The center of the circumscribed circle divides the height of an isosceles triangle drawn to the base into segments, the smaller of which is 8 cm, the base of the triangle is 12 cm. Find the area of this triangle.
Let’s draw a segment OA, which is equal to the radius of the circumscribed circle.
Consider a right-angled triangle AOH, whose angle H is straight, since BH is the height of the triangle, leg OH = 8 cm, and leg AH = AC / 2 = 12/2 = 6 cm, since in an isosceles triangle the height coincides with the median.
Then, by the Pythagorean theorem:
OA ^ 2 = OH ^ 2 + AH ^ 2 = 8 ^ 2 + 6 ^ 2 = 64 + 36 = 100.
ОА = 10 cm.
Segment ОВ = ОА = 10 cm, since both segments are circular radii.
Then the height BH = OB + OH = 10 + 8 = 18 cm.
Determine the area of the triangle ABC.
S = AC * BH / 2 = 12 * 18/2 = 108 cm2.
Answer: The area of the triangle is 108 cm2.
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