A circle is inscribed in an isosceles triangle ABC with base AC. It touches side BC

| May 23, 2021 | education

A circle is inscribed in an isosceles triangle ABC with base AC. It touches side BC at point K. Find the radius of the circle if BK = 2, CK = 8.

Let’s draw the height BH of the triangle ABC. By the property of tangents to a circle drawn from one point, the lengths of these tangents are equal, then CH = CK = 8 cm.

The height BH of the isosceles triangle ABC is also its median, then AH = CH = 8 cm, and AC = 2 * 8 = 16 cm.

In a right-angled triangle BCH, we determine the length of the BH leg.

BH ^ 2 = BC ^ 2 – CH ^ 2 = 100 – 64 = 36.

BH = 6 cm.

Determine the area of ​​the triangle ABC.

Savs = AC * BH / 2 = 16 * 6/2 = 48 cm2.

Let’s define the semi-perimeter of the triangle ABC: p = (10 + 10 + 16) / 2 = 18 cm.

Then the radius of the inscribed circle is: OK = S / p = 48/18 = 8/3 = 2 (2/3) cm.

Answer: The radius of the inscribed circle is 2 (2/3) cm.



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