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

 It is currently 11 Dec 2019, 22:53

### 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
Your Progress

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 above, point B has an x-coordinate of 0.5 and a y-coordi

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59674
In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

27 Nov 2019, 01:56
00:00

Difficulty:

65% (hard)

Question Stats:

64% (02:57) correct 36% (03:03) wrong based on 22 sessions

### HideShow timer Statistics

In the figure above, point B has an x-coordinate of 0.5 and a y-coordinate that is higher than any other point on the semi-circle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units?

A. $$\frac{2\pi}{3}$$

B. $$\pi-\frac{\sqrt{3}}{8}$$

C. $$\pi-1$$

D. $$2\pi-\frac{3}{2}$$

E. $$\frac{3\pi}{8}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:

102Geometry.jpg [ 11.33 KiB | Viewed 430 times ]

_________________
VP
Joined: 20 Jul 2017
Posts: 1145
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

27 Nov 2019, 04:41
1
Bunuel wrote:

In the figure above, point B has an x-coordinate of 0.5 and a y-coordinate that is higher than any other point on the semi-circle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units?

A. $$\frac{2\pi}{3}$$

B. $$\pi-\frac{\sqrt{3}}{8}$$

C. $$\pi-1$$

D. $$2\pi-\frac{3}{2}$$

E. $$\frac{3\pi}{8}$$

Let a perpendicular dropped from Point B meets x-axis at point C
--> ACB is a right angle triangle with ∠C = 90 deg

--> Area of Shaded region = 1/4th the area of circle - Area of the triangle ABC

Let the point B = (0.5, b) & A = (a, 0)
--> y = 2x + 1
--> b = 2*0.5 + 1 = 1 + 1 = 2
--> Point B = (0.5, 2)

For point A,
0 = 2*a + 1
--> 2a = -1
--> a = -0.5
--> Point A = (-0.5, 0)
So, Point C = (0.5, 0)
--> Radius of the circle = BC = 2 - 0 = 2 units
AC = 0.5 - (-0.5) = 1

Area of Shaded region = 1/4*π*$$2^2$$ - 1/2*AC*BC = π - 1/2*1*2 = π - 1

IMO Option C
Manager
Joined: 19 Feb 2019
Posts: 104
Concentration: Marketing, Statistics
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

08 Dec 2019, 22:25
Quote:
Let a perpendicular dropped from Point B meets x-axis at point C
--> ACB is a right angle triangle with ∠C = 90 deg

--> Area of Shaded region = 1/4th the area of circle - Area of the triangle ABC

Shouldn't it be Area of shaded region = Area of semi circle - [Area of 1/4th circle +Area of triangle]??

I did not understand how you got 1/4th area of circle - Area of traingle??
Manager
Joined: 19 Feb 2019
Posts: 104
Concentration: Marketing, Statistics
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

08 Dec 2019, 22:26
Dillesh4096

Never mind
understood my mistake
Thanks
VP
Joined: 20 Jul 2017
Posts: 1145
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

08 Dec 2019, 23:31
devavrat wrote:
Quote:
Let a perpendicular dropped from Point B meets x-axis at point C
--> ACB is a right angle triangle with ∠C = 90 deg

--> Area of Shaded region = 1/4th the area of circle - Area of the triangle ABC

Shouldn't it be Area of shaded region = Area of semi circle - [Area of 1/4th circle +Area of triangle]??

I did not understand how you got 1/4th area of circle - Area of traingle??

Hi devavrat,

See the highlighted part,
Area of shaded region = Area of semi circle - [Area of 1/4th circle +Area of triangle]

We know, Area of semicircle = 1/2 of area of circle!!

So, Area of shaded region = Area of semi circle - [Area of 1/4th circle +Area of triangle]
= 1/2 of area of circle - [Area of 1/4th circle +Area of triangle]
= (1/2 - 1/4) of area of circle - Area of triangle
= 1/4 of area of circle - Area of triangle !!!
Which is what I have mentioned in the formula.

Hope it’s clear!

Posted from my mobile device
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9871
Location: Pune, India
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

09 Dec 2019, 04:43
Bunuel wrote:

In the figure above, point B has an x-coordinate of 0.5 and a y-coordinate that is higher than any other point on the semi-circle. If AB lies on the line y=2x+1, what is the area of the shaded region, in square units?

A. $$\frac{2\pi}{3}$$

B. $$\pi-\frac{\sqrt{3}}{8}$$

C. $$\pi-1$$

D. $$2\pi-\frac{3}{2}$$

E. $$\frac{3\pi}{8}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:
102Geometry.jpg

"point B has an x-coordinate of 0.5"
Also, it lies on y = 2x + 1. So y = 2.
Co-ordinates of B are (0.5, 2) and it is the radius of the circle from the centre, say C (0.5, 0) lying on the x axis.
Co-ordinates of point A: 0 = 2x + 1 (because at A, y coordinate is 0). So A (-0.5, 0)

So area of shaded region is:
Area of quarter of a circle - Area of triangle ABC = $$(1/4)*4*\pi - (1/2)*1*2 = \pi - 1$$

Answer (C)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Joined: 23 May 2019
Posts: 8
Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

10 Dec 2019, 23:41
How did we get Area of 1/4th of circle ? Can somebody please explain this.
Manager
Joined: 16 Feb 2015
Posts: 160
Location: United States
Concentration: Finance, Operations
Schools: INSEAD, ISB
In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

### Show Tags

10 Dec 2019, 23:47
1
aarushisingla wrote:
How did we get Area of 1/4th of circle ? Can somebody please explain this.

Dear aarushisingla ,

As Given in data, we have semicircle. And Point B is at the highest level of semicircle.
Suppose, a perpendicular dropped from Point B meets x-axis at point C
By the Co-ordinates available, We can find BC which is Radius of the circle = 2 units
Area of 1/4 of circle = 1/4 * π * r^2 = π

For finding the area of the shaded region, We are subtracting area of Triangle with 1/4 of circle.

I hope you will get it. Do PM if you didn't understand

Regards,
Rajat Chopra
In the figure above, point B has an x-coordinate of 0.5 and a y-coordi   [#permalink] 10 Dec 2019, 23:47
Display posts from previous: Sort by

# In the figure above, point B has an x-coordinate of 0.5 and a y-coordi

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

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