tricky geometry

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tricky geometry

by Fab » Mon Nov 03, 2008 4:40 pm
Please be clear with the explanations.

THANKS!
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Re: tricky geometry

by logitech » Mon Nov 03, 2008 5:02 pm
Fab wrote:Please be clear with the explanations.

THANKS!
Please post them seperately with the source. This way, people can solve one question at a time, and search it in the future. Thanks.
LGTCH
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by dhanda.arun » Wed Nov 05, 2008 1:09 pm
Q1
consider statement 1
if AB= 3 and BC =2
area of smaller = pi.Ab^2
and area of larger = pi.(AB + BC)^2)
so we can subtract and get the desired area.
Statement b does not help because we can't calculate area of smaller with this.

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by Fab » Wed Nov 05, 2008 1:22 pm
I agree with you, however the OA for question 1 is D (apparently).
I don't know what's the source, a fried of mine passed me the exams.

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by 4meonly » Thu Nov 06, 2008 1:25 am
Q30:
Answer B? B. (sqrt2)/4

Area of BCE = Area of BEO - Area of BOC

(1+(sqrt2)/2)*((sqrt2)/2)*1/2 - (sqrt2)/2)*(sqrt2)/2)*1/2 = (sqrt2)/4
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by muzali » Mon Nov 24, 2008 11:01 pm
1. D

We need the difference in area, i.e. the radii of both circles
First statement: Gives us the radii of both circles: AB (smaller circle), and AB+BC (larger circle) Sufficient
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient

2. Max height of the semicircular tunnel = 20/2 = 10ft
Clearance required = 1/2 ft
Therefore, max height allowed = 10-(1/2) = 9(1/2) ft

3. See the previous post w/ diagram

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by orel » Mon Nov 24, 2008 11:41 pm
for the second question:

I got the max height as 7,5.

i think, the max height without clearance can't be 10, because the width of the vehicle is 12, the width remains the same for the top part of the vehicle as well. well, here we have a rectangilar inscribed in a semicircle. since the radius of the semicircle is 10, and the width if the rectangle 12, its length will be sqrt(10^2-6^2)=8
8-1/2=7 1/2

am i missing smth?

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by Fab » Tue Nov 25, 2008 6:13 am
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient
Are you sure??

Isn't it CD + DE - DE = CD?

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by muzali » Tue Nov 25, 2008 10:15 am
Fab wrote:
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient
Are you sure??

Isn't it CD + DE - DE = CD?
See carefully,
(CD+DE) = radius of larger circle
2*(CD+DE) = diameter of the larger circle
2*(CD+DE)-DE= diameter of smaller circle...enough info to get the area of both circles.

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by Tryingmybest » Fri Nov 28, 2008 10:35 am
Muzali-- Excellent approach to solve these problems...Thanks!!