In an isosceles triangle, the base is 6 cm, and the lateral side is 5 cm. Find the distance from the point

| September 6, 2021 | education

In an isosceles triangle, the base is 6 cm, and the lateral side is 5 cm. Find the distance from the point of intersection of the heights of the triangle to its vertices.

Triangles ABН and AСН are rectangular with a common leg AH. Let BH = X cm, then CH = 5 – X cm.

By the Pythagorean theorem, AH ^ 2 = AB ^ 2 – BH ^ 2 = 25 – X ^ 2.

AH ^ 2 = AC ^ 2 – CH ^ 2 = 36 – (5 – X) ^ 2 = 36 – 25 + 10 * X – X ^ 2 = 11 + 10 * X – X ^ 2.

Then 25 – X ^ 2 = 11 + 10 * X – X ^ 2.

10 * X – 14 = 0.

X = BH = 1.4 cm, then CH = 5 – 1.4 = 3.6 cm.

AH ^ 2 = 25 – 1.96 = 23.04.

AH = 4.8 cm.

The right-angled triangles AOM and BOH are similar in acute angle, then the triangles BOH and AНС are also similar in acute angle.

Then BH / BO = AH / AC.

BО = BH * AC / AH = 1.4 * 6 / 4.8 = 1.75 cm.

In a right-angled triangle, BOH OH ^ 2 = BO ^ 2 – BH ^ 2 = 3.0625 – 1.96 = 1.1025.

OH = 1.05 cm.

Then AO = CO = AH – OH = 4.8 – 1.05 = 3.75 cm.

Answer: From point O to the vertices of the triangle 1.75 cm, 3.75 cm, 3.75 cm.



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