If an isosceles triangle `A B C` in which `A B=A C=6c m` is inscribed in a circle of radius `9c m ,` find the area of the triangle.

Explanation

In a circle, `DeltaABC` is inscribed.
Join OB, OC and OA.
Conside `DeltaABO` and `DeltaACO`
AB=AC [given]
BO=CO [radii of same circle]

AO is common.
`:.DeltaABO~=DeltaACO` [by SSS congruence rule]
`impliesangle1=angle2` [CPOT]
Now, in `DeltaABMand DeltaACM`,
AB=AC [given]
`angle1=angle2` [proved above] AM is common.
`:.DeltaAMB~=DeltaAMC` [by SAS congruence rule]
`impliesangleAMB=angleAMC` [CPCT]
Also, `angleAMB+angleAMC=180^(@) [linear pair]`
`impliesangleAMB=90^(@)`
We know that a perpendicular from centre of circle bisects the chord.
So, OA is perpendicular bisector of BC.
Let AM =x, then OM=9-x [:' OA=radius=9cm]
In right angled `DeltaAMC. AC^(2)=AM^(2)+MC^(2)` [by Pythagoras theorem]
i.e., `("Hypotenuse")^(2)=("Base")^(2)+("Perpendicular")^(2)`
`impliesMC^(2)=6^(2)-x^(2).......(i)`
and in right `DeltaAMC, OC^(2)=OM^(2)+MC^(2)` [by Pythagoras theorem]
`impliesMC^(2)=9^(2)-(9-x)^(2)`
From Eqs. (i) and (ii), `6^(2)-x^(2)=9^(2)-(9-x)^(2)`
`implies36-x^(2)=81-(81+x^(2)-18x)`
`implies36=18ximpliesx=2`
`:.AM=x=2`
In right angled `DeltaABM AB^(2)=BM^(2)+AM^(2)` [by Pythagoras theorem]
`6^(2)=BM^(2)+2^(2)`
`impliesBM^(2)=36-4=32`
`impliesBM=4sqrt2`
`:.BC=2BM=8sqrt2cm`
:.Area of `DeltaABC=(1)/(2)xx"Base"xx"Height"`
`implies(1)/(2)xxBCxxAM`
`implies(1)/(2)8sqrt2xx2=8sqrt2cm^(2)`
Hence, the required area of `DeltaABC" is"8sqrt2cm^(2)`.