In a triangle AB = 28 cm (one side), the other side is divided by the point of contact

| March 24, 2021 | education

In a triangle AB = 28 cm (one side), the other side is divided by the point of contact of the inscribed circle into 12 cm and 14 cm segments. Find the perimeter and area of the triangle.

We construct the radii OK, ОМ, ОН to the points of tangency of the circle and the triangle.

By the property of tangents drawn from one point, the lengths of the tangents are equal.

CM = CH = 14 cm, ВK = ВM = 12 cm, AН = AK.

Then AK = AB – ВK = 28 – 12 = 16 cm, and therefore AH = AK = 16 cm.

Side length BC = CM + BM = 14 + 12 = 26 cm.

Side length AC = AH + CH = 16 + 14 = 30 cm.

The perimeter of the triangle ABC is equal to: Ravs = AB + BC + AC = 28 + 26 + 30 = 84 cm.

The semi-perimeter of the ABC triangle is: p = 84/2 = 42 cm.

Then, according to Heron’s theorem: Sav = √42 * (42 – 28) * (42 – 26) * (42 – 30) = √42 * 14 * 16 * 12 = √112896 = 336 cm2.

Answer: The perimeter of the triangle is 84 cm, the area is 336 cm2.



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