Question

If $\Delta ABO$ is isosceles,the area of the shaded portion in the given figure is $6.125c{m}^2$
If true then enter $1$ and if false then enter $0$

Answer 1

$\Delta AOBisanisosceleseonewithAO=BO=3.5cm.\therefore ThesectorAOBisapartofacircleofradiusr=3.5cm\&\angle O=\theta ={90}^o.SoarsectorAOB=\frac{{\theta }^o}{{360}^o}\times \pi \times r^2=\frac{{90}^o}{{360}^o}\times \frac{22}{7}\times {3.5}^2{cm}^2=9.625{cm}^{2}Now\Delta AOBisanisosceleseonewithAO=BO=3.5cm\&\angle O=\theta ={90}^o.Soitisaright\Delta also.\therefore ar\Delta AOC=\frac{1}{2}\times AO\times CO=\frac{1}{2}\times 3.5\times 2{cm}^2=3.5{cm}^2.\therefore arshadedportion=arsectorAOB-ar\Delta AOC=(9.625-3.5){cm}^2=6.125{cm}^2.Ans-arshadedportion=6.125{cm}^2.$