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25 May 2016, 18:30
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D
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Difficulty:
45%
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Question Stats:
75% (02:34) correct
25%
(02:30)
wrong
based on 75
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There are two squares in a semi-circle as above figure, OC=CE. If the radius of a semi-circle is 4√5, what is the area of the small square?
A. 12
B. 14
C. 15
D. 16
E. 27
*An answer will be posted in 2 days.
A. 12
B. 14
C. 15
D. 16
E. 27
*An answer will be posted in 2 days.
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1.png [ 25.26 KiB | Viewed 3579 times ]
26 May 2016, 22:27
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Let us assume x = OC = OE
Side of larger square = BO + OC = x + x = 2*x or CD = 2*x
Using pythagoras theorem
(2*x)^2 + x^2 = (4*sqrt(5))^2
5*x^2 = 16*5 ==> x = 4
Area of smaller square = x*x = 4*4 = 16
Side of larger square = BO + OC = x + x = 2*x or CD = 2*x
Using pythagoras theorem
(2*x)^2 + x^2 = (4*sqrt(5))^2
5*x^2 = 16*5 ==> x = 4
Area of smaller square = x*x = 4*4 = 16
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File comment: Solution image
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26 May 2016, 23:42
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Say OC = x. Then BC= 2x. This means that CD = 2x.
Consider \(\triangle\) OCD,
\(OC^2\) + \(DC^2\) = Square of (4* \(\sqrt{5}\))
\(x^2\)+4\(x^2\) = 80
5\(x^2\)=80
\(x^2\) = 16.
We want area of smaller square; which is \(x^2\). So our answer is 16.
D is the correct option.
Consider \(\triangle\) OCD,
\(OC^2\) + \(DC^2\) = Square of (4* \(\sqrt{5}\))
\(x^2\)+4\(x^2\) = 80
5\(x^2\)=80
\(x^2\) = 16.
We want area of smaller square; which is \(x^2\). So our answer is 16.
D is the correct option.
