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In the figure shown, what is the area of the circular region

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by mitzwillrockgmat » Sun May 23, 2010 9:34 am
In the figure shown (attached), what is the area of the circular region with center O and diameter BC?

(1) BC/AB = 3/4
(2) BD = 25

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Image

The following is how i solved it. pls confirm i am on the right path!

1) given bc/ab =3/4 we cant conclude anything except bc is 3x and ab 4x. so radius is 1.5x at best according to this stat.

2) given bd = 25 , can i use the pythagorean theorem to determine that bc and cd could be 9 and 16?? although i we wont know which side is 9 and 16.

combined, i know that bc is 3x so it must be the side that is 9.

hence radius is 4.5 and area is (pie)4.5^2

does this make sense??
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Source: — Data Sufficiency |
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by sumanr84 » Sun May 23, 2010 9:45 am
(1) BC/AB = 3/4

Assume BC = 3x and AB = 4x [taking x being the common factor]
BC = AD = Diameter of the circle
Not sufficient to determine the radius
(2) BD = 25
BD ^2 = AB ^2 + AD ^2
No knowledge of AB and AD
Not sufficient to determine the radius

Combining 2,
BD ^2 = AB ^2 + AD ^2
25 ^2 = (4x)^2 + (3x)^2
We can derive x, therefore can get AD - Diameter and thus Radius
Sufficient (C)
I am on a break !!
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by sumanr84 » Sun May 23, 2010 9:50 am
mitzwillrockgmat wrote: 2) given bd = 25 , can i use the pythagorean theorem to determine that bc and cd could be 9 and 16?? although i we wont know which side is 9 and 16.
does this make sense??
You cannot make this assumption here, although it fits here well,

all you can say is X^2 + Y ^2 = 25 ^2,

AB ^2 + AD ^2 = 25 ^2
I am on a break !!
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by mitzwillrockgmat » Sun May 23, 2010 9:58 am
sumanr84 wrote:
mitzwillrockgmat wrote: 2) given bd = 25 , can i use the pythagorean theorem to determine that bc and cd could be 9 and 16?? although i we wont know which side is 9 and 16.
does this make sense??
You cannot make this assumption here, although it fits here well,

all you can say is X^2 + Y ^2 = 25 ^2,

AB ^2 + AD ^2 = 25 ^2
okie got it! thanks!
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by sumanr84 » Sun May 23, 2010 10:52 am
mitzwillrockgmat wrote: okie got it! thanks!
you are always welcome :-)
I am on a break !!
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by kstv » Sun May 23, 2010 8:31 pm
If BD = 25 as in option 2, is there any other value possible for BC and AB other than 20 and 15 ?
The pythagorean triplet is unique.
but option 1 is still needed to known which is BC i.e. the diameter.
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by neha.patni » Sun May 23, 2010 11:44 pm
mitzwillrockgmat wrote:In the figure shown (attached), what is the area of the circular region with center O and diameter BC?

(1) BC/AB = 3/4
(2) BD = 25

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Image

The following is how i solved it. pls confirm i am on the right path!

1) given bc/ab =3/4 we cant conclude anything except bc is 3x and ab 4x. so radius is 1.5x at best according to this stat.

2) given bd = 25 , can i use the pythagorean theorem to determine that bc and cd could be 9 and 16?? although i we wont know which side is 9 and 16.

combined, i know that bc is 3x so it must be the side that is 9.

hence radius is 4.5 and area is (pie)4.5^2

does this make sense??
IMO C and the area of the circular region is pie x 7.5^2[/img]
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