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Geometry - Parallelogram Problem

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by mattnyc15 » Thu Nov 05, 2015 4:13 pm
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OA: 24

Why wouldn't you add the additional triangle length to the area of 21.

When I solve for x: 3x(3)=21, I get 2.3

and when I plug that in for the triangle:
I get an area of 3.45 and add it to 21 for 24.45. It's close, but it's not the answer here.
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by DavidG@VeritasPrep » Thu Nov 05, 2015 4:31 pm
mattnyc15 wrote:Image


OA: 24

Why wouldn't you add the additional triangle length to the area of 21.

When I solve for x: 3x(3)=21, I get 2.3

and when I plug that in for the triangle:
I get an area of 3.45 and add it to 21 for 24.45. It's close, but it's not the answer here.
The problem is that though side EC = 3x, side AB = 4x, so it isn't true that 3x * 3 = 21. (Side DC = x + 3x = 4x. Opposite sides of a parallelogram are equal, so DC = AB = 4x.)

So the shape ABCE is a trapezoid. Area of trapezoid = Height* (Base 1 + Base 2)/2

Base 1 = EC = 3x
Base 2 = AB = 4x
Height = 3

Area = 3* (3x + 4x)/2; we're told that this is 21
3* (3x + 4x)/2 = 21
Solving, we get x = 2.

So Base of parallelogram = 4x = 4*2 = 8
Height = 3

Base * Height = 8*3 = 24

(Or, thinking of it as you did, if x = 2, the area of the triangle would be (2*3)/2 = 3. 21 + 3 = 24.)
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by ahmedshafea » Wed Aug 30, 2017 10:26 am
Is the above reply a correct solution? If not could one explain the steps please
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by Matt@VeritasPrep » Wed Aug 30, 2017 6:02 pm
ahmedshafea wrote:Is the above reply a correct solution? If not could one explain the steps please
You can generally assume that most instructors' replies are either:

1) right, or
2) wrong, but quickly corrected by another instructor looking to score on the first instructor

From those two statements, it more or less follows that an instructor's reply that hasn't been corrected by another instructor for years is probably right. :)
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