A triangle ABC with an obtuse angle B is inscribed in a circle. The altitude AD of the triangle is tangent to the circle.
The side BC has length 12 cm and the segment BD has length 4 cm. Find the area of the triangle ABC (in cm).
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A triangle ABC with an obtuse angle B is inscribed in a circle. The altitude AD of the triangle is tangent to the circle. The side BC has length 12 cm and the segment BD has length 4 cm.
Given,
BC = 12 cm
BD = 4 cm
The secant CD has a total length of,
CD = BC - BD
12 - 4 = 8 cm
It's external part is 4 cm long.
So, |AD|^2 = |BD| × |CD|
4 × 8 = 32
Hence AD = √32 = 5.65 cm
Area of triangle ABC
= bh / 2
= BC × AD / 2
= 12 × 5.65 / 2
= 33.9 ≈ 34 sq. cm
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