In an equal coastal triangle ABC with base AC, the medians intersect at point O. Find the area
In an equal coastal triangle ABC with base AC, the medians intersect at point O. Find the area of triangle ABC if OA = 13 cm. OB = 10 cm.
The medians of the triangle, at the point of their intersection, are divided by a ratio of 2/1 starting from the apex.
Then OB / OM = 2/1.
ОМ = ОВ / 2 = 10/2 = 5 cm.
Then BM = BО + ОМ = 10 + 5 = 15 cm.
In an isosceles triangle, the median drawn to the base also has its height and bisector, then BM is perpendicular to AC, and triangle AOM is rectangular.
By the Pythagorean theorem, in the right-angled triangle AOM, we determine the length of the leg AM.
AM ^ 2 = AO ^ 2 – OM ^ 2 = 169 – 25 = 144.
AM = 12 cm.Then AC = 2 * AM = 2 * 12 = 24 cm.
Determine the area of the triangle ABC.
Savs = AC * BM / 2 = 15 * 24/2 = 180 cm2.
Answer: The area of the triangle is 180 cm2.