Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
03 Apr 2020, 06:48
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:34) correct
15%
(01:44)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 09 Dec 2024
Posts: 3,116
Own Kudos:
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
03 Apr 2020, 07:35
Kudos
Bookmarks
Since the radius of the circle is 13 cm. The length of diagonal is 26cm.
Using Pythagoras theroam.
(Diagonal)^2=(length)^2+(breadth)^2
(26)^2=(L)^2+(B)^2
676=l^2+b^2---------(1)
Now the possible length of length and breadth of rectangle is
1. 25^2+12^2=625+144=769 (not equal to 676)
2. 25^2+10^2=625+100=725 (not equal to 676)
3. 25^2+9^81=625+81=706 (not equal to 676)
4. 24^2+12^2=576+144=720 (not equal to 676)
5. 24^2+10^2=576+100=676 (equal to 676(26)^2)
So OA is E
Posted from my mobile device
Using Pythagoras theroam.
(Diagonal)^2=(length)^2+(breadth)^2
(26)^2=(L)^2+(B)^2
676=l^2+b^2---------(1)
Now the possible length of length and breadth of rectangle is
1. 25^2+12^2=625+144=769 (not equal to 676)
2. 25^2+10^2=625+100=725 (not equal to 676)
3. 25^2+9^81=625+81=706 (not equal to 676)
4. 24^2+12^2=576+144=720 (not equal to 676)
5. 24^2+10^2=576+100=676 (equal to 676(26)^2)
So OA is E
Posted from my mobile device
03 Apr 2020, 08:22
Kudos
Bookmarks
Bunuel
Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
A. 25, 12
B. 25, 10
C. 25, 9
D. 24, 12
E. 24, 10
Are You Up For the Challenge: 700 Level Questions
The rectangle is inside the circle, the diameter of the circle is equal to the diagonal of the rectangle.
Then the rectangle must have a diagonal of exactly 13*2 = 26 cm.
Therefore we want to find the combination of length/width that gives a diagonal of 26 cm.
If you know your Pythagorean integer solutions well, then \(13^2 = 12^2 + 5^2\) should come to mind. We can double it to get \(26^2 = 24^2 + 10^2\), hence the dimensions of the rectangle must be 24 * 10.
Ans: E
