Question

The given figure shows the center of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take $\pi =3.142$)

Answer 1

In right angle triangle $ACB$,

${AB}^2={AC}^2+{BC}^2={\left(12\right)}^2+{\left(16\right)}^2=144+256=400$

$AB=\sqrt{400}=20m=Diameter$

Radius $=\frac{20}{2}=10m$

Arc length ACB $=\frac{1}{2}\times 2\pi r=\frac{1}{2}\times 2\times 3.142\times 10=31.42m$

Perimeter of shaded region $=31.42+12+16=59.42cm$

Area of shaded region $=\frac{1}{2}\pi r^2-\frac{1}{2}\times AC\times CB=\frac{1}{2}(3.142\times {10}^2-12\times 16)$

$=\frac{1}{2}(314.2-192)=\frac{1}{2}\times 122.2=61.1{cm}^2$