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# M06-32

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Math Expert
Joined: 02 Sep 2009
Posts: 44655
M06-32 [#permalink]

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16 Sep 2014, 00:28
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Difficulty:

25% (medium)

Question Stats:

91% (01:34) correct 9% (02:03) wrong based on 44 sessions

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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. $$3(\frac{\pi}{2} - 1)$$
E. $$3 \frac{\pi}{4} - \frac{3}{2}$$
[Reveal] Spoiler: OA

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Math Expert
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Re M06-32 [#permalink]

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16 Sep 2014, 00:28
Expert's post
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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?

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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Joined: 16 Feb 2016
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M06-32 [#permalink]

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26 Apr 2016, 23:38
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?

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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Joined: 29 Sep 2016
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Re: M06-32 [#permalink]

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08 Nov 2016, 11:22
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).
Intern
Joined: 01 Feb 2016
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Re: M06-32 [#permalink]

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16 Dec 2016, 14:03
I would like to know if we can use the 45-45-90 triangle properties here
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Re: M06-32 [#permalink]

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17 Dec 2016, 03:05
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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Re: M06-32 [#permalink]

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23 Mar 2017, 16:11
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 : $$\frac{\text{diagonal}^2}{2}=12$$
Math Expert
Joined: 02 Sep 2009
Posts: 44655
Re: M06-32 [#permalink]

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24 Mar 2017, 04:54
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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M06-32 [#permalink]

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Updated on: 25 May 2017, 09:54
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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Originally posted by susheelh on 25 May 2017, 09:30.
Last edited by susheelh on 25 May 2017, 09:54, edited 1 time in total.
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Re: M06-32 [#permalink]

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25 May 2017, 09:43
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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Concentration: Marketing, Entrepreneurship
Schools: McCombs
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Re: M06-32 [#permalink]

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25 Dec 2017, 16:49
what are steps to reduce 1/4 (6 pi - 12) = 3(pi/2 - 1) ? I just can't see it.
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Math Expert
Joined: 02 Sep 2009
Posts: 44655
Re: M06-32 [#permalink]

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25 Dec 2017, 20:45
RigelKent wrote:
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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Re: M06-32   [#permalink] 25 Dec 2017, 20:45
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# M06-32

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