Question Number 142604 by Engr_Jidda last updated on 02/Jun/21
![](https://www.tinkutara.com/question/20772.png)
Answered by mr W last updated on 02/Jun/21
![a=14 r_1 =b=10.5 r_2 =(√(a^2 +b^2 ))−b r_3 =b−r_2 =2b−(√(a^2 +b^2 )) α=tan^(−1) (b/a) S=((ab)/2)−((r_1 ^2 α)/2)−((r_2 ^2 ((π/2)−α))/2)−((πr_3 ^2 )/4)](https://www.tinkutara.com/question/Q142605.png)
$${a}=\mathrm{14} \\ $$$${r}_{\mathrm{1}} ={b}=\mathrm{10}.\mathrm{5} \\ $$$${r}_{\mathrm{2}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{b} \\ $$$${r}_{\mathrm{3}} ={b}−{r}_{\mathrm{2}} =\mathrm{2}{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}} \\ $$$${S}=\frac{{ab}}{\mathrm{2}}−\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} \alpha}{\mathrm{2}}−\frac{{r}_{\mathrm{2}} ^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\alpha\right)}{\mathrm{2}}−\frac{\pi{r}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by Engr_Jidda last updated on 03/Jun/21
![thank you sir](https://www.tinkutara.com/question/Q142621.png)
$${thank}\:{you}\:{sir} \\ $$