# 10th Class Mathematics Areas Related to Circles Areas Related to Circles

Areas Related to Circles

Category : 10th Class

Areas Related to Circles

Areas Related to Circles

1. Circle: A circle is the locus of a point which moves in such a way that its distance from a fixed point always remains the same.

1. Chord: A line segment joining any two points on a circle is called a chord of the circle.

1. Arc: A continuous piece of a circle is called an arc of the circle. An arc AB is denoted by $\overset\frown{AB}$.

1. Segment: A segment of a circle is the region bounded by an arc and a chord, including the arc and the chord.

1. Sector of a circle: The region enclosed by an arc of a circle and its two bounding radii is called a sector of the             circle.

1. Circumference of a circle = $2\pi r$

1. Area of a circle = $\pi {{r}^{2}}.$

1. Length of an arc of a sector of a circle with radius r and angle with degree measure $\theta$ is$\frac{\theta }{360}\times 2\pi r$.

1. Area of a sector of a circle with radius rand angle with degrees measure $\theta$ is $\frac{\theta }{360}\times \pi {{r}^{2}}$

10.       Area of segment of a circle = Area of the corresponding sector - Area of the corresponding triangle.

Snap Test

1. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

(a) 70 cm                       (b) 50 cm

(c) 60 cm                       (d) 80 cm

(e) None of these

Ans.     (a)

Explanation: Distance covered by the wheel in 1 revolution

$=\,\,\,\frac{total\,dis\operatorname{tance}}{number\,of\,revolution}\,\,=\,\,\left( \frac{1100000}{5000} \right)\,cm\,=\,220\,cm$

$\therefore$        Circumference of the wheel $=\text{ }220\text{ }cm$

$\Rightarrow ~~\,\,\,~2\pi r\,\,=\,\,220cm~~~\Rightarrow \text{ }\pi r\,\,=\,\,110\,cm$

$\Rightarrow ~~\,\,\,~r=\left( \frac{110\times 7}{22} \right)\,cm\,\,=\,\,35\,cm$

Hence, the diameter of the wheel = 70 cm.

1. In the given figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

(a) $8.625\,\,c{{m}^{2}}$

(b) $9.1\,\,c{{m}^{2}}$

(c) $9.625\,\,c{{m}^{2}}$

(d) $7.625\,\,c{{m}^{2}}$

(e) None of these

Ans.     (c)

Explanation: It is given that outer radius $R\text{ }=\text{ }7\text{ }cm$ and inner radius $r\,\,=\,\,3.5\,cm$.

$=\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }R\text{ }=\text{ }7\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,=\text{ }30{}^\circ )$

$-\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }r\text{ }=\text{ }3.5\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,~=\text{ }30{}^\circ )$

= $\frac{\pi {{R}^{2}}\theta }{360}-\frac{\pi {{r}^{2}}\theta }{360}\,\,=\,\,\frac{\pi \theta }{360}({{R}^{2}}-{{r}^{2}})=\frac{\pi \theta }{360}(R+r)(R-r)$

= $\left[ \frac{22}{7}\times \frac{30}{360}\times (7+3.5)\times (7-3.5) \right]\,c{{m}^{2}}$

= $\left[ \frac{22}{7}\times \frac{30}{360}\times 10.5\,\,\times \,\,3.5 \right]\,c{{m}^{2}}\,=9.625\,c{{m}^{2}}$

3.         In the given figure, ABCPA is a quadrant of a circle of radius 14 cm. With AC as diameter, a semicircle is drawn. Find the area of the shaded region.

(a) $94\text{ }c{{m}^{2}}~$

(b) $92\text{ }c{{m}^{2}}$

(c) $95\text{ }c{{m}^{2}}~$

(d) $98\text{ }c{{m}^{2}}$

(e) None of these

Ans.     (d)

Explanation: In right Angled $\Delta \,ABC$:

$AC2\text{ }=\text{ }AB2\text{ }+\text{ }BC2\text{ }=\text{ }142\text{ }+\text{ }142\text{ }=\text{ }196\text{ }+\text{ }196\text{ }=\text{ }392$

$\Rightarrow \,\,\,\,AC=\sqrt{392}\,\,=\,\,14\sqrt{2}\,cm$

= Area of semicircle ACQA - Area of segment ACPA

= (Area of semicircle with diameter AC) - (Area of quadrant ABCPA – Area $\Delta \,ABC$)

= (Area of semicircle with radius                                       ${{r}_{1}}\,=\,\,\frac{1}{2}\,AC\,\,\,=\,\,AC\,\,\,=\,\,7\sqrt{2}~cm)\,\,-\,\,\left( Area\text{ }of\text{ }quadrant\text{ }with\text{ }radius\text{ }{{r}_{2}}=\text{ }14\text{ }cm \right)\text{ }+$

$+\text{ }\left( Area\text{ }of\text{ }triangle\text{ }with\text{ }b\text{ }=\text{ }14\text{ }cm,\text{ }h\text{ }=\text{ }14\text{ }cm \right)$

$=\,\,\,\,\,\left( \frac{1}{2}\times \pi {{r}_{1}}^{2} \right)-\left( \frac{1}{4}\times \pi {{r}_{2}}^{2} \right)+\left( \frac{1}{2}\times b\times h \right)$

= $\left[ \left\{ \frac{1}{2}\times \frac{22}{7}\times {{(7\sqrt{2})}^{2}} \right\}-\left\{ \frac{1}{4}\times \frac{22}{7}\times 14\times 14 \right\}+\left\{ \frac{1}{2}\times 14\times 14 \right\} \right]\,c{{m}^{2}}$

= $\left( 154\text{ }-\text{ }154\text{ }+\text{ }98 \right)\text{ }c{{m}^{2}}=\text{ }98\text{ }c{{m}^{2}}$.

1. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Determine the length of the side of the square.

(a) 24 cm                       (b) 44 cm

(c) 45 cm                       (d) 30 cm

(e) None of these

Ans.     (b)

Explanation: Radius of the circle $r\text{ }=\text{ }28\text{ }cm$.

Length of the wire = circumference of the circle = $2\pi r=\left( 2\times \frac{22}{7}\times 28 \right)\,cm\,\,\,=\,\,176\,\,cm$

$\therefore$  $Perimeter\text{ }of\text{ }the\text{ }square\text{ }=\text{ }length\text{ }of\text{ }the\text{ }wire\text{ }=\text{ }176\text{ }cm$

$4a\text{ }=\text{ }176\text{ }cm\text{ }\left[ where\text{ }a\text{ }=\text{ }side\text{ }of\text{ }the\text{ }square \right]$

$a\,\,=\,\,\left( \frac{176}{4} \right)\,cm\,\,=\,\,44\,cm.$

Hence, the length of the side of the square $=\text{ }44\text{ }cm$.

1. A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

(a) $1246\text{ }{{m}^{2}}$

(b) $2618\text{ }{{m}^{2}}$

(c) $3264\text{ }{{m}^{2}}$

(d) $4460\text{ }{{m}^{2}}$

(e) None of these

Ans.     (b)

Explanation: $Circumference\text{ }of\text{ }the\text{ }circular\text{ }park\text{ }=\text{ }352\text{ }m$

$\Rightarrow \,\,2\pi r\,=\,\,352\,m\,\,\,\Rightarrow \,\,\,r\,\,=\,\,\left( \frac{352}{2\pi } \right)\,m\,\,=\,\,\left( \frac{352\times 7}{2\times 22} \right)m\,\,=\,\,56\,m.$

$\therefore$ $Inner\text{ }radius\text{ }\left( park\text{ }alone \right)\text{ }r\text{ }=\text{ }56\text{ }m.$

$Outer\text{ }radius\text{ }\left( including\text{ }road \right)\text{ }R\text{ }=\text{ }\left( r+7 \right)\text{ }m\text{ }=\text{ }63\text{ }m.$

So, area of the road = $\pi {{R}^{2}}\,\,-\,\,\pi {{r}^{2}}$

$=\,\,\pi ({{R}^{2}}\,\,-\,\,{{r}^{2}})\,\,=\,\,\left[ \frac{22}{7}\times ({{63}^{2}}-{{56}^{2}}) \right]\,{{m}^{2}}$

$=\,\,\,\,\left[ \frac{22}{7}\times (63+56)\,\,\times \,\,(63-56) \right]\,{{m}^{2}}$

$=\,\,\,\left( \frac{22}{7}\times 119\times 7 \right)\,{{m}^{2}}\,\,=\,\,2618\,{{m}^{2}}$

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##### Notes - Areas Related to Circles

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