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Areas Related To Circles

Problems based onAreas of combinations of plane figures

Chapter : Areas Related To Circles Website: www.letstute.com


Q) From each corner of a square of side 8 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Given: Side of the square = 8 cm Quadrant of radius = 1 cm

To Find: Area of the remaining = ? portion of the square

Diameter of circle = 2 cm



Solution: Area of square ABCD

=


= 8 x 8 cm2

= 64 cm2

Area of a quadrant of radius 1 cm

=

=

=

[∵ 900

3600= 14 ]

=


=


Area of four quadrants of radius 1 cm

=

=

= 4 x area of 1 quadrant

=


Area of circle of diameter 2 cm

¿𝛑 𝐫𝟐¿ [𝟐𝟐𝟕 ×𝟏×𝟏]𝐜𝐦𝟐

¿ [𝟐𝟐𝟕 ]𝐜𝐦𝟐


[


Area of the remaining portion of square ABCD

¿¿ [𝟔𝟒−𝟐𝟐𝟕 −

𝟐𝟐𝟕 ]𝐜𝐦𝟐


¿ [𝟒𝟒𝟖−𝟐𝟐−𝟐𝟐𝟕 ]𝐜𝐦𝟐

¿ [𝟒𝟒𝟖−𝟒𝟒𝟕 ]𝐜𝐦𝟐


¿𝟒𝟎𝟒𝟕

𝐜𝐦𝟐

¿Hence, the area of the remaining portion of the square is 57.714 cm2

𝟓𝟕 .𝟕𝟏𝟒𝐜𝐦𝟐



Q) Four equal circles are described about the four corners of a square of side 14 cm so that each circle touches two others as shown in the figure. Find the area of the shaded region

Given: Side of a square = 14 cm

To find: Area of the shaded region = ?


A B

D C

14 cm


Solution: Let ABCD represent the given square of side 14 cm.

Then, radius of each circle = r = = = 7 cm

Area of the shaded region = Area of square ABCD – Area of four quadrants

=

=A 7cm 7cm B

D 7cm 7cm C

7cm

7cm

7cm

7cm


= (196 – 154) cm2

= 𝟒𝟐𝐜𝐦𝟐


Hence, the area of the shaded region is 42 cm2

A 7cm 7cm B

D 7cm 7cm C

7cm

7cm

7cm

7cm



Q) Calculate the area of the shaded part in the figure. [Take

Given: AD = 7 cm DC = 24 cm

To find: Area of the shaded part = ?7 cm

24 cm

A B

D C


ADC =


7 cm

24 cm

A B

D C

Solution: [Given]

𝐀𝐂𝐢𝐬 𝐭𝐡𝐞𝐝𝐢𝐚𝐦𝐞𝐭𝐞𝐫 𝐨𝐟 𝐭𝐡𝐞𝐜𝐢𝐫𝐜𝐥𝐞 .In the right angled triangle ADC, we have,

AC2 = AD2 + DC2 [Pythagoras’ Theorem]

AC2 = (7cm)2 + (24cm)2

AC2 = (49 + 576) cm2

AC2 = 625 cm2

AC = 25 cm



Let r be the radius of the circle. Then, r = = = cm

= 12.5 cm

Area of the shaded part

=

= [3.14

= Area of the circle – Area of the rectangle

7 cm

24 cm

A B

D C


= (3.14)


= (490.625 – 168)

= 𝟑𝟐𝟐 .𝟔𝟐𝟓𝐜𝐦𝟐

Hence, the area of the shaded part is 322.625 cm2

7 cm

24 cm

A B

D C



Q) The given figure is a cross section which consists of a rectangle and two semi circles. From the information given, finda) Perimeter of the cross-sectionb) Area of the cross-section.

Given: Length of the rectangle = 28cm Breadth of the rectangle = 12cm

To find: a) Perimeter of the cross-sectionb) Area of the cross-section.

l=28cm

b=12cm



Solution: Let the length and breadth of the rectangle be l and b respectively,

Let r be the radius of each semicircular part of the cross-section.

Then, l = 28 cm, b = 12 cm

And r = = cm = 14 cm


l=28cm

b=12cm


a) Perimeter of the cross-section =

= 2

= 2(

= 2

= 2 (44 + 12) cm

=𝟏𝟏𝟐𝐜𝐦


l=28cm

b=12cm


b) Area of the cross-section

= Area of the 2 semicircles + area of the rectangle= 2

=

= cm2

= 616 + 336

= 𝟗𝟓𝟐𝐜𝐦𝟐


l=28cm

b=12cm


Hence, a) The perimeter of the cross-section is 112 cm

b) The area of the cross-section is


l=28cm

b=12cm


Now we know…

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