A. \(\dfrac{5\sqrt2\pi}2\)
B. \(\dfrac{25\pi}4\)
C. \(\dfrac{25\pi}2\)
D. \(5\sqrt2\pi\)
E. \(25\pi\)
Answer: C
Source: Princeton Review
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
In the figure above, square \(ABCD\) has an area of \(25.\) What is the area of the circle with center \(O?\)
This topic has expert replies
-
Gmat_mission
- Legendary Member
- Posts: 1622
- Joined: Thu Mar 01, 2018 7:22 am
- Followed by:2 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
In the figure above, square \(ABCD\) has an area of \(25.\) What is the area of the circle with center \(O?\)
A. \(\dfrac{5\sqrt2\pi}2\)
B. \(\dfrac{25\pi}4\)
C. \(\dfrac{25\pi}2\)
D. \(5\sqrt2\pi\)
E. \(25\pi\)
Answer: C
Source: Princeton Review
A. \(\dfrac{5\sqrt2\pi}2\)
B. \(\dfrac{25\pi}4\)
C. \(\dfrac{25\pi}2\)
D. \(5\sqrt2\pi\)
E. \(25\pi\)
Answer: C
Source: Princeton Review
-
ravinikhil
- Junior | Next Rank: 30 Posts
- Posts: 11
- Joined: Thu Sep 10, 2020 6:03 am
ABCD is a square with area 25.
now let side length be a.
hence a^2 = 25
ie a = 5.
Now, a= AD = 5.
Imagine drawing a line from O to X which is on AD and bisects it in half.
Now consider AOX triangle
here AX is 5/2 in length, since ABCD is a square , hence angle OAD will be 45 degrees.
Now OA cos 45 = AX. { understood right }
OA / sqrt ( 2 ) = 5/2
OA = 5 * sqrt (2) / 2
Area of the circle is pie radius^2.
radius is OA.
hence the overall area is = pie * 25 / 2
now let side length be a.
hence a^2 = 25
ie a = 5.
Now, a= AD = 5.
Imagine drawing a line from O to X which is on AD and bisects it in half.
Now consider AOX triangle
here AX is 5/2 in length, since ABCD is a square , hence angle OAD will be 45 degrees.
Now OA cos 45 = AX. { understood right }
OA / sqrt ( 2 ) = 5/2
OA = 5 * sqrt (2) / 2
Area of the circle is pie radius^2.
radius is OA.
hence the overall area is = pie * 25 / 2
