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

Question

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 ?

https://uploads.tapatalk-cdn.com/20180823/a9e3149c79d2f9b2db98ad613951ff7c.jpg

TheGraceful · Answer

Answer for Q1: 
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.

TheGraceful · Answer

Answer for Q2: 
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))=  /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 =   +  -  
=================>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.

DyingNut · Answer

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.
Thank you for your reply again.

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

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