GRE Prep 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.
ABCD is a rectangle
[#permalink]
19 Oct 2017, 14:45
19 Oct 2017, 14:45
1
Expert Reply
2
Bookmarks
00:00
Question Stats:
62% (00:53) correct
37% (01:11) wrong based on 135 sessions
Hide Show timer Statistics
Attachment:
#GREpracticequestion ABCD is a rectangle.jpg [ 36 KiB | Viewed 7061 times ]
ABCD is a rectangle
R is the midpoint of AD
Quantity A |
Quantity B |
The area of Triangle APD |
Twice the area of Triangle AQR |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: ABCD is a rectangle
[#permalink]
07 May 2020, 01:40
07 May 2020, 01:40
5
Any 2 triangles constrained within 2 parallel lines with a common base WILL HAVE THE SAME AREA. No matter how skewed the triangle is to either side, because the height remains constant.
Using this logic, you can figure out that since the base of one is half the base of other, the area of the double base triangle will be two times the area of the triangle with half base. Since you are doubling the area of the smaller triangle, the 2 quantities are equal.
Using this logic, you can figure out that since the base of one is half the base of other, the area of the double base triangle will be two times the area of the triangle with half base. Since you are doubling the area of the smaller triangle, the 2 quantities are equal.
General Discussion
Re: ABCD is a rectangle
[#permalink]
20 Oct 2017, 06:34
20 Oct 2017, 06:34
3
Given that R is the midpoint of AD, AR = RD. Then, the area of triangle APD is \(\frac{AD*AB}{2}\), where AB is the height of the triangle corresponding to the shortest side of the rectangle. The double area of triangle AQR, instead, is equal to \(2*\frac{AR*AB}{2} = AR*AB\). Then, given AR = RD and AR+RD = AD, we can rewrite AD as 2AR. Thus, the are of APD becomes \(\frac{2AR*AB}{2} = AR*AB\).
Answer C
Answer C
