Question

A circle passes through the 3 vertices of an isosceles triangle that has equal sides of 3cm and base of 2 cm. find the area of the circle.

Answer 1

Let ABC an isosceles triangle with side AB = AC = 3cm and BC = 2cm.
Then Also AD is the perpendicular height of the triangle ABC.
=>OA = radius of the circle.
Also BD = CD = 2cm ( line perpendicular to a chord of a circle bisects it)
Then , by Pythagoras Theorem,
In ∆ABD,
AB^2 = AD^2 + BD^2
3^2 = AD^2 +2^2
9-4 = AD^2
√5cm = AD.
Now Let OA be x.
Then OD = (x - √5)cm
Now Join OB.Then OB = r
=>OB = OA = x.
By Pythagoras Theorem,
In ∆OBD
OB^2 = OD^2 + BD^2
x^2 = (x-√5)^2 + 2^2
x^2 = x^2 + 5 - √10x + 4
√10x = 5+4
10x = 9^2
x = 81/10
x = 8.1
Therefore, the radius of the circle is 8.1cm.
Now area of the circle = πr^2
= 3.14 × 8.1 × 8.1
= 3.14 × 65.61
= 206.0154
Therefore, the area of the circle is 206.02cm.(approximately)
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Answer 2

Hope this helps.

Sorry for the bad handwriting