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

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

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New post 27 Nov 2019, 01:56
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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

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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 27 Nov 2019, 04:41
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Bunuel wrote:
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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
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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 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??
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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 08 Dec 2019, 22:26
Dillesh4096

Never mind
understood my mistake
Thanks
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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 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!

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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 09 Dec 2019, 04:43
Bunuel wrote:
Image
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)
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Re: In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 10 Dec 2019, 23:41
How did we get Area of 1/4th of circle ? Can somebody please explain this.
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In the figure above, point B has an x-coordinate of 0.5 and a y-coordi  [#permalink]

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New post 10 Dec 2019, 23:47
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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

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