# 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 - Mathematics

Sum

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,

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) cm2

Hence, the required area of ∆ABC is 8sqrt(2) cm2

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 9 Circles
Exercise 9.4 | Q 13 | Page 112
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