| In figure, AOB is a flower bed in the shape of a sector of a circle of radius 40 m and\[\angle AOB=60{}^\circ \]. Also, a 15 m wide concrete track is made as shown in the figure. Flower bed is made at the rate of Rs 2.40 per \[{{m}^{2}}\] and rate of making the concrete track is Rs 20 per \[{{m}^{2}}\], then the total amount spent for the job is |
| [take \[\pi =3.14\]] |
 |
A) Rs 2009
B) Rs 53034.60
C) Rs 5303.460
D) None of the above
Correct Answer: B
Solution :
Given, radius of circle = 40 m and angle of sector = \[60{}^\circ \] Area of the sector \[AOB=\frac{60{}^\circ }{360{}^\circ }\times 3.14\times {{\left( 40 \right)}^{2}}\] \[=\frac{1}{6}\times 3.14\times 1600=\frac{2512}{3}{{m}^{2}}\] Amount spent for making the flower bed at the rate of Rs 240 per \[{{m}^{2}}\] \[=Rs\,2.40\times \frac{2512}{3}=Rs\,2009.60\] Angles for the major sectors of both the circle at O is same, i.e. \[300{}^\circ \]. Radius of inner circle = 40 -15 = 25 m Area of concrete track \[=\left\{ \frac{300{}^\circ }{360{}^\circ }\times \pi \times {{\left( 40 \right)}^{2}}-\frac{300{}^\circ }{360{}^\circ }\times \pi \times \left( {{25}^{2}} \right) \right\}\] [\[\because \]area of track = area of outer sector - area of inner sector] \[=\frac{5}{6}\times \pi \times \left( 1600-625 \right)=\frac{5}{6}\times 3.14\times 975\] \[=2551.25{{m}^{2}}\] Amount spent for making the concrete track at the rate of Rs 20 per \[{{m}^{2}}\] \[=\text{ }20\text{ }\times \text{ }2551.25\text{ }=Rs\,51025\] Thus, the total amount spent for both the jobs = Rs 2009.60 + Rs 51025 = Rs 53034.60
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