In figure , AB is a diameter of the circle, AC = 6cm and BC = 8cm . Find the area of the shaded region . (use `pi= 3.14`)

Explanation

Given, AC = 6cm and BC = 8cm
we know that, triangle in a semi-circle with hypotenuse as diameter is right angled triangle.
`:. angleC= 90^(@)`
In right angled `DeltaACB`, use pythagoras theorem,
`:.AB^(2) = AC^(2)+CB^(2)`
`rArr AB^(2)= 6^(2) + 8^(2) = 36 + 64`
`rArrAB^(2)= 100`
`rArr` AB = 10cm [since, side cannot be negative]
`:.` Area of `DeltaABC = (1)/(2) xxBC xx AC = (1)/(2)xx8xx6= 24cm^(2)`
Here , diameter of circle, AB = 10cm
`:.` Radius of circle, r = `(10)/(2)= 5cm`
Area of circle = `pir^(2) = 3.14xx(5)^(2)`
= `3.14xx 25= 78.5cm^(2)`
`:.` Area of the shaded region = Area of circle - Area of `DeltaBC`
= `78.5-24= 54.5cm^(2)`