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The diameter of a circle equals the diagonal of a square who
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21 Oct 2018, 08:16
21 Oct 2018, 08:16
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Question Stats:
80% (01:00) correct
19% (00:30) wrong based on 76 sessions
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The diameter of a circle equals the diagonal of a square whose side length is 4.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The circumference of the circle |
\(20 \sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Re: The diameter of a circle equals the diagonal of a square who
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31 Jan 2019, 09:09
31 Jan 2019, 09:09
1
Solution,
Let us recall the area formula of a square and circumference of a circle.
Circumference of a circle ( C ) = 2 π r = πd ( where, r = radius and d = diameter of a circle)
i.e. d = C/ π
Area of a square = 1/2 d^2 ( where, d = diagonal of a square and l= length of a square )
l^2 = 1/2 d^2
4^2= 1/2 d^2
d^2 = 16 *2
d^2 = 32
d= √32
d = 4√2
Given that,
The diameter of a circle = the diagonal of a square
C/ π = 4√2
C= 4√2 * π
= 4√2 * 3.14
= 12.56 √2
Since, Quantity B is greater than Quantity A, i.e. 12.56 √2 < 20 √2, the answer is B.
Let us recall the area formula of a square and circumference of a circle.
Circumference of a circle ( C ) = 2 π r = πd ( where, r = radius and d = diameter of a circle)
i.e. d = C/ π
Area of a square = 1/2 d^2 ( where, d = diagonal of a square and l= length of a square )
l^2 = 1/2 d^2
4^2= 1/2 d^2
d^2 = 16 *2
d^2 = 32
d= √32
d = 4√2
Given that,
The diameter of a circle = the diagonal of a square
C/ π = 4√2
C= 4√2 * π
= 4√2 * 3.14
= 12.56 √2
Since, Quantity B is greater than Quantity A, i.e. 12.56 √2 < 20 √2, the answer is B.
The diameter of a circle equals the diagonal of a square who
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04 Jan 2024, 22:04
04 Jan 2024, 22:04
The diameter of a circle equals the diagonal of a square whose side length is \(4\).
The diagonal length will be \(4\sqrt{2}
\)
The radius of the circle = \(2\sqrt{2}
\)
Its circumference will be \(2 \times \pi \times 2\sqrt{2} = 4 \times \pi \times \sqrt{2} = 4 \times 3.14 \times \sqrt{2} = 12.56 \times \sqrt{2}\)
Now cancelling \(\sqrt{2}\) found in both quantities whose values we are comparing, we have
Quantity A = \(12.56\)
Quantity B = \(20\)
Quantity B is greater.
The diagonal length will be \(4\sqrt{2}
\)
The radius of the circle = \(2\sqrt{2}
\)
Its circumference will be \(2 \times \pi \times 2\sqrt{2} = 4 \times \pi \times \sqrt{2} = 4 \times 3.14 \times \sqrt{2} = 12.56 \times \sqrt{2}\)
Now cancelling \(\sqrt{2}\) found in both quantities whose values we are comparing, we have
Quantity A = \(12.56\)
Quantity B = \(20\)
Quantity B is greater.
