Question

In a circle with center O and radius 5 cm AB is a chord of length `5sqrt(3)` cm. FInd the area of `triangleAOB`

Answer 1

Here, we are given,
`AB = 5sqrt3cm, OA = 5cm`
Let `OX` is the perpendicular to `AB`.
then, `AX = XB = (5sqrt3)/2cm`
So, in right angle triangle `OAX`,
`OA^2=OX^2 +AX^2`
`=>5^2 = OX^2+((5sqrt3)/2)^2`
`=>25 = OX^2+75/4`
`=>OX^2 = 25-75/4 = 25/4`
`=>OX = sqrt(25/4) = 5/2 = 2.5cm`
`:.` Area of `Delta AOB(A) = 1/2**OX**AB`
`=>A = 1/2**2.5**5sqrt3 = (25sqrt3)/4 cm^2`