Attachment 24296

I don't know how to do this question and the correct answer isA.

Can anybody tell me how to proof that theAnswer Ais correct?

And how to find82?π

Thanks for answering.

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- Jul 14th 2012, 08:36 PMwilliam24The problem of shaded region's area
Attachment 24296

I don't know how to do this question and the correct answer is**A**.

Can anybody tell me how to proof that the**Answer A**is correct?

And how to find**82**?**π**

Thanks for answering. - Jul 14th 2012, 09:40 PMhemvaneziRe: The problem of shaded region's area
HAVE NAMED THE RECTANGLE AS ABCD

A IS THE VERTEX AT LEFT HAND CORNER, B AT RIGHT HAND CORNER, C AND D ARE THE VERTICES ON THE DIAMETER OF THE SEMI-CIRCLE

Let the center of the circle be at O

Radius of circle=OA

NOW OA=SQRT(100+64)=SQRT(164)=RADIUS

AREA OF SEMICIRCLE-1/2(PI*OA*OA)=82PI

SHADED AREA=AREA OF SEMICIRCLE-AREA OF RECTANGLE

=82*PI-160 - Jul 14th 2012, 10:20 PMProve ItRe: The problem of shaded region's area
To evaluate the area of a circle or semicircle, you need to know what the radius is.

Draw a segment from the centre of the circle (on the diameter) to one of the corners of the rectangle that touches the semicircle. This is the radius of the circle and semicircle. It also creates a right angle triangle, so you can use Pythagoras to evaluate the length of the radius. Then you can evaluate the area of the semicircle and subtract the area of the rectangle.