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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 \angle QPR = 90^{\circ}, then length of PQ is 
Option: 1 3 cm
Option: 2 4 cm
Option: 3 2cm
Option: 4 2\sqrt{2}cm

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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Safeer PP

In Figure-2, PQ is tangent to the circle with centre at O, at the point B. If \angle AOB = 100^{\circ} , then \angle ABP is equal to 
Option: 1 50°
Option: 2 40°
Option: 3 60°
Option: 4 80°

\\ $ In triangle AOB $ \\ OA = OB \\ \therefore \angle OAB = \angle OBA \\ \angle OAB + \angle OBA + \angle AOB = 180^0 \\ \angle OBA + \angle OBA + \angle AOB = 180^0\\ 2 \angle OBA + 100^0 = 180^0 \\ \angle OBA = 40^0 \\ \angle OBP = 90^0 \\ \angle OBA + \angle ABP = 90^0 \\ 40^0 + \angle ABP = 90^0 \\ \angle ABP = 50^0 \\

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Safeer PP

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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 \sqrt{39\: cm}
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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Safeer PP

In Figure-2, TP and TQ are tangents drawn to the circle with centre at O. If \angle POQ = 115^{\circ} then \angle PTQ is 
Option: 1 115°
Option: 2 57.5°
Option: 3 55°
Option: 4 65°

\\\angle PTQ + \angle POQ = 180^0 \\ \angle PTQ + 115^0 = 180^0 \\ \angle PTQ = 180^0 - 115^0 \\ \angle PTQ = 65^0 \\

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Safeer PP

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In Fig.1, \bigtriangleup ABC is circumscribing a circle, the length of BC is _______cm.          
 

4cm

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Rakshanaa

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

Answer [462 cm2]

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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infoexpert24

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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.

Answer [32.16cm2]

Solution

            

By using Pythagoras in \triangleABC
   \\(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

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infoexpert24

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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infoexpert24

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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}

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infoexpert24

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.

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infoexpert24

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