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#### In Figure-3, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If , then length of PQ is  Option: 1 Option: 2 Option: 3 Option: 4

Length of the tangent is equal to radius of the circle if tangents from external points are perpendicular.

Hence the length PQ = 4 cm

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#### From an external point Q, the length of the tangent to a circle is 5 cm and the distance of Q from the centre is 8 cm. The radius of the circle is Option: 1 39 cm Option: 2 3 cm Option: 3 Option: 4 7 cm

Tangents are perpendicular to the radius of the circle.

$Radius = \sqrt{8^2-5^2} = \sqrt{39} \ cm$

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4cm

#### Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Solution

Radius = 21 cm

Angle = 120°

Area of circle = $\pi r^{2}$

$\\=\frac{22}{7}\times 21\times 21\\ =1386 cm^{2}$

Area of minor sector with angle 120° OABO =$\frac{\pi r^{2} \theta}{360}$                                   $\left [ \theta =120^{\circ} \right ]$

$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times120\\ =462cm^{2}$

Area of major sector AOBA= Area of circle – area of minor sector

= 1386-462=924cm2

Required area =924-462=462cm2

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#### Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Solution

By using Pythagoras in $\triangle$ABC
$\\(AB)^{2}+(BC)^{2}=(AC)^{2}\\ r^{2}+r^{2}=25\\ 2r^{2}=25\\ r=\frac{5}{\sqrt{2}}cm$

Area of circle =$\pi r^{2}=\frac{22}{7} \times \frac{5}{\sqrt{2}}\times\frac{5}{\sqrt{2}}=39.28cm^{2}$

Area of sector =$\frac{\pi r^{2} \theta}{360}$
$\\=3.14 \times \frac{25}{2}\times\frac{90}{360}\\ =9.81cm^{2}$

Area of $\triangle ABC=\frac{1}{2}\times base \times height$

$\\=\frac{1}{2}\times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\\ =\frac{25}{4} =6.25cm^{2}$

Area of minor segment = Area of sector – Area of DABC

=9.81-6.25

=3.56cm2

Area of major segment = Area of circle – Area of minor segment

=39.28-3.56=35.72cm2

Required difference = Area of major segment – Area of minor segment

=35.72-3.56=32.16cm2

#### Find the number of revolutions made by a circular wheel of area 1.54 m2 in Rolling a distance of 176 m.

[40] revolutions

Solution

Circumference of circle =$2\pi r$

Area of wheel = 1.54m2

Distance = 176 m

$\\\pi r^{2}=1.54\\\\ r^{2}=\frac{154}{100}\times \frac{7}{22}\\\\ r=\sqrt{\frac{539}{1100}}$

r  =  0.7m

Circumference   =$2\pi r$

$\\=2 \times \frac{22}{7}\times0.7\\ =2 \times \frac{22}{7}\times \frac{7}{10}\\ =\frac{44}{10}\\ =4.4m$

Number of revolution $=\frac{\text{distance}}{\text{circumference}}\\$

$=\frac{176}{44}\times 10=40\ revolutions$

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#### Find the area of the shaded region given in Figure.

[Area of shaded area=154.88cm2 ]

Solution

Area of square PQRS =(side)2=(14)2

=196 cm2

Area of ABCD (let side a) =(side)2=(a)2

Area of 4 semi circle $\left ( r=\frac{a}{2} \right )=4 \times \frac{1}{2}\pi\left ( \frac{a}{2} \right )^{2}$

Area of semi-circle=$\frac{1}{2}\times \pi \times r^{2}$

$\\=\frac{2\pi a^{2}}{4}\\ =\frac{\pi a^{2}}{2}$

Total inner area = Area of ABCD + Area of 4 semi circles

$\\=a^{2}+\frac{\pi a^{2}}{2}\\$

$\\EF=8cm\\ EF=\frac{a}{2}+a+\frac{a}{2}\\ 8=\frac{a+2a+a}{2}\\ 4a=16\\ a=4cm$

Area of inner region =$4^{2}+\frac{\pi 4^{2}}{2}\\$
$\\=16+\frac{ 16 \pi}{2}\\ =16+8 \pi$

Area of shaded area = Area of PQRS – inner region area

$\\=196-16-8 \pi\\ =180-8 \times 3.14\\ =180-25.12\\ =154.88 cm^{2}$

#### The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

$\because$ Area of sector =$\frac{\pi r^{2} \theta}{360}$

Radius of first sector (r1) = 7 cm

Angle ($\theta_{1}$ ) = 120°

Area of first sector(A1) =$\frac{\pi r_{1}^{2} \theta}{360}$

$\\=\frac{22 \times 7 \times 7 \times 120^{\circ} }{360^{\circ}}\\ =\frac{154}{3}cm^{2}$

Radius of second sector (r2) = 21 cm

Angle ($\theta_{2}$ ) = 40°

Area of sector of second circle (A2)=$\frac{\pi r_{2}^{2} \theta}{360}$
$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times40\\\\ =154 cm^{2}$

Corresponding arc length of first circle =$\frac{2\pi r_{1} \theta}{360}$

=$\frac{\pi r_{1} \theta}{180}$

$\\=\frac{22}{7}\times \frac{7\times 120}{180}\\=\frac{44}{3}cm$

Corresponding arc length of second circle =$\frac{2\pi r_{2}\theta}{360}$

=$\frac{\pi r_{2}\theta}{180}$

$\\=\frac{22}{7}\times \frac{21\times 40}{180}\\=\frac{44}{3}cm$

We observe that the length of the arc of both circles is equal.