Question

In fig.4, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take $\pi =3.14$)

Answer 1

Given, AB is a diameter of length 13 cm and AC = 12 cm.

Then, by pythagoras theorem,

$(BC{)}^2=(AB{)}^2-(AC{)}^2$

$(BC{)}^2=(13{)}^2-(12{)}^2$

$BC=\sqrt{169-144}$

$BC=\sqrt{25}$

$\therefore BC=5cm$

Now, Area of shaded region = Area of semi circle - Area of $\Delta ABC$

= $\frac{\pi r^2}{2}-\frac{1}{2}\times BC\times AC$

= $\frac{1}{2}\times 3.14\times {\left(\frac{13}{2}\right)}^2-\frac{1}{2}\times 5\times 12$

= $\frac{1.57\times 169}{4}-30$

= $66.33-30$

= $36.33c{m}^2$

So, area of shaded region is $36.33c{m}^2$.