It is currently 22 Apr 2019, 14:21

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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

In the figure below, square ABCD is inscribed in circle O. I

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 6215
Followers: 99

Kudos [?]: 1201 [0], given: 5727

In the figure below, square ABCD is inscribed in circle O. I [#permalink]  02 Feb 2019, 02:27
Expert's post
00:00

Question Stats:

42% (01:08) correct 57% (02:06) wrong based on 7 sessions
In the figure below, square ABCD is inscribed in circle O. If the perimeter of ABCD is 24, what is the area of the shaded region?

Attachment:

#GREpracticequestion In the figure below, square ABCD.jpg [ 13.34 KiB | Viewed 642 times ]

A) $$18 \pi - 36$$

B) $$18 \pi - 24$$

C) $$12 \pi - 36$$

D) $$9 \pi - 36$$

E) $$9 \pi - 24$$
[Reveal] Spoiler: OA

_________________
Director
Joined: 20 Apr 2016
Posts: 855
WE: Engineering (Energy and Utilities)
Followers: 11

Kudos [?]: 631 [1] , given: 134

Re: In the figure below, square ABCD is inscribed in circle O. I [#permalink]  02 Feb 2019, 04:15
1
KUDOS
Carcass wrote:
In the figure below, square ABCD is inscribed in circle O. If the perimeter of ABCD is 24, what is the area of the shaded region?

Attachment:
#GREpracticequestion In the figure below, square ABCD.jpg

A) $$18 \pi - 36$$

B) $$18 \pi - 24$$

C) $$12 \pi - 36$$

D) $$9 \pi - 36$$

E) $$9 \pi - 24$$

Explanation::

Perimeter of the square = 24

i.e. 4*Side of square = 24

or side of the square = 6

Area of square = $$6^2 = 36$$

Now Area of square = $$\frac{{diagonal}^2}{2}$$

or $$36 * 2 = diagonal^2$$

or diagonal = diameter of the circle = $$6\sqrt{2}$$

Area of the circle = $$\pi * radius^2 = \pi * (3\sqrt{2})^2 = 18\pi$$

Required area = Area of circle - Area of square

= $$18\pi − 36$$
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html

Re: In the figure below, square ABCD is inscribed in circle O. I   [#permalink] 02 Feb 2019, 04:15
Display posts from previous: Sort by