In Fig. 3, APB and AQO are semicircles, and \[AO=OB\]. If the perimeter of the figure is 40 cm, find the area of the shaded region. \[\left[ \text{Use}\,\pi =\frac{22}{7} \right]\] |
Answer:
Given, \[OA=OB=r\](say) We have, perimeter of the figure \[=\pi r+\frac{\pi r}{2}+r\] \[\therefore 40=\frac{22}{7}\times r+\frac{22}{7}\times \frac{r}{2}+r\] \[280=22r+11r+7r\] \[40r=280\] \[\therefore r=7\] Now, area of the shaded region \[=\frac{\pi {{r}^{2}}}{2}+\frac{\pi }{2}{{\left( \frac{r}{2} \right)}^{2}}\] \[=\frac{1}{2}\times \frac{22}{7}\times 7\times 7+\frac{1}{2}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\] \[=77+\frac{77}{4}\] \[=\frac{77\times 5}{4}=\frac{385}{4}=96\frac{1}{4}c{{m}^{2}}\]
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