Last visit was: 31 Oct 2024, 16:01 It is currently 31 Oct 2024, 16:01

Close

GRE Prep 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
Your Progress

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

Practice
Pays

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
avatar
Manager
Manager
Joined: 12 Jan 2016
Posts: 142
Own Kudos [?]: 185 [3]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 12 Apr 2018
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 02 Apr 2018
Posts: 9
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
avatar
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 133 [0]
Given Kudos: 0
Send PM
Re: ABCD is a square ......... [#permalink]
Answer should be C.

Let side of Square = \(a\)

If AEC is 1/4 of circumference it means that Angle ADC is 90 as the circle is cut into 4 quarters. (also evident from the fact that ADC is the angle of a square)
Since AD is the radius as well, we can find the area of the arc AEC. Similarly we can find the area of the Arc AFC. Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square. This gives us the area of the shaded region.

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Area of Unshaded portion = Area of Square - Area of Shaded portion

= \(a^2 - (2*\frac{\pi*a^2}{4}-a^2)\)\(=\) \(2*a^2 - \frac{\pi*a^2}{2}=\frac{4*a^2 - \pi*a^2}{2}\)

\(\frac{Area of Unshaded portion}{Area of Shaded portion}=\frac{4*a^2 - \pi*a^2}{\pi*a^2-2*a^2}=\frac{4 - \pi}{\pi-2}\)

Putting \(\pi = \frac{22}{7}\) in above ratio , we get \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)

Answer C
avatar
Intern
Intern
Joined: 14 Jan 2019
Posts: 31
Own Kudos [?]: 21 [0]
Given Kudos: 0
Send PM
Re: ABCD is a square ......... [#permalink]
AE wrote:
Answer should be C.

Let side of Square = \(a\)

If AEC is 1/4 of circumference it means that Angle ADC is 90 as the circle is cut into 4 quarters. (also evident from the fact that ADC is the angle of a square)
Since AD is the radius as well, we can find the area of the arc AEC. Similarly we can find the area of the Arc AFC. Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square. This gives us the area of the shaded region.

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Area of Unshaded portion = Area of Square - Area of Shaded portion

= \(a^2 - (2*\frac{\pi*a^2}{4}-a^2)\)\(=\) \(2*a^2 - \frac{\pi*a^2}{2}=\frac{4*a^2 - \pi*a^2}{2}\)

\(\frac{Area of Unshaded portion}{Area of Shaded portion}=\frac{4*a^2 - \pi*a^2}{\pi*a^2-2*a^2}=\frac{4 - \pi}{\pi-2}\)

Putting \(\pi = \frac{22}{7}\) in above ratio , we get \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)

Answer C


Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)


Please can you simplify this,it's quite confusing.
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 467 [2]
Given Kudos: 0
Send PM
Re: ABCD is a square ......... [#permalink]
1
Expert Reply
1
Bookmarks
Sonalika42 wrote:
ABCD is a square and AEC and AFC are one fourth of the circumference of the circle whose radius is equal to the length of the side of the square ABCD. Ratio of unshaded to shaded region? image is attached below



(A) 22 : 3
(B) 4 : 3
(C) 3 : 4
(D) 7 : 4
(E) cannot be determined

Join AC, it divides the square into 2 equal parts..
Let us take one part ABFC, ABC is isosceles right angled triangle with sides, say 2, so area = (1/2)*2*2=2
Area of ABFC.. ABFC is 1/4th of a circle with side 2, so area = \(\frac{\pi*2^2}{4}=\pi\)
so shaded portion = area of ABFC - area of ABC = \(\pi-2\)
But there are two shade portion, one each side of AC, so area = 2(\(\pi-2\))=\(2\pi-4\)

therefore, the area of unshaded portion = area of square - area of shaded portion = \(2*2-2(\pi-2)=4-2\pi+4\)=\(8-2\pi\)

Ratio unshaded to shaded = \(8-2\pi\):\(2\pi-4\)=\(4-\pi\):\(\pi-2\)=\(\frac{6}{7}:\frac{8}{7}=6:8=3:4\)
C
Attachments

1111.png
1111.png [ 13.63 KiB | Viewed 2864 times ]

avatar
Intern
Intern
Joined: 14 Jan 2019
Posts: 31
Own Kudos [?]: 21 [0]
Given Kudos: 0
Send PM
Re: ABCD is a square ......... [#permalink]
Many thanks! @chetan2u
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5000
Own Kudos [?]: 73 [0]
Given Kudos: 0
Send PM
Re: ABCD is a square ......... [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: ABCD is a square ......... [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
222 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne