In the Diagram Above, ABCD is a Parallelogram and the Areas of Yellow GMAT Problem Solving

Question: In the diagram above, ABCD is a parallelogram and the areas of yellow regions are 8, 10, 72, and 79. What is the area of the red triangle?

1. 7
2. 8
3. 9
4. 10
5. 12

“In the diagram above, ABCD is a parallelogram and the areas of yellow" - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been borrowed from the book “GMAT Official Guide Quantitative Review”.

To understand GMAT Problem Solving questions, applicants must possess fundamental qualitative skills. Quant tests a candidate's aptitude in reasoning and mathematics. The GMAT Quantitative test's problem-solving phase consists of a question and a list of possible responses. By using mathematics to answer the question, the candidate must select the appropriate response. The problem-solving section of the GMAT Quant topic is made up of very complicated math problems that must be solved by using the right math facts.

Answer: C

Solution and Explanation:

Approach Solution 1:

Recall that the area of a parallelogram is bhbh, where bb is the base of the parallelogram and hh is its height (altitude to the base).

Consider the blue triangle below:

It's area is 12∗ AD∗(height to AD)12∗AD∗(height to AD) but AD∗(height to AD)AD∗(height to AD) is the area of the parallelogram, so the area of the blue triangle is half of the area of the parallelogram.

Now, consider two green triangles. Their combined area is
1/2 ∗ AE ∗ (height to AB) + 1/2 ∗ EB ∗ (height to AB) =
1/2∗(height to AB)∗(AE+EB)=1/2∗(height to AB)∗AB

The same way as above, (height to AB)∗AB(height to AB)∗AB is the area of the parallelogram, so the combined area of two green triangle is half of the area of the parallelogram.
Thus, the area of blue triangle is equal to the combined area of two green triangles (both are half of the area of the parallelogram):

x+79+y+10=x+(red region)+72+y+8
(red region) = 89−80 = 9

Hence C is the correct answer.

Approach Solution 2:

You see such a question on a parallelogram, you should realize that the area within same height and base will be equal..

Let us give the areas a variable as shown in the sketch.

If you look at parallel sides AB and CD.

The areas of △DEF+△BFC=△ADE+△BEF
B + 79 + c + e + 10 + f = x + a + 72 + d + 8 .......b + c + e + f + 89 = x + a + d + 80 ........b + c + e + f + 9 = x + a + d ....(I)

Now, if you look at the parallel sides AD and CB..
The areas of △ABG+△CDG=△ADG
X + b + 72 + e + c + 8 + f = a + 79 + d + 10 ....... b + c + e + f + 80 = a + d + 89 ........ x + b + c + e + f − 9 = a + d...(II)

Substitute the value of a+d from (II) in (I)
B + c + e + f + 9 = x + a + d
B +c + e + f + 9 = x + x + b + c + e + f − 9
2 x = 18
x=9

Hence, C is the correct answer.

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