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# Which is greater?

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Senior Manager
Joined: 20 May 2014
Posts: 282
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Kudos [?]: 49 [0], given: 220

Which is greater? [#permalink]  12 Nov 2017, 01:40
00:00

Question Stats:

33% (01:03) correct 66% (00:43) wrong based on 6 sessions

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.

[Reveal] Spoiler:
Attachment:

gqgpp_img3.png [ 6.58 KiB | Viewed 893 times ]

Attachment:

AAAA-QC-for-circle.jpg [ 3.6 KiB | Viewed 889 times ]
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 327 [2] , given: 66

Re: Which is greater? [#permalink]  12 Nov 2017, 09:57
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Since triangle ABC is built on the diameter of the circle, the angle in C is 90°. Thus we can use the Pythagorean theorem to get side AC as $$AB^2-BC^2 = 16-9 = 7$$ so that Ac is equal to $$\sqrt7$$. Then, the area of the square ACDE is $$AC^2 = 7$$

Answer C
Manager
Joined: 03 Dec 2017
Posts: 65
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Kudos [?]: 15 [0], given: 20

Re: Which is greater? [#permalink]  07 Dec 2017, 19:56
Bunuel please correct how wrong I am,
OB=2, OC=2, BC=3 ; OA=2, OC=2, AC=3 similar triangle AOC~BOC
AC*AC = 3*3 = 9
Senior Manager
Joined: 20 May 2014
Posts: 282
Followers: 14

Kudos [?]: 49 [0], given: 220

Re: Which is greater? [#permalink]  10 Dec 2017, 04:22
wongpcla wrote:

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Bunuel please correct how wrong I am,
OB=2, OC=2, BC=3 ; OA=2, OC=2, AC=3 similar triangle AOC~BOC
AC*AC = 3*3 = 9

No. Triangles AOC and BOC are not similar. Only two side there are equal. To find AC you shoul recall the following property:

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle (the reverse is also true: if the diameter of a circle is also the triangle’s side, then that triangle is a right triangle).

Thus, ABC is a right tringle right angled at C. Now, using Pythagoras theorem:
AC^2 + BC^2 = AB^2;
AC^2 + 9 = 16;
$$AC =\sqrt{7}$$

The area of the square $$side^2=(\sqrt{7})^2=7$$.

The two quantities are equal

Answer: C.
Re: Which is greater?   [#permalink] 10 Dec 2017, 04:22
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# Which is greater?

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