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.
If the area of the quadrilateral ABCD in the above figure is 180 and
[#permalink]
14 Jan 2025, 17:01
14 Jan 2025, 17:01
Expert Reply
00:00
Question Stats:
100% (02:40) correct
0% (00:00) wrong based on 5 sessions
Hide Show timer Statistics
Attachment:
GRE geometry polgon question.png [ 26.53 KiB | Viewed 617 times ]
If the area of the quadrilateral ABCD in the above figure is 180 and TD, the shortest distance of the side AB from vertex D, is parallel to BC, what is the value of X?
A. 10
B. 15
C. 20
D. 25
E. 28
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: If the area of the quadrilateral ABCD in the above figure is 180 and
[#permalink]
23 Feb 2025, 01:09
23 Feb 2025, 01:09
Expert Reply
Area of the quadrilateral ABCD is same as the sum of the area of trapezium BTDC and the area of right triangle
$\(\mathrm{ATD}=\frac{1}{2} \times(\mathrm{TD}+\mathrm{BC}) \times \mathrm{TB}+\frac{1}{2} \times \mathrm{AT} \times \mathrm{TD}=\frac{1}{2} \times(10+5) \times \mathrm{x}+\frac{1}{2} \times 6 \times 10\)$ which is same as $\(7.5 x+30=180\)$, so we get $\(x=20\)$
$\(\mathrm{ATD}=\frac{1}{2} \times(\mathrm{TD}+\mathrm{BC}) \times \mathrm{TB}+\frac{1}{2} \times \mathrm{AT} \times \mathrm{TD}=\frac{1}{2} \times(10+5) \times \mathrm{x}+\frac{1}{2} \times 6 \times 10\)$ which is same as $\(7.5 x+30=180\)$, so we get $\(x=20\)$
Re: If the area of the quadrilateral ABCD in the above figure is 180 and
[#permalink]
06 Aug 2025, 10:11
06 Aug 2025, 10:11
Expert Reply
Bump for further discussions
