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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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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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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}

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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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Area of a sector of central angle 200° of a circle is 770 cm2. Find the length of the corresponding arc of this sector.

\left [ 73\frac{1}{3}cm \right ] 

Solution

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

Angle = 200°

  Area of sector = 770 cm2    

\\\frac{\pi r^{2}\theta}{360}=770\\\\ \pi r^{2} \times 200^{\circ}=770 \times 360 ^{\circ}\\\\ r^{2}=\frac{770 \times 360 ^{\circ} \times 7}{200^{\circ \times 22}} \;\;\; \left [ here \pi=\frac{22}{7} \right ]\\\\ r^{2}=49 \times 9\\ r=\sqrt{49 \times 9}=7 \times 3=21cm

Length of the corresponding arc =\frac{\theta \times 2\pi r}{360}
\\=\frac{200^{\circ} \times 2 \times \pi \times 21}{360^{\circ}}\\\\ =\frac{10 \times 7}{3}\times\frac{22}{7}\\\\ =\frac{220}{3}\\\\ =73\frac{1}{3}cm

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The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 a m.

\left [ 45\frac{5}{6}cm^{2} \right ]

Solution

We know that minute hand revolving in 60 min =360^{\circ}

In 1 minute it is revolving =\frac{360}{60}=6^{\circ}

Time difference =(6:40am -6:05am) =35 min

In 6:05 am and 6.40 am there is 35 minutes

In 35 minutes angle between min hand and hour hand =\left (6\times 35 \right )^{\circ} =210^{\circ}

Length of minute hand (r)=5cm

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

\\=\frac{22}{7}\times \frac{5 \times 5 \times 210^{\circ}}{360}\;\;\;\left \{\because \theta = 210^{\circ} \right \}\\ =\frac{11\times 5\times 5}{6}\\ =\frac{275}{6}\\ =45\frac{5}{6}cm^{2}                                                        

Hence required area is 45\frac{5}{6}cm^{2}

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An archery target has three regions formed by three concentric circles as shown in Figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions

[1 : 3 : 5]

Solution

 

d1:d2:d3   =   1: 2 : 3 [multiplying by s]

  =   s : 2s  :  3s       

 Radius of inner circle (r1)=\frac{s}{2}

Radius of middle circle (r2)=\frac{2s}{2}=s

Radius of outer circle (r3)=\frac{3s}{2}

Area of region enclosed between second and first circle   

 \\=\pi r_{2}^{2}-\pi r_{1}^{2}\\ =\pi s^{2}-\frac{\pi s^{2}}{4}\\ =\frac{3 \pi s^{2}}{4}

 Area of region enclosed between third and second circle   

\\=\pi r_{3}^{2}-=\pi r_{2}^{2}\\ =\frac{\pi 9s^{2}}{4}-\pi s^{2}\\ =\frac{ 5\pi s^{2}}{4}

Area of first circle =\pi r_{1}^{2}=\frac{\pi s^{2}}{4}

Ratio of area of three regions

\\=\frac{\pi s^{2}}{4}:\frac{3\pi s^{2}}{4}:\frac{5\pi s^{2}}{4}\\ =\pi s^{2}:3 \pi s^{2}:5\pi s^{2}\\ =1:3:5

           

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All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm2. (Use \pi = 3.14).

Answer [800 cm2]

Solution

 

Given that area of circle =1256cm2
 \\\pi r^{2}=1256\\\\ r^{2}=\frac{1256}{314}\times 100\\\\ r^{2}=400 \; \; \; \; \; \; \; \; \left ( \pi=3.14 \right )\\\\ r=\sqrt{400}\\\\ r=20cm

Diameter of circle = 40 cm

As we know that the diameter of circle is equal

Diagonals of rhombus = Diameters of circle = 40 cm

Each diagonals of rhombus = 40 cm

Area of rhombus =\frac{1}{2}  \times product of digonals

                            = \frac{1}{2}  \times 40 \times 40

                            = 800cm2

Hence required area of rhombus is =800cm2

           

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Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Figure. Find the area of floor that remains uncovered with tiles. (Use \pi = 3.14)

[4.3 m2]

Solution

Diameter of tile =50cm=0.5m      (1m = 100cm)

Radius =\frac{50}{2}=25=0.25m

Number of tiles lengthwise =\frac{5}{0.5}=10  tiles

Number of tiles widthwise =\frac{4}{0.5}=8  tiles

Total tiles =10 \times 8=80

Area of floor not covered by tiles = Area of rectangular floor – Area of 80 tiles 

                                            \\=5 \times 4-80\pi r^{2}\\ =20-80 \times \pi \times 0.25 \times 0.25\\ =20-\frac{8 \times 314 \times 25 \times 25}{100 \times100\times100}\\ =20-\frac{157}{10}\\ =20-15.7\\ =4.3m^{2}

           

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