Question

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.

Answer 1

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,

AC² = AM² + MC²  ...[By Pythagoras theorem]

⇒ MC² = 6² – x² …(i)

In right angle ∆OMC,

OC² = OM² + MC²  ...[By Pythagoras theorem]

⇒ MC² = 9² – (9 – x)²

From equation (i) and (ii),

6² – x² = 9² – (9 – x)²

⇒ 36 – x² = 81 – (81 + x² – 18x)

⇒ 36 = 18x

⇒ x = 2

∴ AM = 2 cm

From equation (ii),

MC² = 9² – (9 – 2)²

⇒ MC² = 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`.