Help PAT Question

Find the area of the shaded region of the circle in the gure below, as a
function of the radius R.
studentroom question.JPG

What is try to do is make an equilateral triangle such as this : studentroom question 2.jpg

Where the height of the equilateral triangle is

\frac{3}{2}

pi
, bit can't seem to be able to get the height of this triangle. If I did get it then it would just be the area of the circle, minus the area of the triangle , and then divide that by 3.

Thanks in advance!

Reply 1

journeyinwards

1

Consider the upper half of the shaded region. You could set up an integral for y=sqrt(r^2-x^2) and integrate it from x=r/2 to r. (Note that y is positive throughout this region)
This would give you the area of the upper half. For the total area, multiply that by 2.

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Reply 2

BobLoblawLawBlog

16

Original post by Zoffmann

Find the area of the shaded region of the circle in the gure below, as a
function of the radius R.
studentroom question.JPG

What is try to do is make an equilateral triangle such as this : studentroom question 2.jpg

Where the height of the equilateral triangle is

\frac{3}{2}

pi
, bit can't seem to be able to get the height of this triangle. If I did get it then it would just be the area of the circle, minus the area of the triangle , and then divide that by 3.

Thanks in advance!

Try looking at the sector of the circle going from the centre of the circle to each end of the shaded area.

Reply 3

journeyinwards

1

Or you could also draw 2 radii from the centre to the endpoints of the chord.
That would give you 2 right angled triangles with height=r/2 and hypotenuse=r. Find the base using Pythagoras' theorem and hence find the area of each of the triangles.
Also find the angle subtended by the chord at the centre using trigonometry (it'll be 60°). Use this angle to find the area Of the sector.
Area of shaded region=area of sector-(2*area of right angled triangles found above)

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Reply 4

Zoffmann

OP

Original post by journeyinwards

Or you could also draw 2 radii from the centre to the endpoints of the chord.
That would give you 2 right angled triangles with height=r/2 and hypotenuse=r. Find the base using Pythagoras' theorem and hence find the area of each of the triangles.
Also find the angle subtended by the chord at the centre using trigonometry (it'll be 60°). Use this angle to find the area Of the sector.
Area of shaded region=area of sector-(2*area of right angled triangles found above)

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I love you

Reply 5

journeyinwards

1

Original post by Zoffmann

I love you

Lol, best of luck with your application!

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Reply 6

Zoffmann

OP

Original post by journeyinwards

Lol, best of luck with your application!

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hahah thanks! help is much appreciated!

Reply 7

ZafarS

5

I had trouble with this problem too, but it was because I didn't know the area for a circle sector, which is

\dfrac{1}{2}R^2\theta

So draw 2 lines from the centre to the edges of the shaded area, those are obviously

R

. Use pythagoras to the base of the triangle formed, which is

\sqrt{3}

. We get that the area of triangle is

\dfrac{1}{4}\sqrt{3}R^2

. The area of the segment you want is

\dfrac{1}{2}R^2\theta - \dfrac{1}{4}\sqrt{3}R^2

. Voila.. except that you need to find what

\theta

is. Luckily this is trivial: You just use trig to get the area of the triangle once more, you'll get

\dfrac{1}{2}R^2 \sin\theta

. Now that you have 2 expressions for the area, you just solve the equality and get that

\theta = \dfrac{2}{3} \pi

. And voila, you're done.

Reply 8

sarjan100

1

I got a final answer of (1/2)R^2((1/3)pie - 1)

To the person above I'm pretty sure the angle for a sector is ( 1/2 )( r^2 )(Angle in radians)<===(Not sin(theta))

(edited 10 years ago)

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