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Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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Difficulty:   45% (medium)

Question Stats: 72% (01:40) correct 28% (01:40) wrong based on 60 sessions

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A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximate area of the square that is not occupied by the circle?

A. 1.7
B. 2.7
C. 3.4
D. 5.4
E. 8

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Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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Official Solution:

A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximate area of the square that is not occupied by the circle?

A. 1.7
B. 2.7
C. 3.4
D. 5.4
E. 8

The area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{4^2}{2}=8$$;

Next, since the diagonal of the square equals to 4 centimeters then the side of the square equals to $$2\sqrt{2}$$ (you can find it for example using Pythagorean theorem: $$a^2+a^2=4^2$$);

Now, the radius of inscribed circle will be half of the side of the square, so $$\sqrt{2}$$, which makes its area equal to $$\pi{r^2}=2\pi$$;

Hence, the approximate area of the square that is not occupied by the circle is $$8- 2\pi \approx 8-6.28 = 1.72$$.

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Intern  Joined: 05 Jul 2016
Posts: 5

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Hello,

if the diagonal is of the square is 4, should the side not be 4/2sqrt? Not 2/2sqrt. This is based on the 45-45-90 special triangle rule.

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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abhilash53 wrote:
Hello,

if the diagonal is of the square is 4, should the side not be 4/2sqrt? Not 2/2sqrt. This is based on the 45-45-90 special triangle rule.

Thanks

Solve for x:
x^2 + x^2 = 4^2.
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Director  B
Status: I don't stop when I'm Tired,I stop when I'm done
Joined: 11 May 2014
Posts: 534
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Top Contributor
Bunuel wrote:
A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximate area of the square that is not occupied by the circle?

A. 1.7
B. 2.7
C. 3.4
D. 5.4
E. 8

Side of the square =4/$$\sqrt{2}$$

So area of the square =$$(\frac{4}{\sqrt{2}})^2$$=8
and
area of the circle=$$\pi$$*$$(\frac{4}{(\sqrt{2}*2)})^2$$=4$$\pi$$

So the approximate area of the square that is not occupied by the circle=8-4$$\pi$$=8-4*(>1.5)=8-(>6)
Only answer Choice A meets the substruction requirement to be less than 2,which is 1.7

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Intern  B
Joined: 04 Sep 2016
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I don't agree with the explanation. Answer is wrong here.

it should be 4*pie - 8, which is around 4.56.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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MountainGMAT wrote:
I don't agree with the explanation. Answer is wrong here.

it should be 4*pie - 8, which is around 4.56.

My friend, the answer is correct here. Please provide your detailed solution and I'll try to find an error you are making there.
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Joined: 02 May 2016
Posts: 77
Location: India
Concentration: Entrepreneurship
GRE 1: Q163 V154 WE: Information Technology (Computer Software)

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MountainGMAT wrote:
I don't agree with the explanation. Answer is wrong here.

it should be 4*pie - 8, which is around 4.56.

Hi!
1. The question says circle is inscribed in a square, and not the other way round!
and,
2. No way it can be 4*pie. Area of circle will be 2*pie
Intern  Joined: 11 Aug 2017
Posts: 5
WE: Account Management (Energy and Utilities)

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Hello,

Why wouldn't the circumference of the cirlce be the same as the diagonal of the square? Thus, the radios would be half of the diagonal, or 4...
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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1
DouglassJensen wrote:
Hello,

Why wouldn't the circumference of the cirlce be the same as the diagonal of the square? Thus, the radios would be half of the diagonal, or 4...

Attachment:
circle-in-square.png

The circumference of a circle is the distance around it.
The diagonal or a square is the distance between two opposite vertices.

How are those two equal???
>> !!!

You do not have the required permissions to view the files attached to this post.

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Intern  Joined: 11 Aug 2017
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WE: Account Management (Energy and Utilities)

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Oh my gosh. Clearly, the end of a long day and got mixed up. What I meant was Diameter, not circumference.

Wouldnt the Diagonal of the square equal the diameter of the circle? So then the radius would be 4?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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DouglassJensen wrote:
Oh my gosh. Clearly, the end of a long day and got mixed up. What I meant was Diameter, not circumference.

Wouldnt the Diagonal of the square equal the diameter of the circle? So then the radius would be 4?

Please check the image above. The side of the square = the diameter of the circle.
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Joined: 10 Sep 2014
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Bunuel wrote:
Official Solution:

A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximate area of the square that is not occupied by the circle?

A. 1.7
B. 2.7
C. 3.4
D. 5.4
E. 8

The area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{4^2}{2}=8$$;

Next, since the diagonal of the square equals to 4 centimeters then the side of the square equals to $$2\sqrt{2}$$ (you can find it for example using Pythagorean theorem: $$a^2+a^2=4^2$$);

Now, the radius of inscribed circle will be half of the side of the square, so $$\sqrt{2}$$, which makes its area equal to $$\pi{r^2}=2\pi$$;

Hence, the approximate area of the square that is not occupied by the circle is $$8- 2\pi \approx 8-6.28 = 1.72$$.

Bunuel please explain this part "Next, since the diagonal of the square equals to 4 centimeters then the side of the square equals to $$2\sqrt{2}$$ (you can find it for example using Pythagorean theorem: $$a^2+a^2=4^2$$);"
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

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1
Bunuel wrote:
Official Solution:

A circle is inscribed in a square with the diagonal of 4 centimeters. What is the approximate area of the square that is not occupied by the circle?

A. 1.7
B. 2.7
C. 3.4
D. 5.4
E. 8

The area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{4^2}{2}=8$$;

Next, since the diagonal of the square equals to 4 centimeters then the side of the square equals to $$2\sqrt{2}$$ (you can find it for example using Pythagorean theorem: $$a^2+a^2=4^2$$);

Now, the radius of inscribed circle will be half of the side of the square, so $$\sqrt{2}$$, which makes its area equal to $$\pi{r^2}=2\pi$$;

Hence, the approximate area of the square that is not occupied by the circle is $$8- 2\pi \approx 8-6.28 = 1.72$$.

Bunuel please explain this part "Next, since the diagonal of the square equals to 4 centimeters then the side of the square equals to $$2\sqrt{2}$$ (you can find it for example using Pythagorean theorem: $$a^2+a^2=4^2$$);"

$$side^2 + side^2 = 4^2$$;

$$2*side^2=16$$;

$$side^2=8$$;

$$side=\sqrt{8}=2\sqrt{2}$$.
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# M07-05

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