question_answer

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 \[=\frac{\pi {{r}^{2}}}{2}+\frac{\pi }{2}{{\left( \frac{r}{2} \right)}^{2}}\]

\[=\frac{1}{2}\times \frac{22}{7}\times 7\times 7+\frac{1}{2}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\]

\[=77+\frac{77}{4}\]

\[=\frac{77\times 5}{4}=\frac{385}{4}=96\frac{1}{4}c{{m}^{2}}\]