Question

$ABC \: is \: an \: isosceles \: triangle \: inscribed \: in \: a \: circle. \: If AB = AC = 12 \sqrt{5} \: cm \: and \: BC = 24 cm, find \: the \: radius \: of \: the \: circle.$
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Answer 1

Step-by-step explanation:

Given :-

ABC is an isosceles triangle

AB = AC = 12√5 cm

BC = 24 cm

To Find :-

Radius of circle

Steps of construction :-

• Make a circle and name the center as O
• Make a triangle ABC inside the circle
• Join OB,OC and OA
• From centre O
• Make AD ⊥ BC

Solution :-

Let the radius be r

We know that

Perpendicular drawn to a chord always bisects the centre of circle

Since OD bisects BC through D

So,

BC = DC/2 = BD/2

DC = 24/2

DC = 12 cm

BD = 24/2

BD = 12 cm

In right ∆ABD

=> AD² + BD² = AB²

=> AD² = AB² - BD²

=> AD² = (12√5)² - (12)²

=> AD² = (12)²(√5)² - (12)²

=> AD² = 720 - 144

=> AD² = 576

=> AD = √(576)

=> AD = √(24 × 24)

=> AD = 24 cm

Now,

AO = BO = CO [Since AD ⊥ BC]

AO = BO = CO = r

OD + AO = AD

OD = AD - AO

OD = 24 - r

In right ∆OBD

=> OD² + BD² = OB²

=> (24 - r)² + 144 = r²

• (a - b)² = a² + b² - 2ab

=> (24)² + r² - 2(24)(r) + 144 = r²

=> 576 + r² - 48r + 144 = r²

=> 576 + 144 - 48r = 0

=> 720 - 48r = 0

=> 720 = 0 + 48r

=> 720 = 48r

On dividing both sides by 48

=> 720/48 = 48r/48

=> 15 = r

Hence,

Radius is 15 cm

Answer 2

HOPE THIS HELPS YOU..

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