In a triangle abc, ab=12, bc=18, ca=25. A semi circle is inscribed in abc such that the diameter of the circle lies on ac.?

Let R be the radius of the semi-cicle.

The semi-circle is inscribed in ΔABC with diameter lying on AC.
Then, AB and BC are the tangent of the semi-circle.

For a circle, the radius perpendicular to the tangent at the point of contact.
Area of ΔABO = (1/2) × AB × R ...... [1]
Area of ΔBOC = (1/2) × BC × R ...... [2]

For two triangle with the same altitude, their ratio of area is equal to the ratio of the base.
Area of ΔABO : Area of ΔBOC = OA : OC ...... [3]

Substitute [1] and [2] into [3] :
[(1/2) × AB × R] : [(1/2) × BC × R] = OA : OC
AB : BC = OA : (CA - OA)
12 : 18 = OA : (25 - OA)
18 * OA = 12 * (25 - 0A)
18 OA = 300 - 12 OA
30 OA = 300
OA = 10 (units)

...... The answer is : 3. 10 units

OA:OC = b1:b2 = c:a = 12:18 = 2:3

b1 + b2 = b = 25 --> b1 = 10 --> OA = 10

AB:BC = 12:18 = 2:3, so AO:OC = 2:3 = 10:15
AO=10

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