The lateral side AB of the isosceles triangle ABC is 5 cm, the base of the AC is 6 cm, the center of the inscribed

| May 22, 2021 | education

The lateral side AB of the isosceles triangle ABC is 5 cm, the base of the AC is 6 cm, the center of the inscribed circle lies at the height of BH and is 2.5 cm away from the apex B. Find the radius of the inscribed circle.

Let’s denote the center of the inscribed circle O, then VO = 2.5 cm, OH – the radius of the inscribed circle.
The height BH, drawn to the base of the AC, is at the same time the median, which means it divides the base in half. Therefore, AH = CH = AC / 2 = 6/2 = 3 cm.
Consider a right-angled triangle ABH: AB – hypotenuse, AH and BH – legs.
The sum of the squares of the legs is equal to the square of the hypotenuse: AB ^ 2 = AH ^ 2 + BH ^ 2, hence BH ^ 2 = AB ^ 2-AH ^ 2 = 5 ^ 2-3 ^ 2 = 25-9 = 16, BH = √16 = 4.
BH = BO + OH;
The radius of a circle inscribed in a triangle ABC is: OH = BH-BO = 4-2.5 = 1.5 cm.



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