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 2312:30 AM EDT
-01:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Sep 2401:30 AM EDT
-01:30 AM EDT
Tejas studied hard but kept repeating the same mistakes. After 3 attempts at 645, a focused 4-week strategy helped him identify and fix what was holding him back—leading to a 695 with just 12–15 hours of study per week.
Sep 2606:00 PM PDT
-07:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
04 May 2020, 07:02
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (02:37) correct
33%
(02:10)
wrong
based on 109
sessions
History
Date
Time
Result
Not Attempted Yet
[GMAT math practice question]
The figure shows a quadrangle \(ABCD\) with \(∠ADC = 90°, ∠DCB = 90°\) and \(AD = 6, AB = 8\) and \(BC = 10.\) What is the length \(BD\)?
5.4ps.png [ 14.35 KiB | Viewed 2665 times ]
A. \(2\sqrt{37}\)
B. \(10 \)
C. \(3\sqrt{39 }\)
D. \(14\)
E. \(5\sqrt{41}\)
The figure shows a quadrangle \(ABCD\) with \(∠ADC = 90°, ∠DCB = 90°\) and \(AD = 6, AB = 8\) and \(BC = 10.\) What is the length \(BD\)?
Attachment:
5.4ps.png [ 14.35 KiB | Viewed 2665 times ]
A. \(2\sqrt{37}\)
B. \(10 \)
C. \(3\sqrt{39 }\)
D. \(14\)
E. \(5\sqrt{41}\)
04 May 2020, 07:14
Kudos
Bookmarks
Draw a perpendicular from A to BC and make a square. Call it P. Now AB is the hypotenuse and BP is the base. AP is the height.
Now AP = \(\sqrt{AB^2-BP^2}\) = \(\sqrt{64-16} = 4\sqrt{3}\)
Now, AP = DC = 4\(\sqrt{3}\)
Now, BCD is a right angled triangle. Hypotenuse BD = \(\sqrt{100+48} = 2\sqrt{37}\)
A is the answer.
Now AP = \(\sqrt{AB^2-BP^2}\) = \(\sqrt{64-16} = 4\sqrt{3}\)
Now, AP = DC = 4\(\sqrt{3}\)
Now, BCD is a right angled triangle. Hypotenuse BD = \(\sqrt{100+48} = 2\sqrt{37}\)
A is the answer.
Attachments
5.4ps.png [ 15.66 KiB | Viewed 2572 times ]
06 May 2020, 01:21
Kudos
Bookmarks
=>
5.4PS(A).png [ 18.47 KiB | Viewed 2498 times ]
From the above figure, since we have \(EC = 6\), we have \(BE = 4\). Then we use Pythagorean Theorem to solve for \(AE\), giving us\( AE = \sqrt{8^2- 4^2} = \sqrt{64-16} = \sqrt{48} = 4√3\) and \(AE = DC = 4√3.\)
Thus, we have \(x = \sqrt{10^2+(4√3)^2} = \sqrt{100+48} = 2√37.\)
Therefore, A is the answer.
Answer: A
Attachment:
5.4PS(A).png [ 18.47 KiB | Viewed 2498 times ]
From the above figure, since we have \(EC = 6\), we have \(BE = 4\). Then we use Pythagorean Theorem to solve for \(AE\), giving us\( AE = \sqrt{8^2- 4^2} = \sqrt{64-16} = \sqrt{48} = 4√3\) and \(AE = DC = 4√3.\)
Thus, we have \(x = \sqrt{10^2+(4√3)^2} = \sqrt{100+48} = 2√37.\)
Therefore, A is the answer.
Answer: A
