The lateral side of an isosceles triangle is 40 cm, and the height drawn to the base is 4√91 cm. Find the distance between

| February 6, 2021 | education

The lateral side of an isosceles triangle is 40 cm, and the height drawn to the base is 4√91 cm. Find the distance between the points of intersection of the bisectors of the angles at the base of the triangle with its lateral sides.

Since BH is height, then triangle ABH is rectangular, then, according to the Pythagorean theorem, AH ^ 2 = AB ^ 2 – BH ^ 2 = 1600 – 1456 = 144.

BH = 12 cm.

Since the triangle ABC is isosceles, then ВН is also a median and bisectors, then AC = 2 * AH = 24 cm.

By the property of the bisector of a triangle, AK / AC = ВK / BC.

Let AK = X cm, then ВK = 40 – X cm.

X / 24 = (40 – X) / 40.

960 – 24 * X = 40 * X.

64 * X = 960. X = 960/64 = 15 cm.

AK = 15 cm, then ВK = 40 – 15 = 25 cm.

Triangles AKC and AMC are equal in side and two adjacent angles, then AK = CM, which means KM is parallel to AC.

Triangles ABC and ВKM are similar in two angles with a similarity coefficient K = ВK / AB = 25/40 = 5/8.

Then KM / AC = 5/8.

KM = 5 * AC / 8 = 5 * 24/8 = 15 cm.

Answer: The length of the KM segment is 15 cm.



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