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16 Jun 2022, 04:15
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E
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79% (02:04) correct
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Circle O is inscribed with square ABCD. Square EFGH is inscribed within circle O. What is the ratio of the area of ABCD to the area of EFGH?
A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
16 Jun 2022, 09:48
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Bunuel
Circle O is inscribed with square ABCD. Square EFGH is inscribed within circle O. What is the ratio of the area of ABCD to the area of EFGH?
A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
Bunuel, getting the ratio 8:1.A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
Help, please!
16 Jun 2022, 11:03
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ShahadatHJ
Bunuel
Circle O is inscribed with square ABCD. Square EFGH is inscribed within circle O. What is the ratio of the area of ABCD to the area of EFGH?
A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
Bunuel, getting the ratio 8:1.A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
Help, please!
ShahadatHJ
A characteristic of a circle inscribed in a square is that that the squares diagonal and the circle's diameter are equal (this is square ABCD).
A characteristic of square inscribed in a circle is that the circle's diameter and the squares side are equal (this is square EFGH).
To make things easier assign values:
Let the circle have a diameter of \(2\).
As square ABCD will have sides equal to the diameter of the circle, the area of ABCD is: \(2*2 = 4\)
Now as the the EFGH's diagonal is equal to the diameter of the circle (2), and we know that the diagonal of a square makes two 45-45-90 special right-angle triangles with sides \(x, x\)&\(x\sqrt{2}\), respectively, we know then that \(x\sqrt{2}=2\) and thus \(x = \frac{2}{\sqrt{2}}\).
The area for the EFGH will then be: \(\frac{2}{\sqrt{2}} * \frac{2}{\sqrt{2}}\)
\(=\frac{4}{2}\)
\(=2\).
So the ratio of the area of ABCD to the area of EFGH is: \(4:2\) which simplified is \(2:1\)
Answer E
