# Given an isosceles triangle with base 16 and lateral side 10. Find the median drawn to the lateral side.

1) For the condition, an isosceles triangle is given. In order to find the value of the median, you need to draw a ray from the angle from which the median is drawn, where the point of intersection of the median with the lateral side will be the middle. Connect the end point of the ray to the vertices of the other two corners of the triangle. As a result, we get a figure ABCD – parallelograms (since in this quadrangle BO = OC, since AO is the median of triangle ABC).

2) Since the diagonals in ABCD at the intersection point are halved, based on the properties). Then, by the property of the diagonals, it will be equal to: AD ^ 2 + BC ^ 2 = 2 * (AB ^ 2 + AC ^ 2). AD ^ 2 + 10 ^ 2 = 2 * (10 ^ 2 + 16 ^ 2). AD ^ 2 = 2 * 100 + 2 * 256 – 100. AD ^ 2 = 612. AD = 6 * √ 17. 3) Based on this, then the median AO = 1/2 AD = 1/2 * 6 * √ 17 = 3 * √ 17.

Answer: 3 * √17.

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