In a triangle with sides 25 cm, 25 cm, 14 cm, find the distance from the point of intersection

| March 12, 2021 | education

In a triangle with sides 25 cm, 25 cm, 14 cm, find the distance from the point of intersection of the medians to the vertices of the triangle.

Since, by condition, AB = BC = 25 cm, then the ABC triangle is isosceles, and the median BH is also the height of the triangle.

The median BH divides the base of the AC in half, then AH = CH = AC / 2 = 14/2 = 7 cm.

In a right-angled triangle ABН, we determine the length of the ВН leg.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 625 – 49 = 576.

BH = 24 cm.

The medians of the triangle, at the point of their intersection, are divided by a ratio of 2/1, starting from the vertex.

Then ВO = 2 * OH.

BH = 24 = OH + 2 * OH = 3 * OH.

OH = 24/3 = 8 cm.

ВO = 24 – 8 = 16 cm.

In a right-angled triangle AOН, AO ^ 2 = OH ^ 2 + AH ^ 2 = 64 + 49 = 113.

AO = CO = √113 cm.

Answer: The distance from the point of intersection of the medians to the vertices of the triangle is 8 cm and √113 cm.



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