In an isosceles triangle, the base is 48 cm, and the bisector drawn to the base is 18 cm.

| July 16, 2021 | education

In an isosceles triangle, the base is 48 cm, and the bisector drawn to the base is 18 cm. Find the median drawn to the lateral side.

Since triangle ABC is isosceles, then the bisector BH is also the height and median of the triangle, then AH = CH = AC / 2 = 48/2 = 24 cm, and triangles ABH and CBH are rectangular.

By the Pythagorean theorem AB ^ 2 = AH ^ 2 + BH ^ 2 = 576 + 324 = 90.

AB = BC = 30 cm.

Let us define in triangle BAH the cosine of angle A.

CosA = AH / AB = 24/30 = 4/5.

In the AMC triangle, the segment AM = AB / 2 = 30/2 = 15 cm.

From the triangle AMC, by the cosine theorem, we determine the length of the side of the CM.

CM ^ 2 = AM ^ 2 + AC ^ 2 – 2 * AM * AC * CosA = 225 + 2304 – 2 * 15 * 48 * 4/5 = 2529 – 1152 = 1377.

CM = 9 * √17 cm.

Answer: The length of the median CM is 9 * √17 cm.



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