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# Circles

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this Question is based on cirlces

Oct 12, 2018

#1
+7601
+2

Let the radius of the smaller circle  =  s

Let the radius of the bigger circle  =  b

And let's label the point of tangency " P ".

Side  s  is drawn from the center of the circle to  P ,  so side  s  meets  AB  at a right angle.

And by the hypotenuse-leg theorem, the two triangles are congruent.  And  AB =  80  So...

AP + PB  =  80

And we know  AP  =  PB

PB + PB  =  80

2 * PB  =  80

PB  =  40

By the Pythagorean theorem....

402  +  s2  =  b2

1600 + s2  =  b2

1600  =  b2 - s2

area of shaded region  =  area of bigger circle - area of smaller circle

area of shaded region  =   π b2   -   π s2

area of shaded region  =  π( b2 - s2 )

area of shaded region  =  1600π     (sq units)

Oct 12, 2018

#1
+7601
+2

Let the radius of the smaller circle  =  s

Let the radius of the bigger circle  =  b

And let's label the point of tangency " P ".

Side  s  is drawn from the center of the circle to  P ,  so side  s  meets  AB  at a right angle.

And by the hypotenuse-leg theorem, the two triangles are congruent.  And  AB =  80  So...

AP + PB  =  80

And we know  AP  =  PB

PB + PB  =  80

2 * PB  =  80

PB  =  40

By the Pythagorean theorem....

402  +  s2  =  b2

1600 + s2  =  b2

1600  =  b2 - s2

area of shaded region  =  area of bigger circle - area of smaller circle

area of shaded region  =   π b2   -   π s2

area of shaded region  =  π( b2 - s2 )

area of shaded region  =  1600π     (sq units)

hectictar Oct 12, 2018
#2
+100595
+2

Very nice, hectictar.....!!!

CPhill  Oct 12, 2018
#3
+464
+1

Thank you Hectictar

Oct 13, 2018