|
|
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.
01 Mar 2022, 03:00
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:31) correct
27%
(03:43)
wrong
based on 37
sessions
History
Date
Time
Result
Not Attempted Yet
In the circle above, if the area of the rectangle set inside the circle is 200 and b = 8a, what is the circumference of the circle?
A. \(25\pi \sqrt{65}\)
B. \(5 \sqrt{65}\)
C. \(5\pi \sqrt{13}\)
D. \(5\pi \sqrt{65}\)
E. \(5\pi\)
Attachment:
circle.png [ 7.6 KiB | Viewed 1176 times ]
01 Mar 2022, 11:58
Kudos
Bookmarks
Bunuel
Breaking Down the Info:
We are given a rectangle, so the area is simply \(a*b = 200\), which is \(a*(8a) = 200\) and \(a = 5\).
Then \(b = 8a = 40\). Finally to find the circumference, we start with the diameter of the circle which can be calculated from the Pythagorean theorem.
\(D = \sqrt{5^2 + 40^2} = \sqrt{5^2*(1 + 8^2)} = 5\sqrt{65}\).
Multiply that by \(\pi\) to find the circumference.
Answer: D
kungfury42
Joined: 07 Jan 2022
Last visit: 31 May 2023
Posts: 580
Own Kudos:
Given Kudos: 724
Schools: NUS '25 (A)
GMAT 1: 740 Q51 V38
GPA: 4
Products:
01 Mar 2022, 14:32
Kudos
Bookmarks
Area of the rectangle \(= L*B = a*b\)
But since \(b=8a\)
Area \(= a*8a = 8a^2 = 200\)
This gives us \(a^2=25\) or \(a=5\) and correspondingly \(b=8*5=40\)
Now we know that every angle in a rectangle is equal to \(90°\) and we also know that an angle in a semicircle is also equal to \(90°\) hence we can conveniently say that the diagonal of this rectangle will be the diameter of the circle.
Diagonal
\(= \sqrt{a^2+b^2}\\
= \sqrt{5^2+40^2}\\
= \sqrt{25+1600}\\
= \sqrt{1625}\\
= \sqrt{25*65}\\
= 5*\sqrt{65}\)
Hence, Diameter of circle
\(= 5*\sqrt{65}\)
Circumference of circle
\(= 2*\pi*r\\
= \pi*d\\
= \pi*5*\sqrt{65}\\
= 5*\pi*\sqrt{65}\)
Hence, option D.
Posted from my mobile device
But since \(b=8a\)
Area \(= a*8a = 8a^2 = 200\)
This gives us \(a^2=25\) or \(a=5\) and correspondingly \(b=8*5=40\)
Now we know that every angle in a rectangle is equal to \(90°\) and we also know that an angle in a semicircle is also equal to \(90°\) hence we can conveniently say that the diagonal of this rectangle will be the diameter of the circle.
Diagonal
\(= \sqrt{a^2+b^2}\\
= \sqrt{5^2+40^2}\\
= \sqrt{25+1600}\\
= \sqrt{1625}\\
= \sqrt{25*65}\\
= 5*\sqrt{65}\)
Hence, Diameter of circle
\(= 5*\sqrt{65}\)
Circumference of circle
\(= 2*\pi*r\\
= \pi*d\\
= \pi*5*\sqrt{65}\\
= 5*\pi*\sqrt{65}\)
Hence, option D.
Posted from my mobile device
