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M07-05

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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
[Reveal] Spoiler: OA

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Re M07-05 [#permalink]

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New post 16 Sep 2014, 00:34
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 \(8- 2\pi \approx 8-6.28 = 1.72\).


Answer: A
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Re: M07-05 [#permalink]

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New post 16 Jul 2016, 15:29
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
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M07-05 [#permalink]

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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


Correct answer A
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Re M07-05 [#permalink]

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New post 05 Jun 2017, 17:04
I don't agree with the explanation. Answer is wrong here.

it should be 4*pie - 8, which is around 4.56.
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New post 06 Jun 2017, 09:37
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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Re: M07-05 [#permalink]

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New post 08 Jun 2017, 22:52
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
Answer is correct!
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Re: M07-05 [#permalink]

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New post 06 Dec 2017, 22:43
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...
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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...


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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Re: M07-05 [#permalink]

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New post 07 Dec 2017, 10:10
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?
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New post 07 Dec 2017, 10:13
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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Re: M07-05 [#permalink]

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New post 27 Mar 2018, 21:40
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 \(8- 2\pi \approx 8-6.28 = 1.72\).


Answer: A

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: \(a^2+a^2=4^2\));"
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Re: M07-05 [#permalink]

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New post 28 Mar 2018, 02:44
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sadikabid27 wrote:
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 \(8- 2\pi \approx 8-6.28 = 1.72\).


Answer: A

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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M07-05   [#permalink] 28 Mar 2018, 02:44
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