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# There are two squares in a semi-circle as above figure, OC=CE. If the

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There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink]

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25 May 2016, 18:30
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45% (medium)

Question Stats:

78% (02:06) correct 22% (04:54) wrong based on 45 sessions

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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.
[Reveal] Spoiler: OA

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Re: There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink]

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26 May 2016, 22:27
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
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There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink]

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26 May 2016, 23:42
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.

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Re: There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink]

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28 May 2016, 19:39
If OC=GC=DG=a, then a^2+(2a)^2=(4√5)^2=80. From 5a^2=80, we get a^2=16. So the area of the small square is a^2=16. Hence, the correct answer is D.
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Re: There are two squares in a semi-circle as above figure, OC=CE. If the [#permalink]

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15 Sep 2017, 08:36
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Re: There are two squares in a semi-circle as above figure, OC=CE. If the   [#permalink] 15 Sep 2017, 08:36
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