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EC is the diameter of the semicircle shown above. Is the

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EC is the diameter of the semicircle shown above. Is the [#permalink] New post 25 Aug 2006, 11:54
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EC is the diameter of the semicircle shown above. Is the area of the right isosceles triangle ABC greater than 80 square units?

(1) The area of the semicircle is greater than 40 square units
(2) |BE| is greater than 2 units
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Last edited by kevincan on 25 Aug 2006, 13:11, edited 1 time in total.
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 [#permalink] New post 25 Aug 2006, 12:44
if stmt (2) says |BE| instead of BC, then answer is (C).
Else answer is (E).
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 [#permalink] New post 25 Aug 2006, 13:58
prashrash wrote:
if stmt (2) says |BE| instead of BC, then answer is (C).
Else answer is (E).


Would you please explain the steps ? I also get E earlier but am not sure how you can be certain of C after the correction.
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 [#permalink] New post 25 Aug 2006, 14:21
1. The area of the semicircle = 1/2 * pi * (EC/2)^2 > 40
EC > 8 * sqrt (5/pi)

Area of the triangle = 1/2 * AC * BC = 1/2 BC^2 ( isoceles triangle)

BC = EC + ? say 1
then area of the triangle = 1/2 * (8* sqrt(5/pi) + 1)^2 < 80
if BC = EC + 3 then area of triangle > 80 INSUFF

1. BE is > 2 . by itself INSUFF

combined: SUFF so C

Kevincan, is there an easier way to approach the math?
It took me more than 2 minutes to do the calculation..
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 [#permalink] New post 25 Aug 2006, 14:39
Answer: A
Q: Area of ABC > 80

As isosceles triangle :
AC= BC = a

1/2 x a x a > 80

or a^2 > 160

S1: Given Area of semicircle > 40

pi x r x r > 2 x 40
or r x r > 2 x 40 x 7/22 (approx) (pi = 22/7)

r^2 > 280/11

or d^2 > 1120/11

or EC^2 > 1120/11

We know BC = a > EC

For area to be greater than 80,
=> a^2 > EC^2
=> 160 > 1120/11
=> 1760 > 1120 True...

Sufficient.

S2: |BE| > 2

Not sufficient to determine.

Answer: A
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 [#permalink] New post 26 Aug 2006, 08:29
I think the answer is E.

ABC = 45, OD is a tangent, so ODB=90, so BOD=45
OD=R, so BD=R, so BO=R sqrt(2)

Area of triangle ABC

=1/2AC.BC
=0.5(R+Rsqrt(2))^2
=0.5R^2(2.414)^2 =2.9R^2

From 1:
Area of ½ circle = 0.5pi R^2>40
or R^2>80/3.14 ...(1)

Now, we are looking for ABC > 80
or 2.9R^2 > 80
or R^2>80/2.9 (2)

As we can see 1 does not prove 2. INSUFF

From 2:
0.414R>2
R>4.8
R^2 > 23.04
2.9R^2 > 69 max, not 80 hence insuff.

Of course, I may be totally off here, everyone else has a different answer :)
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 [#permalink] New post 26 Aug 2006, 13:50
haas_mba07 wrote:
Answer: A
For area to be greater than 80,
=> a^2 > EC^2
=> 160 > 1120/11
=> 1760 > 1120 True...


Can u pls explain what u are doing in this step ? IMO this step takes 80 > 40 and then writes this in terms of a and EC and then converts back to numbers to try and prove the assumption. Pls let me know if this is not the case.
IMO that becomes a circular fallacy!! :-)

I have tried by a lengthy geometric analysis to see if area of triangle is greater than 80 based on the first statement and I am not getting a satisfactory answer, i.e. what I am getting is that it may or may not be greater than 80.

In conjunction with st (2) it will become greater than 80.
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  [#permalink] 26 Aug 2006, 13:50
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