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 ?
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))=
[(pi)(r^2)]/2Now
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.
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The Graceful
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