Question

In figure there is a chord $AB$ of a circle with centre $O$ and radius $10cm$, that subtends a right angle at the centre of the circle. Find the area of the minor segment $AQBP$. Hence find the area of major segment $ALBQA$.

Answer 1

$\angle ALB=\frac{1}{2}\angle AOB=\frac{90}{2}={45}^0$

$LQ$ is perpendicular bisection on $AB$. Hence by isoceles triangles property.

$LA=LB$

$OB=10cm$, & $OA=10cm$

$AB=\sqrt{{10}^2+{10}^2}=10\sqrt{2}cm$

$QB=\frac{AB}{2}=5\sqrt{2}cm$

$OQ=\sqrt{{OB}^2-{QB}^2}=\sqrt{100-50}=\sqrt{50}=5\sqrt{2}cm$

$LQ=LO+OQ=(10+5\sqrt{2})cm$

Area of $ALBQA=\frac{1}{2}\times AB\times LQ=\frac{1}{2}\times 10\sqrt{2}\times (10+5\sqrt{2})$ $=50(1+\sqrt{2}){cm}^2$

Area of $AQBPA=\frac{{\pi r}^2}{4}-Ar.ofAOB$

$=\frac{\pi \times {10}^2}{4}-\frac{1}{2}\times {10}^2\Rightarrow 28.54{cm}^2$