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 Post subject: GRE Geometry (Select One)
PostPosted: Thu May 26, 2011 4:15 pm 
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Joined: Sun May 30, 2010 3:15 am
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In the diagram above, BD = 8, AB = 6, and ED = 5. What is the length of AE?
A. 3
B. 4
C. 5
D. 7
E. 10

(C) Use this link for a flash video web page explanation: http://www.800score.com/explanations/GMAT_MATH_T1_Q6_Easy.html
We are given that BD = 8, AB = 6, and ED = 5. Let's break this problem down into a few steps.

1) Since triangle ABD is a right triangle where lengths of the two shorter sides are in the ratio of 4 to 3, the triangle must be a 3-4-5 right triangle (a common right-triangle type).

2) We can find the length of AD.
AB = 3 × 2 = 6
BD = 4 × 2 = 8
Therefore, AD = 5 × 2 = 10.

3) Since AE + ED = AD, AE + 5 = 10

4) AE must equal 5.

The correct answer is choice (C).
----------
I used this formula. Area of a triganle will be A = (1/2)bh. And we know BD = 8, AB = 6, ED = 5.
Therefore Area = 1/2 × 8 × 6 = 24.
We need the parameter of the diagram which is P = BD + AB + ED + AE, AE is unknown.
P = 8 + 6 + 5 + AE.
However we know the area of the entire diagram. From adding the
parimteter p = 8 + 6 + 5 = 19
Subtracting parimeter from Area 24 – 19 = 5 this will represent AE. Therefore, AE = 5.

I know this can be represented in a better way or correct way..
Thank you...


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 Post subject: Re: math (t.1, qt.6): geometry
PostPosted: Thu May 26, 2011 4:29 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
Subtracting parimeter from Area 24 – 19 = 5 this will represent AE. Therefore, AE = 5.
That is NOT correct.
A perimeter is measured in units, while an area is measured in square units. You CANNOT add or subtract values that have different units. A perimeter does NOT equal to an area.


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 Post subject: Re: math (t.1, qt.6): geometry
PostPosted: Fri Jul 22, 2011 3:06 pm 
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 Post subject: Re: math (t.1, qt.6): geometry
PostPosted: Thu Aug 18, 2011 6:35 pm 
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Could this answer not have been found through calculating the hypotenuse of ABD, and then subtract ED from the result?


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 Post subject: Re: math (t.1, qt.6): geometry
PostPosted: Thu Aug 18, 2011 6:43 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
Could this answer not have been found through calculating the hypotenuse of ABD, and then subtract ED from the result?
That's exactly what we did. We calculated the hypotenuse of ABD using the property of the 3-4-5 right triangle (You can use the Pythagorean theorem instead). It equals 10. Then we subtracted ED, which equals 5, and calculated the length of AE: 10 – 5 = 5.


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