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length of segment

by ranjeet75 » Sun Mar 11, 2012 9:49 pm
What is the length of segment BC?

(1) Angle ABC is 90 degrees. (2) The area of the triangle is 30.

OA is [spoiler]A[/spoiler] but I think the answer should be D because if area and two sides are given then by the formula: Area = sqrt s(s-a)(s-b)(s-c) we can find the 3rd side.

where i am wrong?
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by Anurag@Gurome » Sun Mar 11, 2012 10:17 pm
ranjeet75 wrote:What is the length of segment BC?

(1) Angle ABC is 90 degrees. (2) The area of the triangle is 30.

OA is A but I think the answer should be D because if area and two sides are given then by the formula: Area = sqrt s(s-a)(s-b)(s-c) we can find the 3rd side.

where i am wrong?
(1) Angle ABC is 90 degrees.
Since ABC is right angled at B, we can easily find the length of BC using Pythagoras Theorem; SUFFICIENT.

(2) The area of the triangle is 30.
Although Heron's formula is out of GMAT scope, but still let me clarify it here that why statement 2 is NOT sufficient.

Let us assume that BC = x
Then s = (5 + 13 + x)/2 = (18 + x)/2 = 9 + (x/2)
Now area of triangle, A = √[s(s - a)(s - b)(s - c)] or A² = [s(s - a)(s - b)(s - c)]
(30)² = [9 + (x/2)] * [9 + (x/2) - 5] * [9 + (x/2) - 13] * [9 + (x/2) - x]
900 = [9 + (x/2)] * [4 + (x/2)] * [-4 + (x/2)] * [9 - (x/2)]
900 = [81 - (x/2)²][(x/2)² - 16]
Let (x/2)² = y
900 = [81 - y][y - 16]
900 = 81y - y² + 16y - 1296
y² - 97y + 2196 = 0
y² - 36y - 61y + 2196 = 0
y(y - 36) - 61(y - 36) = 0
(y - 61)(y - 36) = 0
y = 36, 61
(x/2)² = 36, (x/2)² = 61, which clearly implies that we are getting 2 values of x, which means 2 values of BC. So, statement 2 is NOT sufficient.

The correct answer is A.

I hope that helps.
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by vinaypat » Thu Mar 15, 2012 3:27 am
What if we draw a perpendicular from point B to line AC. This perpendicular touches AC at D. Assume the length of this line is x

from 2) 1/2 * x * 13 = 30
i.e. x = 60/13

in right angle triangle ABD (ADB is 90) we can find AD = sqrt ( (5*5) - (60/13 * 60/13)) == assume y

So, DC = 13 - y

Now in right angle triangle BDC. BC and be found by
sqrt ( (60/13 * 60/13) + (13-y)(13-y)) == So we can find the solution

I know this is complicated calculation but in DS we don't have to solve the equations
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by sam2304 » Thu Mar 15, 2012 5:58 am
vinaypat wrote:What if we draw a perpendicular from point B to line AC. This perpendicular touches AC at D. Assume the length of this line is x

from 2) 1/2 * x * 13 = 30
i.e. x = 60/13

in right angle triangle ABD (ADB is 90) we can find AD = sqrt ( (5*5) - (60/13 * 60/13)) == assume y
AD = 25 - 21.6 ~= 4. Sqrt of 4 we get + or - 2

so AD or y = + or - 2.
So, DC = 13 - y

Now in right angle triangle BDC. BC and be found by
sqrt ( (60/13 * 60/13) + (13-y)(13-y)) == So we can find the solution

I know this is complicated calculation but in DS we don't have to solve the equations
Now BC will also have two values based on y. You cannot arrive at a single solution. Hope I am right and this helps !!
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by vinaypat » Thu Mar 15, 2012 3:29 pm
There is no such thing of taking -ve value when calculating length of the Triangle. Only +ve values are valid as any side of a triangle can't be less then difference of other 2 sides
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by [email protected] » Thu Mar 15, 2012 11:55 pm
Anurag@Gurome


well I do not think that the Heron's formula is out of scope in GMAT. I really do not think so...
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