\sqrt{12} * \sqrt{12} = 12\)

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New post 08 Nov 2016, 10:22
maipenrai

The area of the shaded region = 1/4 (there are part equal divided parts around the circle) of the area of the circle - area of the square. Must subtract the area of the square in order to find the area of the piece we are looking for.

Thus,
Quote:
14(6π−12)=3(π2−1)14(6π−12)=3(π2−1) .


Perhaps, for better understanding, let's see a more cumbersome but detailed approach.

Since it's square, we can safely divide the central point of the circle into 4 equal sections. 360/4 = 90. Since we found out that the area of the circle must be 6π, (90/360 = x/6π) cross multiplied x = 3π/2. Subtract the area of the 45-45-90 triangle of the square, 3, = 3π/2 - 3, "take" 3 out would give me 3(π/2 - 1).

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New post 16 Dec 2016, 13:03
I would like to know if we can use the 45-45-90 triangle properties here

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New post 17 Dec 2016, 02:05
aqeelnaqvi10 wrote:
I would like to know if we can use the 45-45-90 triangle properties here


There is no 45-45-90 triangle here so you can't use it
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New post 23 Mar 2017, 15:11
Bunuel wrote:
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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New post 24 Mar 2017, 03:54
Prostar wrote:
Bunuel wrote:
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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New post 25 May 2017, 08:30
HI nguyendinhtuong!

Why do you feel that there is no 45-45-90 triangle here? I give my reasoning here. Correct me if I am wrong.

I think a diagonal to a square will ALWAYS divide the square into two isosceles right angle triangle. Hence the inner square consists of two 45-45-90 triangles.

nguyendinhtuong wrote:
aqeelnaqvi10 wrote:
I would like to know if we can use the 45-45-90 triangle properties here


There is no 45-45-90 triangle here so you can't use it

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Last edited by susheelh on 25 May 2017, 08:54, edited 1 time in total.

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New post 25 May 2017, 08:43
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 wrote:
I would like to know if we can use the 45-45-90 triangle properties here

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New post 25 Dec 2017, 15:49
what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.
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Re: M06-32   [#permalink] 25 Dec 2017, 19:45
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