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In the Following Figure, Ab = 36 Cm and M is Mid-point of Ab. Semi-circles Are Drawn on Ab, Am and Mb as Diameters. a Circle with Centre C Touches All the Three Circles. Find the - Mathematics

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Question

In the following figure, AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on ABAM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.

 

Sum

Solution

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AB − area of two semicircles with diameters AM and MB - area of circle ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

`"Area of semi-circle with AB as a diameter "= pir^2/2`

`∴ "Area of semi-circle with AB as a diameter "=(pi(36/2)^2)/2`

`"∴Area of semi-circle with AB as a diameter"=(pixx18^2)/2` 

Now we will find the area of the semi-circle with AM as a diameter.

`"∴Area of semi-circle with AM as a diameter"=(pi r^2)/2`

`"∴Area of semi-circle with AM as a diameter"=pi(18/2)^2`

`"∴Area of semi-circle with AM as a diameter"=(pixx9^2)/2`

with diameter with AM as a diameter.

Now we will find the area of the circle with centre C.

Area of circle=`pir^2`

We know that radius of the circle is one sixth of AB

Area of circle=`pixx6^2`

Now we will substitute all these values in equation (1).

`∴"Area of shaded region"=(pixx18^2)/2-(pixx9^2)/2-(pixx9^2)/2-36pi`

`∴"Area of shaded region"=pixx18^2/2-pixx9^2-36pi`

`"∴Area of shaded region"=(pixx18^2)/2-81 pi-36pi`

`"∴ Area of shaded region"=(pixx18^2)/22-117pi`

`"∴Area of shaded region"=(162-117)pi`

`∴"Area of shaded region"=45pi`

Therefore, area of shaded region is `45 pi cm^2`

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Chapter 13: Areas Related to Circles - Exercise 13.4 [Page 63]

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RD Sharma Mathematics [English] Class 10
Chapter 13 Areas Related to Circles
Exercise 13.4 | Q 43 | Page 63

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