| A memento is made as shown in the figure. Its base PBCR is silver plated from the front side at the rate of Rs 20 per\[c{{m}^{2}}\]. Then, the total cost of the silver plating is \[\left[ take\,\,\pi =\frac{22}{7} \right]\] |
 |
A) Rs 460
B) Rs 115
C) Rs 230
D) None of these
Correct Answer: C
Solution :
From the figure, we have \[A5=7+3=10cm\] \[AC=\text{ }7+3\text{ }=10\text{ }cm\] \[\angle BAC=90{}^\circ \] Also, radius of circle = AP = AR = 7 cm \[\therefore \]Area of right angled \[\Delta BAC\] \[=\frac{1}{2}\times AB\times AC\] \[=\frac{1}{2}\times 10\times 10=50c{{m}^{2}}\] Area of the sector \[APR=\frac{90{}^\circ }{360{}^\circ }\times \pi {{r}^{2}}=\frac{1}{4}\times \pi \times {{\left( 7 \right)}^{2}}\] \[\left[ \because \,\,\,area\,of\,\sec tor=\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}} \right]\] \[=\frac{1}{4}\times \frac{22}{7}\times 49=\frac{11\times 7}{2}=38.5\,c{{m}^{2}}\] Then, area of base PBCR (shaded) which is to be silver plated = Area of the right angled \[\Delta BAC\] - Area of the sector APR \[=50-38.5=\text{ }11.5c{{m}^{2}}\] \[\therefore \] Total cost of silver plating at the rate of Rs \[20\text{ }per\text{ }c{{m}^{2}}\] \[=Rs\,\,20\times 11.5=Rs\,230\]
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