If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle

#### Solution

Join OB, OC and OA.

In ∆ABO and ∆ACO,

AB = AC ......[Given]

BO = CO ......[Radii of same circle]

AO = AO ......[Common side]

∴ ∆ABO ≅ ∆ACO ......[By SSS congruence criterion]

⇒ ∠1 = ∠2 ......[CPCT]

Now, In ∆ABM and ∆ACM,

AB = AC ......[Given]

∠1 = ∠2 ......[Proved above]

AM = AM ......[Common side]

∴ ∆AMB ≅ ∆AMC ......[By SAS congruence criterion]

⇒ ∠AMB = ∠AMC ......[CPCT]

Also, ∠AMB + ∠AMC = 180° ......[Linear pair]

⇒ ∠AMB = 90°

We know that a perpendicular from the centre of circle bisects the chord.

So, OA is a perpendicular bisector of BC.

Let AM = x, then OM = 9 – x ......[∵ OA = radius = 9 cm]

In right angle ∆AMC,

AC^{2} = AM^{2} + MC^{2} .......[By Pythagoras theorem]

⇒ MC^{2} = 6^{2} – x^{2} …(i)

In right angle ∆OMC,

OC^{2} = OM^{2} + MC^{2} .......[By Pythagoras theorem]

⇒ MC^{2} = 9^{2} – (9 – x)^{2}

From equation (i) and (ii),

6^{2} – x^{2} = 9^{2} – (9 – x)^{2}

⇒ 36 – x^{2} = 81 – (81 + x^{2} – 18x)

⇒ 36 = 18x

⇒ x = 2

∴ AM = 2 cm

From equation (ii),

MC^{2} = 9^{2} – (9 – 2)^{2}

⇒ MC^{2} = 81 – 49 = 32

⇒ MC = `4sqrt(2)` cm

∴ BC = 2 MC = `8sqrt(2)` cm

∴ Area of ∆ABC = `1/2` × Base × Height

= `1/2 xx BC xx AM`

= `1/2 xx 8sqrt(2) xx 2`

= `8sqrt(2)` cm^{2}

Hence, the required area of ∆ABC is `8sqrt(2)` cm^{2}