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13 Jun 2022, 08:15
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E
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Difficulty:
25%
(medium)
Question Stats:
83% (01:42) correct
17%
(01:11)
wrong
based on 23
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Square ABCD and a circle with center C intersect as shown. If point E is at the center of ABCD and if the radius of circle C is k, then what is the area of ABCD, in terms of k ?
A. \(\frac{k^2}{(2π)}\)
B. \(\frac{πk^2}{2}\)
C. \(πk^2\)
D. \(k^2\)
E. \(2k^2\)
Attachment:
c18_r106.png [ 6.68 KiB | Viewed 3841 times ]
13 Jun 2022, 22:50
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Point E is at the center of ABCD and the radius of circle C is k.
CE = radius of the circle = k
AE = CE (because E is the center of ABCD) ; Hence, AE = CE = k
We can conclude that the diagonal of the square (d) = 2k
We can use the formula:
d = √2 x side (We know that the diagonal = 2k)
2k/√2 = side
Therefore, side = k√2
We need to find the area of square ABCD = side^2
Area of square = (k√2)^2 = 2(k)^2
Answer choice (E)
CE = radius of the circle = k
AE = CE (because E is the center of ABCD) ; Hence, AE = CE = k
We can conclude that the diagonal of the square (d) = 2k
We can use the formula:
d = √2 x side (We know that the diagonal = 2k)
2k/√2 = side
Therefore, side = k√2
We need to find the area of square ABCD = side^2
Area of square = (k√2)^2 = 2(k)^2
Answer choice (E)
14 Jun 2022, 02:57
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aarushi1710
Another way this question can be solved is using Pythagoras theorem. We know that AE = EC = k.
Thus, AC = 2k (which will be the diagonal of triangle ABC).
Let side AB = x (since this is a square we know that AB = BC = x)
Using Pythagoras, x^2 + x^2 = (2k)^2
2x^2 = 4(k)^2
x^2 = 2(k)^2 (We also know that x^2 = area of a square).
Answer choice (E)
