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Areas Related to Circles - Class 10


Notes for areas related to circles chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

 

(1) For a circle of a radius r, we have
(i) Circumference = 2\pi r
(ii) Area = \pi {r^2}
(iii) Area of semi-circle = {{\pi {r^2}} \over 2}
(iv) Area of a quadrant = {{\pi {r^2}} \over 4}

For Example: Find circumference and area of a circle of radius 4.2 cm
Solution: We know that the circumference C and area A of a Circle of radius r given by, C = 2\pi r and A = \pi {r^2} respectively.
(i) Circumference of the circle C = 2\pi r
= 2 \times {{22} \over 7} \times (4.2)= 26.4cm
(ii) Area of the circle A = \pi {r^2}
= {{22} \over 7} \times {(4.2)^2} \Rightarrow 55.4c{m^2}
Hence, Circumference of the circle and area of the circle and area of the circle are 26.4 cm  and 55.4c{m^2} respectively.
(iii) Area of a semi-circle = = {{\pi {r^2}} \over 2} = {{22 \times {{(4.2)}^2}} \over {7 \times 2}} = 27.2 c{m^2}
(iv) Area of a quadrant = {{\pi {r^2}} \over 4} = {{22 \times {{(4.2)}^2}} \over {7 \times 4}} = 13.86 c{m^2}

(2) If R and r are the radii of two concentric circles such that R > r then, Area enclosed by the two circles = \pi {R^2} - \pi {r^2} = \pi \left( {{R^2} - {r^2}} \right)
For Example: The area enclosed between the concentric circle is 770 {cm^2}. If the radius of the outer circle is 21 cm, find the radius of the inner circle.Solution: Let the radius of inner and outer radius be {r_1} and {r_2} respectively.
It is given that area that area enclosed between concentric circles is 770 {cm^2}

Radius of the outer circle is 21 cm
Then, area enclosed between the concentric circle  = \pi r_2^2 - \pi r_1^2


 \Rightarrow \pi r_2^2 - \pi r_1^2 = 770
 \Rightarrow \pi ((21) - r_1^2) = 770
 \Rightarrow (441 - r_1^2) = {{770 \times 7} \over {22}}
 \Rightarrow r_1^2 = 441 - 245 = 196
 \Rightarrow {r_1} = 14
Hence, the radius of the inner circle is 14 cm.

(3) If a sector of a circle of radius r contains an angle of \theta Then,
(i) Length of the arc of the sector = {\theta \over {360}} \times 2\pi r = {\theta \over{360}}x(Circumference)
For Example: Find the Length of the arc of the sector that subtends an angle of {30^ \circ } at the centre of a circle of radius 4 cm.
Solution: The length of the arc is given by l = {\theta \over {360}} \times 2\pi r

Here, r = 4 cm and \theta = {30^ \circ }
 \Rightarrow l = \left( {{{30} \over {360}} \times 2\pi \times 4} \right)
 \Rightarrow l = {{2\pi } \over 3} cm
Hence, the length of the arc is {{2\pi } \over 3} cm

(ii) Perimeter of the sector= 2r + {\theta \over {360}} \times 2\pi r
For Example: The cross section of railway tunnel the radius of the circular part is 2m. if \angle AOB = {90^ \circ } calculate the perimeter of the cross section.
Solution:We have OA = 2m

Now using Pythagoras theorem in \Delta AOB, AB = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 m
Let the height of the tunnel be h
Area of \Delta AOB= {1 \over 2} \times 2 \times 2= {1 \over 2} \times 2\sqrt 2 \times OM = 2
OM = \sqrt 2
h = (2 + \sqrt 2 )cm
Perimeter of cross-section is = major arc ABAB\left( {2\pi \times 2 \times {3 \over 4}} \right) + 2\sqrt 2 (3\pi + 2\sqrt 2 )cm


(iii) Area of the sector= {\theta \over {360}} \times \pi {r^2} = {\theta \over {360}} \times (Area of the circle)
For Example: AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divided the circle into two segments find the area of the minor segment
Solution:It is given that chord AB divides the circle into two segments

In \Delta AOB
OA = OB = 4cm
AM = {{AB} \over 2} = 2cm
Let \angle AOB = 2\theta , then
\angle AOM = \angle BOM = \theta
In \angle OAM we have
\sin \theta = {{AM} \over {AO}} = {2 \over 4} = {1 \over 2}
\theta = {\sin ^{ - 1}}{1 \over 2}
= {30^ \circ }
Hence, \angle AOB = 2\theta = 2 \times {30^ \circ } = {60^ \circ }
We know that the area of minor segment of angle \theta in a circle of radius r is
A = \left\{ {{{\pi \theta } \over {360}} - sin{\theta \over 2}\cos {\theta \over 2}} \right\}{r^2}
Now, using the value of r and \theta we can find the area of minor segment
A = \left\{ {{{\pi \theta } \over {360}} - sin{{{{60}^ \circ }} \over 2}\cos {{{{60}^ \circ }} \over 2}} \right\}{(4)^2}
 \Rightarrow A = \left\{ {{\pi \over 6} - {1 \over 2} \times {{\sqrt 3 } \over 2}} \right\}{(4)^2}
A = \left\{ {{{8\pi } \over 3} - 4\sqrt 3 } \right\} c{m^2}
Hence, area of minor segment is  \left\{ {{{8\pi } \over 3} - 4\sqrt 3 } \right\} c{m^2}

(iv) Area of the segment = Area of the corresponding sector - Area of the corresponding triangle
={\theta \over {360}} \times \pi {r^2} - {r^2}\sin {\theta \over 2}\cos {\theta \over 2} = \left\{ {{{\pi \theta } \over {360}} - \sin {\theta \over 2}\cos {\theta \over 2}} \right\}{r^2}

For Example: The radius of a circle with centre O is 5 cm. two radii OA and OB are drawn at right angles to each other. Find the areas of segment made by chord AB.
Solution: Radius of the circle = 5 cm
Area of the minor segment AB = \left( {{{\pi \theta } \over {{{360}^ \circ }}} - \sin {\theta \over 2}\cos {\theta \over 2}} \right){(r)^2}
AB = \left( {{{3.14 \times {{90}^ \circ }} \over {{{360}^ \circ }}} - \sin {{45}^ \circ }\cos {{45}^ \circ }} \right){(5)^2}
 = \left( {{{282.6} \over {{{360}^ \circ }}} - {1 \over {\sqrt 2 }} \times {1 \over {\sqrt 2 }}} \right){(5)^2}
=7.125 c{m^2}
Area of minor segment = area of circle – area of minor segment = \pi {r^2} - 7.125
=3.14 - 25 - 7.125 = 71.37 c{m^2}



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