Chapter 11: Areas Related to Circles

Ans:

Given that, 

Radius of the circle = \[r = 6cm\]

Angle made by the sector with the center, \[\theta  = {60^ \circ }\]

Let OACB be a sector of the circle making \[{60^ \circ }\] angle at center O of the circle.

We know that area of sector of angle, \[ = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}\]

Thus, Area of sector OACB \[ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times {(6)^2}\]

\[ = \dfrac{1}{6} \times \dfrac{{22}}{7} \times 6 \times 6\]

\[ = \dfrac{{132}}{7}c{m^2}\]

Therefore, the area of the sector of the circle making 60° at the center of the circle is \[\dfrac{{132}}{7}c{m^2}\].

Ans:

Given that,

Circumference = 22 cm

Let the radius of the circle be \[r\].

According to the given condition,

\[2\pi r = 22\]

\[ \Rightarrow r = \dfrac{{22}}{{2\pi }}\]

\[ = \dfrac{{11}}{\pi }\]

We know that, quadrant of a circle subtends \[{90^ \circ }\] angle at the center of the circle.

Area of such quadrant of the circle \[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi  \times {r^2}\]

\[ = \dfrac{1}{4} \times \pi  \times {\left( {\dfrac{{11}}{\pi }} \right)^2}\]

\[ = \dfrac{{121}}{{4\pi }}\]

\[ = \dfrac{{77}}{8}c{m^2}\]

Hence, the area of a quadrant of a circle whose circumference is 22 cm is \[ = \dfrac{{77}}{8}c{m^2}\].

Ans:

Given that,

Radius of clock or circle = \[r\] = 14 cm.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates \[{360^ \circ }\]. 

Thus, in 5 minutes, minute hand will rotate \[ = \dfrac{{{{360}^ \circ }}}{{{{60}^ \circ }}} \times 5\]

\[ = {30^ \circ }\]

Now, 

the area swept by the minute hand in 5 minutes = the area of a sector of \[{30^ \circ }\] in a circle of 14 cm radius.

 Area of sector of angle \[\theta  = \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}\] 

 Thus, Area of sector of \[{30^ \circ } = \dfrac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \dfrac{{22}}{7} \times 14 \times 14\]

\[ = \dfrac{{11 \times 14}}{3}\]

\[ = \dfrac{{154}}{3}c{m^2}\]

Therefore, the area swept by the minute hand in 5 minutes is \[\dfrac{{154}}{3}c{m^2}\].

Ans:

Given that, 

Radius of the circle \[ = r = 10cm\]

Angle subtended by the cord = angle for minor sector\[ = {90^ \circ }\]

Angle for major sector \[ = {360^ \circ } - {90^ \circ } = {270^ \circ }\]

Ans: It is evident from the figure that, 

Area of minor segment ACBA = Area of minor sector OACB − Area of ΔOAB

Thus,

Area of minor sector OACB \[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\] \[ = \dfrac{1}{4} \times 3.14 \times {(10)^2}\] \[ = 78.5c{m^2}\]

Area of ΔOAB \[ = \dfrac{1}{2} \times OA \times OB = \dfrac{1}{2} \times {(10)^2} = 50c{m^2}\]

Area of minor segment ACBA \[ = 78.5 - 50 = 28.5c{m^2}\]

Hence, area of minor segment is \[28.5c{m^2}\]

Ans: It is evident from the figure that,

Area of major sector OADB \[ = \dfrac{{{{270}^ \circ }}}{{{{360}^ \circ }}} = \dfrac{3}{4} \times 3.14 \times {(10)^2} = 235.5c{m^2}\].

Hence, the area of the major sector is \[235.5c{m^2}\].

Ans:

Given that, 

Radius of circle = \[r = \] 21 cm 

Angle subtended by the given arc = \[\theta  = {60^ \circ }\]

Ans: We know that, Length of an arc of a sector of angle \[\theta  = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r\]

Thus, Length of arc ACB \[ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 2 \times \dfrac{{22}}{7} \times 21\]

\[ = 22cm\] 

Hence, the length of the arc of a given circle is \[22cm\].

Ans: We know that, Area of sector OACB \[ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = 231c{m^2}\]

Hence, the area of the sector formed by the arc of the given circle is \[231c{m^2}\].

Ans: In \[OAB\],

As radius \[OA = OB\]

\[ \Rightarrow \angle OAB = \angle OBA\]

\[\angle OAB + \angle AOB + \angle OBA = {180^ \circ }\]

\[2\angle OAB + {60^ \circ } = {180^ \circ }\]

\[\angle OAB = {60^ \circ }\]

Therefore, \[OAB\] is an equilateral triangle.

Now, area of \[OAB\] \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]

\[ = \dfrac{{\sqrt 3 }}{4} \times {\left( r \right)^2}\]

\[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {21} \right)^2}\]

\[ = \dfrac{{441\sqrt 3 }}{4}c{m^2}\]

We know that, Area of segment ACB = Area of sector OACB − Area of $\Delta AOB$

\[ = \left( {231 - \dfrac{{441\sqrt 3 }}{4}} \right)c{m^2}\].

Ans:

Given that, 

Radius of circle = \[r = \] 15 cm 

Angle subtended by chord \[ = \theta  = {60^ \circ }\]

Area of circle \[ = \pi {r^2}\] \[ = 3.14{\left( {15} \right)^2}\]

\[ = 706.5c{m^2}\]

Area of sector OPRQ \[ = \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

 \[ = \dfrac{1}{6} \times 3.14{(15)^2} = 117.75c{m^2}\]

Now, for the area of major and minor segments, 

In \[OPQ\],

Since, \[OP = OQ\]

\[ \Rightarrow \angle OPQ = \angle OQP\]

\[\angle OPQ = {60^ \circ }\]

Thus, \[OPQ\] is an equilateral triangle.

Area of \[OPQ\] \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]

\[ = \dfrac{{\sqrt 3 }}{4} \times {(r)^2}\] 

\[ = \dfrac{{225\sqrt 3 }}{4}\] \[ = 97.3125c{m^2}\].

Now, 

Area of minor segment PRQP = Area of sector OPRQ − Area of \[OPQ\] 

\[ = 117.75 - 97.3125\]

\[ = 20.4375c{m^2}\]

Area of major segment PSQP = Area of circle − Area of minor segment PRQP 

\[ = 706.5 - 20.4375\]

\[ = 686.0625c{m^2}\]

Therefore, the areas of the corresponding minor and major segments of the circle are \[20.4375c{m^2}\] and \[686.0625c{m^2}\] respectively.

Ans: 

Draw a perpendicular OV on chord ST bisecting the chord ST such that SV = VT

Now, values of OV and ST are to be found.

Therefore, 

In \[OVS\],

\[\cos {60^ \circ } = \dfrac{{OV}}{{OS}}\]

\[ \Rightarrow \dfrac{{OV}}{{12}} = \dfrac{1}{2}\] 

\[ \Rightarrow OV = 6cm\]

Also, \[\dfrac{{SV}}{{SO}} = \sin {60^ \circ }\]

\[ \Rightarrow \dfrac{{SV}}{{12}} = \dfrac{{\sqrt 3 }}{2}\] 

\[ \Rightarrow SV = 6\sqrt 3 \]

Now, \[ST = 2SV\]

\[ = 2 \times 6\sqrt 3  = 12\sqrt 3 cm\]

Area of \[OST = \dfrac{1}{2} \times ST \times OV\]

\[ = \dfrac{1}{2} \times 12\sqrt 3  \times 6\]

\[ = 62.28c{m^2}\]

Area of sector OSUT \[ = \dfrac{{{{120}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(12)^2}\]

 \[ = 150.42c{m^2}\]

Area of segment SUTS = Area of sector OSUT − Area of \[OVS\]

                                     \[ = 150.72 - 62.28\]

                                      \[ = 88.44c{m^2}\]

Hence, the area of the corresponding segment of the circle is \[88.44c{m^2}\].

Ans:

From the above figure, it is clear that the horse can graze a sector of \[{90^ \circ }\] in a circle of 5 m radius.

Hence, 

\[\theta \text{ }\!\!~\!\!\text{ }={{90}^{{}^\circ }}\]

\[r=5m\]

Ans: It is evident from the figure,

Area that can be grazed by horse = Area of sector OACB

                                                       \[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

                                                        \[ = \dfrac{1}{4} \times 3.14 \times {(5)^2}\] \[ = 19.625{m^2}\]

Ans: It is evident from the figure,

Area that can be grazed by the horse when length of rope is 10 m long

\[ = \dfrac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi  \times {(10)^2}\]

\[ = 78.5{m^2}\]

Therefore, the increase in grazing area for horse \[ = (78.5 - 19.625){m^2} = 58.875{m^2}\].

Ans:

Given that, 

Radius of the circle \[ = r\] \[ = \dfrac{{diameter}}{2}\]\[ = \dfrac{{35}}{2}mm\]

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° (i.e., 360°/10=36°) at the center of the circle.

Ans:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Circumference of brooch \[ = 2\pi r\]

                                         \[ = 2 \times \dfrac{{22}}{7} \times \left( {\dfrac{{35}}{2}} \right)\]

                                         \[ = 110mm\]

Length of wire required \[ = 110 + \left( {5 \times 35} \right)\]

                                        \[ = 285mm\]

Therefore, The total length of the silver wire required is \[285mm\].

Ans:

Area of each sector \[ = \dfrac{{{{36}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

                                 \[ = \dfrac{1}{{10}} \times \dfrac{{22}}{7} \times {\left( {\dfrac{{35}}{2}} \right)^2}\] 

                                 \[ = \dfrac{{385}}{4}m{m^2}\]

Hence, The area of each sector of the brooch is \[\dfrac{{385}}{4}m{m^2}\].

Ans: 

Given that, 

Radius of the umbrella \[ = r\]\[ = 45cm\]

There are 8 ribs in an umbrella. 

The angle between two consecutive ribs is subtending \[\dfrac{{{{360}^ \circ }}}{8} = {45^ \circ }\] at the center of the assumed flat circle. 

Area between two consecutive ribs of the assumed circle \[ = \dfrac{{{{45}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

                                                                                            \[ = \dfrac{1}{8} \times \dfrac{{22}}{7} \times {\left( {45} \right)^2}\]

                                                                                            \[ = \dfrac{{22275}}{{28}}c{m^2}\] 

Hence, the area between the two consecutive ribs of the umbrella is \[\dfrac{{22275}}{{28}}c{m^2}\].