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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?
A. 7
B. 8
C. 9
D. 10
E. 12
Are You Up For the Challenge: 700 Level Questions
M37-105
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2020-05-20_2336.png
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09 Dec 2021, 07:30
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Official Solution:
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?
A. \(7\)
B. \(8\)
C. \(9\)
D. \(10\)
E. \(12\)
Recall that the area of a parallelogram is \(bh\), where \(b\) is the base of the parallelogram and \(h\) is its height (altitude to the base).
Consider the blue triangle below:
It's area is \(\frac{1}{2}*AD*(height \ to \ AD)\) but \(AD*(height \ to \ AD)\) 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
\(\frac{1}{2}*AE*(height \ to \ AB)+\frac{1}{2}*EB*(height \ to \ AB) =\)
\(\frac{1}{2}*(height \ to \ AB)*(AE+EB)= \frac{1}{2}*(height \ to \ AB)*AB\). The same way as above, \((height \ to \ AB)*AB\) 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):
\(x+79+y+10=x+(red \ region)+72+y+8\);
\((red \ region)=89-80=9\).
Answer: C
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?
A. \(7\)
B. \(8\)
C. \(9\)
D. \(10\)
E. \(12\)
Recall that the area of a parallelogram is \(bh\), where \(b\) is the base of the parallelogram and \(h\) is its height (altitude to the base).
Consider the blue triangle below:
It's area is \(\frac{1}{2}*AD*(height \ to \ AD)\) but \(AD*(height \ to \ AD)\) 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
\(\frac{1}{2}*AE*(height \ to \ AB)+\frac{1}{2}*EB*(height \ to \ AB) =\)
\(\frac{1}{2}*(height \ to \ AB)*(AE+EB)= \frac{1}{2}*(height \ to \ AB)*AB\). The same way as above, \((height \ to \ AB)*AB\) 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):
\(x+79+y+10=x+(red \ region)+72+y+8\);
\((red \ region)=89-80=9\).
Answer: C
General Discussion
20 May 2020, 20:04
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Bunuel
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 \(\triangle DEF + \triangle BFC=\triangle ADE+\triangle BEF\)
\(b+79+c+e+10+f=x+a+72+d+8.......b+c+e+f+89=x+a+d+80........b+c+e+f+9=x+a+d\)....(I)
Now, if you look at the parallel sides AD and CB..
The areas of \(\triangle ABG + \triangle CDG=\triangle ADG\)
\(x+b+72+e+c+8+f=a+79+d+10.......b+c+e+f+80=a+d+89........x+b+c+e+f-9=a+d\)...(II)
Substitute the value of a+d from (II) in (I)
\(b+c+e+f+9=x+a+d\)
\(b+c+e+f+9=x+x+b+c+e+f-9\)
\(2x=18\)
\(x=9\)
C
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