Chapter 11: Areas Related to Circles

Ans:

Given that, 

Each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of sector \[ = \dfrac{{{{115}^ \circ }}}{{{{360}^ \circ }}} \times \pi  \times {\left( {25} \right)^2}\]

                        \[ = \dfrac{{158125}}{{252}}c{m^2}\]

Area swept by 2 blades \[ = 2 \times \dfrac{{158125}}{{252}}\]

                                      \[ = \dfrac{{158125}}{{126}}c{m^2}\].

Therefore, the total area cleaned at each sweep of the blades is \[\dfrac{{158125}}{{126}}c{m^2}\].

Ans: 

Given that, 

The lighthouse spreads light across a sector (represented by the shaded part in the figure) of \[{80^ \circ }\] in a circle of 16.5 km radius. 

Area of sector OACB \[ = \dfrac{{{{80}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

                                    \[ = \dfrac{2}{9} \times 3.14 \times {(16.5)^2}\]

                                    \[ = 189.97k{m^2}\]

Hence, the area of the sea over which the ships are warned is \[189.97k{m^2}\].

Ans:

Given in the figure,

The designs are segments of the circle. 

Radius of the circle is 28cm.

Consider segment APB and chord AB is a side of the hexagon. 

Each chord will substitute at \[\dfrac{{{{360}^ \circ }}}{6} = {60^ \circ }\] at the center of the circle.

In \[OAB\],

Since, \[OA = OB\]

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

\[2\angle OAB = {180^ \circ } - {60^ \circ } = {120^ \circ }\]

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

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

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

                        \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {28} \right)^2}\]

                        \[ = 333.2c{m^2}\]

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

                                   \[ = \dfrac{1}{6} \times \dfrac{{22}}{7} \times {(28)^2}\]

                                   \[ = \dfrac{{1232}}{3}c{m^2}\] 

Area of segment APBA = Area of sector OAPB − Area of ∆OAB

                                       \[ = \left( {\dfrac{{1232}}{3} - 333.2} \right)c{m^2}\]

Therefore, area of designs \[ = 6 \times \left( {\dfrac{{1232}}{3} - 333.2} \right)c{m^2}\]

                                           \[ = 464.8c{m^2}\]

Now, given that Cost of making 1 \[c{m^2}\] designs = Rs 0.35 

Cost of making 464.76 \[c{m^2}\] designs \[ = 464.8 \times 0.35 = \]162.68 

Therefore, the cost of making such designs is Rs 162.68.

Ans: 

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

So, Area of sector of angle \[P = \dfrac{P}{{{{360}^ \circ }}}\left( {\pi {R^2}} \right)\] 

                                                \[ = \left( {\dfrac{P}{{{{720}^ \circ }}}} \right)\left( {2\pi {R^2}} \right)\]

Hence, (D) is the correct answer.