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 = `7sqrt2`

Radius of semi-circle = `"AC"/2 = (7sqrt2)/2`

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

Area of semi-circle = `1/2 pir^2 = 1/2 pi((7sqrt2)/2)^2`
= `(98pi)/8 = (49pi)/4`
= `49/4`
= `49/4 xx 22/7`
= `77/2 "cm"^2` 

Area of ΔABC = `1/2` x BC x AB
= `1/2` x 7 x 7
= `49/2 "cm"^2 `

Thus, Area of the shaded region = `77/2 "cm"^2 - 49/2 "cm"^2 = 28/2 "cm"^2 = 14 "cm"^2`.