GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 10 Dec 2018, 18:44

### 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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
• ### Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.

# ABCD is a square and AEC and AFC are one fourth of the circumference

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51072
ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

03 Jul 2018, 08:25
00:00

Difficulty:

95% (hard)

Question Stats:

40% (04:02) correct 60% (02:50) wrong based on 40 sessions

### HideShow timer Statistics

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

Attachment:

Image50.gif [ 2.19 KiB | Viewed 911 times ]

_________________
Senior Manager
Joined: 22 Feb 2018
Posts: 411
Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

03 Jul 2018, 11:44
OA: C

Let side of Square = $$a$$

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 $$\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}$$
_________________

Good, good Let the kudos flow through you

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3325
Location: India
GPA: 3.12
Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

03 Jul 2018, 12:03

Let's assume the side of the square to be 7. The area of this square is 49.

Similarly, as the length of the sector is also 7, the area of one sector $$\frac{1}{4}*\frac{22}{7}*7^2 = \frac{77}{2}$$
There are 2 such sectors AEC and AFC each having the same area, making the total area $$77$$

We use the following diagram to find the area of the shaded region
Attachment:
Image50.gif

Area of AEC = 1 + 2 | Area of AFC = 2 + 3 | Area of Square = 1 + 2 + 3

The area of shaded region = Area of AEC + Area of AFC - Area of Square = 77 - 49 = 28
Area of unshaded region = Area of square - Area of shaded region = 49 - 28 = 21

Therefore, the ratio of the unshaded to the shaded region is 21:28 or 3:4(Option C)
_________________

You've got what it takes, but it will take everything you've got

Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

03 Jul 2018, 12:14
Bunuel 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

Attachment:
Image50.gif

Area of unshaded region=Area of square-2*Area of segment AEC or AFC

Let the side of the square be 'a' unit.

Area of segment= Area of sector-Area of triangle=$$\frac{1}{2}\pi*a-\frac{1}{2} \sqrt{2}a*\frac{a}{\sqrt{2}}$$
=$$\frac{\pi a^2}{4}-\frac{a^2}{2}$$
=$$\frac{a^2(\pi-2)}{4}$$

Now, the required ratio=$$\frac{A_{shaded}}{A_{unshaded}}$$=$$\frac{Area of square}{2*Area of segments AEC or AFC}-1$$
=$$\frac{a^2}{a^2(\pi-2)/2}-1$$
=$$\frac{2}{\pi-2}-1$$
=$$\frac{4-\pi}{\pi-2}$$
=$$\frac{3}{4}$$

Ans. C
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Intern
Joined: 15 Aug 2012
Posts: 42
Schools: AGSM '19
Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

04 Jul 2018, 11:51
Princ wrote:
OA: C

Let side of Square = $$a$$

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 $$\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}$$

I'm confused how you get the first equation. Could you please elaborate? It says AEC and AFC are 1/4 of circumference i.e. 1/4*2*pi*a

I fail to see how this came up: $$2*\frac{\pi*a^2}{4}-a^2$$
Intern
Joined: 23 Feb 2018
Posts: 11
ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

04 Jul 2018, 12:07
2
Quote:
I'm confused how you get the first equation. Could you please elaborate? It says AEC and AFC are 1/4 of circumference i.e. 1/4*2*pi*a

I fail to see how this came up: $$2*\frac{\pi*a^2}{4}-a^2$$

Hi,

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.

Go through the solution posted by Princ now, should make things clearer.
Senior Manager
Joined: 22 Feb 2018
Posts: 411
Re: ABCD is a square and AEC and AFC are one fourth of the circumference  [#permalink]

### Show Tags

05 Jul 2018, 10:34
1
rajudantuluri

munzuto has already explained how I got to first equation.
In this question, Double counting concept is similar to two overlapping set problem.

_________________

Good, good Let the kudos flow through you

Re: ABCD is a square and AEC and AFC are one fourth of the circumference &nbs [#permalink] 05 Jul 2018, 10:34
Display posts from previous: Sort by