Question

Question 13
If an isosceles $\Delta$ ABC in which AB = AC = 6cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Answer 1

In a circle, $\Delta$ ABC is inscribed.
Join OB, OC, and OA
Consider $\Delta$ ABO and $\Delta$ ACO
AB = AC [given]
BO = CO [ radii of same circle]

AO is common
$\therefore \Delta ABO\cong \Delta ACO$ [ by SSS congruence rule ]
$\Rightarrow \angle 1and\angle 2$ [CPCT]
Now, in $\Delta$ ABM and $\Delta$ ACM , AB=AC [given]
$\angle BAMand\angle CAM$ [proved above]
AM is common
$\Rightarrow \Delta AMB\cong \Delta AMC$ [ by SAS congruence rule]
Also $\angle AMB+\angle AMC={180}^{\circ }$ [ linear pair]
$\Rightarrow \angle AMB={90}^{\circ }$
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 [$\because$ OA = radius = 9 cm]
In right angled $\Delta$ AMC. $A{C}^2=A{M}^2+M{C}^2$ [ by Pythagoras theorem]
i.e $(Hypotenuse{)}^2=(base{)}^2+(perpendicular{)}^2$
$\Rightarrow M{C}^2=6^2-x^2$ (i)
And in right $\Delta$ OMC, $O{C}^2=O{M}^2+M{C}^2$ [ by Pythagoras theorem]
$\Rightarrow M{C}^2=9^2-(9-x{)}^2$ (ii)
From Eqs (i) and (ii) $6^2-x^2=9^2-(9-x{)}^2$
$\Rightarrow 36-x^2=81-(81+x^2-18x)$
$\Rightarrow 36=18x\Rightarrow x=2$
In right angled $\Delta$ ABM, $A{B}^2=B{M}^2+A{M}^2$ [ By Pythagoras theorem]
$6^2=B{M}^2+2^2\Rightarrow B{M}^2=36-4=32\Rightarrow BM=4\sqrt{2}\therefore BC=2BM=8\sqrt{2}cm$
$\therefore$ Area of $\Delta$ ABC = $\frac{1}{2}\times BC\times AM$
$=\frac{1}{2}\times 8\sqrt{2}\times 2=8\sqrt{2}c{m}^2$
Hence , the required area of $\Delta$ ABC is $8\sqrt{2}c{m}^2$