Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1312:30 AM EDT
-01:30 AM EDT
Struggling to find the right strategies to score a 99 %ile on GMAT Focus? Riya (GMAT 715) boosted her score by 100-points in just 15 days! Discover how the right mentorship, tailored strategies, and an unwavering mindset can transform your GMAT prep.
Sep 1301:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 1808:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
20 May 2020, 12:42
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
35% (01:57) correct
65%
(02:01)
wrong
based on 75
sessions
History
Date
Time
Result
Not Attempted Yet
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
Attachment:
2020-05-20_2336.png
Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
09 Dec 2021, 07:30
Kudos
Bookmarks
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
Kudos
Bookmarks
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
Attachments
2020-05-20_2336.png [ 107.43 KiB | Viewed 10507 times ]
