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Re M07-05 [#permalink]
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 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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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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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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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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Re: M07-05 [#permalink]
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06 Jun 2017, 09:37
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Re: M07-05 [#permalink]
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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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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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Re: M07-05 [#permalink]
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06 Dec 2017, 22:54
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Re: M07-05 [#permalink]
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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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Re: M07-05 [#permalink]
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07 Dec 2017, 10:13
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Re: M07-05 [#permalink]
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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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28 Mar 2018, 02:44
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
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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