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ABC is 1/12 the area of triangle ADE

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by goyalsau » Sun Oct 17, 2010 5:33 am
In the figure above, AC = 3, CE = x, and BC is parallel to DE.

If the area of triangle ABC is 1/12 the area of triangle ADE, then x =

6 + 2√3
12√3 + 3
6√3 - 3
33
10√2
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by shovan85 » Sun Oct 17, 2010 5:57 am
IMO C

ABC is similar to ADE

so similarity of triangle says BC/DE = AC/AE = height of ABC/height of ADE = 3/(3+x).
as AC = 3 and AE = 3+x

Area ABC/Area ADE = 1/12
=> [(1/2) * (BC) * (height of ABC)]/[(1/2) * (DE) * (height of ADE)] = 1/12
=> [BC/DE]*[height of ABC/height of ADE] = 1/12
=> [3/(3+x)] ^2 = 1/12
=> x^2 + 6x - 99 = 0

Solving this we get x = -3+6sqrt(3) and -3-6sqrt(3).

We cannot take the -ve only +ve one is the answer.

Because of your previous complement I have to be more cautious while answering. Please dont do that anymore ;)
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by GMATGuruNY » Sun Oct 17, 2010 6:34 am
goyalsau wrote:In the figure above, AC = 3, CE = x, and BC is parallel to DE.

If the area of triangle ABC is 1/12 the area of triangle ADE, then x =

6 + 2√3
12√3 + 3
6√3 - 3
33
10√2
Let B = base of ADE and H = height of ADE.
Let b = base of ABC and h = height of ABC.
As noted above, ADE is similar to ABC. This means that all corresponding sides must yield the same proportion: B:b = H:h.
Since ADE is 12 times as big as ABC, BH = 12bh.
This means that B = √12b and H = √12h.
Now we can see that the proportion between corresponding sides of ADE and ABC is √12:1.
Thus, AE = √12(AC) = 3√12 = 6√3.
x = AE - AC = 6√3 - 3.

The correct answer is C.
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by gmatter2012 » Sun Jul 15, 2012 10:20 pm
shovan85 wrote:IMO C

ABC is similar to ADE

so similarity of triangle says BC/DE = AC/AE = height of ABC/height of ADE = 3/(3+x).
as AC = 3 and AE = 3+x

Area ABC/Area ADE = 1/12
=> [(1/2) * (BC) * (height of ABC)]/[(1/2) * (DE) * (height of ADE)] = 1/12
=> [BC/DE]*[height of ABC/height of ADE] = 1/12
=> [3/(3+x)] ^2 = 1/12
=> x^2 + 6x - 99 = 0

Solving this we get x = -3+6sqrt(3) and -3-6sqrt(3).

We cannot take the -ve only +ve one is the answer.

Because of your previous complement I have to be more cautious while answering. Please dont do that anymore ;)
Can somebody explain how this quadratic was solved?

middle term factorization will obviously not work
tried -b(+-)square root(b^2-4ac)/2a couldn't do it

so either I made a mistake while solving or this was solved by some other method. Please assist in the factorization of x^2 + 6x - 99 = 0
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by eagleeye » Sun Jul 15, 2012 10:36 pm
gmatter2012 wrote:
shovan85 wrote:IMO C

ABC is similar to ADE

so similarity of triangle says BC/DE = AC/AE = height of ABC/height of ADE = 3/(3+x).
as AC = 3 and AE = 3+x

Area ABC/Area ADE = 1/12
=> [(1/2) * (BC) * (height of ABC)]/[(1/2) * (DE) * (height of ADE)] = 1/12
=> [BC/DE]*[height of ABC/height of ADE] = 1/12
=> [3/(3+x)] ^2 = 1/12
=> x^2 + 6x - 99 = 0

Solving this we get x = -3+6sqrt(3) and -3-6sqrt(3).

We cannot take the -ve only +ve one is the answer.

Because of your previous complement I have to be more cautious while answering. Please dont do that anymore ;)
Can somebody explain how this quadratic was solved?

middle term factorization will obviously not work
tried -b(+-)square root(b^2-4ac)/2a couldn't do it

so either I made a mistake while solving or this was solved by some other method. Please assist in the factorization of x^2 + 6x - 99 = 0
Hi gmatter2012:

We can in fact do this by using the formula:

x = (-6 +/- sqrt(6^2-4*1*(-99)))/2 = 1/2*(-6 +/- sqrt(36+396)) = 1/2* ( -6 +/- sqrt(432))
=> x = 1/2*( -6 +/- (sqrt(144*3)) = 1/2 * ( -6 +/- 12*sqrt(3)) = -6/2 +/- 12*sqrt(3)/2
= -3 +/- 6*sqrt(3).

Hence the answer.
In case you were interested, there is a lot simpler way to solve the question using properties of similar triangles.

Since the triangles ABC and ADE are similar, ratio of areas = square of ratio of corresponding sides.
=> 1/12 = 3^2/(x+3)^2
=> (x+3)^2 = 9*12 = 3^2*2^2*3
=> x+3 = 3*2*sqrt(3) (I only took the positive root since x+3 can't be negative)
=> x = 6*sqrt(3) - 3.

:)
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by gmatter2012 » Sun Jul 15, 2012 10:57 pm
Thank you eagle eye both for the solution of the quadratic and the simpler solution

My mistake was not taking c as negative in b^2- 4ac as c = -99 so this expression will be positive
36+396 = 432

but I was taking 36-396 so I was getting a negative, never really solved quad's this equation way. so was a bit rusty. But thanks to you now I am more enlightened.

also your simpler way to solve was amazing.

1/12 = 9/(x+3)^2

(x+3)^2= 12*9
= 2^2 * 3^2 *3
x+3 =2*3root(3)
x= 6 root(3)-3

This is a great way to solve the quadratic rather than reducing it to x^2 + 6x - 99 = 0! Great
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