Question

ABC is an isosceles right-angled triangle with ∠ABC = 90°. A semi-circle is drawn with AC as the diameter. If AB = BC = 7 cm, find the area of the shaded region. [Take π = 22/7]

Answer 1

In ΔABC, using Pythagoras theorem

AC² = AB² + BC²
AC² = (7)² + (7)²
AC² = 49 + 49
AC² = 98 
⇒ AC = ${\displaystyle 7\sqrt{2}}$

Radius of semi-circle = ${\displaystyle \frac{\text{AC}}{2}=\frac{7\sqrt{2}}{2}}$

∴ Area of the shaded region = Area of semi-circle – Area of ΔABC

Area of semi-circle = ${\displaystyle \frac{1}{2}\pi r^2=\frac{1}{2}\pi {\left(\frac{7\sqrt{2}}{2}\right)}^2}$
= ${\displaystyle \frac{98\pi }{8}=\frac{49\pi }{4}}$
= ${\displaystyle \frac{49}{4}}$
= ${\displaystyle \frac{49}{4}\times \frac{22}{7}}$
= ${\displaystyle \frac{77}{2}{\text{cm}}^2}$ 

Area of ΔABC = ${\displaystyle \frac{1}{2}}$ x BC x AB
= ${\displaystyle \frac{1}{2}}$ x 7 x 7
= ${\displaystyle \frac{49}{2}{\text{cm}}^2}$

Thus, Area of the shaded region = ${\displaystyle \frac{77}{2}{\text{cm}}^2-\frac{49}{2}{\text{cm}}^2=\frac{28}{2}{\text{cm}}^2=14{\text{cm}}^2}$.