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# Show that the area of the shaded region = (pi+1)r cm

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Intern
Joined: 22 Aug 2018
Posts: 3
Show that the area of the shaded region = (pi+1)r cm  [#permalink]

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23 Aug 2018, 00:21
1
[Refer to the picture below]

Given the triangle QRT is equilateral and the radius of both circles is r cm, show that the area of the shaded region = (pi+1)r cm.

I tried multiple times but always end up with (pi)(r^2)/6 - (1/2)sin60(r^2). Is there any solutions for this question ?

Manager
Joined: 10 Apr 2018
Posts: 130
GMAT 1: 600 Q44 V28
GPA: 3.56
WE: Engineering (Computer Software)
Re: Show that the area of the shaded region = (pi+1)r cm  [#permalink]

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23 Aug 2018, 01:23
DyingNut wrote:
[Refer to the picture below]

Given the triangle QRT is equilateral and the radius of both circles is r cm, show that the area of the shaded region = (pi+1)r cm.

I tried multiple times but always end up with (pi)(r^2)/6 - (1/2)sin60(r^2). Is there any solutions for this question ?

Since radius is 'r' and both circles pass through center of each other, distance between two centres is 'r' i.e. QR=r.
QT is radius of first circle (centered at 'Q') and therefor has a value of 'r'
Similarly RT is radius of second circle (circle centred at R) and therefor has a value of 'r'

SInce, all the three sides are having a value of 'r', it's an equilateral triangnle with eavey angle having a value 60 degree.
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The Graceful
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Every EXPERT was a beginner once...
Don't look at the clock. Do what it does, keep going
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To achieve great things, two things are needed:a plan and not quite enough time - Leonard Bernstein.
Manager
Joined: 10 Apr 2018
Posts: 130
GMAT 1: 600 Q44 V28
GPA: 3.56
WE: Engineering (Computer Software)
Re: Show that the area of the shaded region = (pi+1)r cm  [#permalink]

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23 Aug 2018, 01:45
DyingNut wrote:
[Refer to the picture below]

Given the triangle QRT is equilateral and the radius of both circles is r cm, show that the area of the shaded region = (pi+1)r cm.

I tried multiple times but always end up with (pi)(r^2)/6 - (1/2)sin60(r^2). Is there any solutions for this question ?

Area of shaded region = Area of semicircle QRSTQ - Area covered between points QRT

Area of circle with cenre at R= (pi)(r^2)
Therefore Area of semicircle is=1/2(Area of circle with cenre at R)=1/2((pi)(r^2))=[(pi)(r^2)]/2

Now
Since angle QRT is 60 degree, we can get the area of sector QRT (however, this is without area beyond line TR) using formula:
A(Arc QRT)/Area of circle with center of R =Central anagle QRT/360
==>A(Arc QRT)/(pi)(r^2)=60/360
==>A(Arc QRT)/(pi)(r^2)=1/6
==>A(Arc QRT)=(pi)(r^2)/6

Now, similarly, we can get the area of sector TQR (however, this is without area beyond line QT) using formula:
A(Arc TQR)/Area of circle with center of Q =Central anagle QRT/360
==>A(Arc TQR)/(pi)(r^2)=60/360
==>A(Arc TQR)/(pi)(r^2)=1/6
==>A(Arc TQR)=(pi)(r^2)/6

Area covered by QTR = Area of sector QTR + Area of sector TQR - Area of traianle QTR
==>Area covered by QTR = [(pi)(r^2)/6] +[(pi)(r^2)/6] - [(r^2)Sq.root of 3/4]
=================>2[(pi)(r^2)/6]-[(r^2)Sq.root of 3/4]
=================>[(pi)(r^2)/3]-[(r^2)Sq.root of 3/4

......
intent of reply was to give you the idea, hope you can calculate and see that it gives the answer stated in question.
_________________

The Graceful
----------------------------------------------------------
Every EXPERT was a beginner once...
Don't look at the clock. Do what it does, keep going
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To achieve great things, two things are needed:a plan and not quite enough time - Leonard Bernstein.
Intern
Joined: 22 Aug 2018
Posts: 3
Show that the area of the shaded region = (pi+1)r cm  [#permalink]

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23 Aug 2018, 03:11
Thank for your help. I think the problem itself has the problem. I will refer the question to my teacher because I found the same question in one of my reference books. The question is the same but the answer (pi+1)r cm is referred to the perimeter of the shaded region :/ ..... Both reference book and the paper have the same mistake. Not to mention that your answer teaches me to look the question at a different perspective.

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Show that the area of the shaded region = (pi+1)r cm   [#permalink] 23 Aug 2018, 03:11
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