Question

(iv) Ois the circumcentre of isosceles triangle ABC and ZABC=120°; if the length of the radius of
the circle is 5cm, let us find the value of the side AB.
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Answer 1

Answer:

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Answer 2

Given :- The circumcentre of the isosceles triangle ABC is on and ∠ABC = 120°. If the radius of the circle be 5 cm, then determine the length of AB ?

Solution :-

from image we get,

→ Reflex(∠AOC) = 2∠ABC

→ 360° - ∠AOC = 2 * 120°

→ ∠AOC = 360° - 240°

→ ∠AOC =120° --------- Eqn.(1)

now in ∆OAB and ∆OCB we have,

→ AB = CB ( given that, ∆ABC is isosceles ∆.)

→ OA = OC ( radius .)

→ OB = OB ( common .)

so,

→ ΔOAB ≅ ΔOCB ( By SSS congruence rule. )

then,

→ ∠AOB = ∠BOC ( By CPCT .) ------- Eqn.(2)

→ ∠ABO = ∠CBO ( By CPCT .) ------- Eqn.(3)

then,

→ ∠ABO + ∠CBO = ∠ABC

→ 2∠ABO = 120° { from Eqn.(3) }

→ ∠ABO = 60°

and,

→ ∠AOB + ∠BOC = ∠AOC

→ 2∠AOB = 120° {from Eqn.(2)}

→ ∠AOB = 60°

therefore, in ΔOAB we have,

→ ∠AOB = 60°

→ OA = OB (radius)

→ ∠OAB = ∠OBA (Angle opposite to equal sides are equal.)

→ ∠AOB + ∠OAB + ∠OBA = 180° (By angle sum property.)

→ 60° + 2∠OAB = 180°

→ 2∠OAB = 180° - 60°

→ ∠OAB = 60° = ∠OBA .

hence, we can conclude that, ∆OAB is an equaliteral ∆ .

we have given that, radius of circle is 5 cm.

so,

→ OA = OB = AB = 5 cm. { All sides of an equaliteral ∆ are equal in length. }

∴ Length of AB is 5 cm.

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