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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:Â Find circumference and area of a circle of radius 4.2 cm
Solution:Â 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 $26.4 cm$ Â and $55.4c{m^2}$ respectively.
(iii) Area of a semi-circle = = ${{\pi {r^2}} \over 2}$ = ${{22 \times {{(4.2)}^2}} \over {7 \times 2}}$ = $27.2$Â $c{m^2}$
(iv) Area of a quadrant = ${{\pi {r^2}} \over 4} = {{22 \times {{(4.2)}^2}} \over {7 \times 4}}$ = $13.86$Â $c{m^2}$

(2) If R and r are the radii of two concentric circles such that $R > r$ then,Â Area enclosed by the two circles = $\pi {R^2} - \pi {r^2} = \pi \left( {{R^2} - {r^2}} \right)$
For Example:Â 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:Â 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 = ${\theta \over {360}} \times 2\pi r$Â = ${\theta \over{360}}$x(Circumference)
For Example:Â 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:Â 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:Â 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 arcÂ $AB$ +Â $AB$ =Â $\left( {2\pi \times 2 \times {3 \over 4}} \right) + 2\sqrt 2$ =Â $(3\pi + 2\sqrt 2 )cm$

(iii) Area of the sector= ${\theta \over {360}} \times \pi {r^2} = {\theta \over {360}} \times$ (Area of the circle)
For Example:Â 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:Â 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:Â 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 â€“ area of minor segment = $\pi {r^2} - 7.125$
=$3.14 - 25 - 7.125$ = $71.37$ $c{m^2}$

### 5 Comments

• Anonymous

Nyc

• Sarvam

Exellent notes for class 10th examination

• Sarvam

Good

• Sameer

Best notes I have ever seen....thanks for this..

• Anonymous

nice work

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