# In an isosceles triangle, the length of the lateral side is 5, and the area of the triangle is 12.

**In an isosceles triangle, the length of the lateral side is 5, and the area of the triangle is 12. Point M is taken from the base of the triangle. Find the sum of the distances from point M to the lateral sides of the triangle.**

Dan △ ABC: AB = BC = 5, S △ ABC = 12.

1. Draw a segment BM from the vertex B to the point M. BM divides the original △ ABC into 2 triangles △ ABM and △ CBM.

In △ ABM we draw the height MH (MH is the distance from point M to the side AB), and in △ CBM we draw the height MK (MK is the distance from point M to the side BC).

The area of a triangle is found by the formula:

S = ah / 2,

where a is the side of the triangle, h is the height drawn to the side a.

Area △ ABM is equal to:

S △ ABM = AB * MH / 2 = 5MH / 2.

Area △ CBM is equal to:

S △ CBM = BC * MK / 2 = 5MK / 2.

2. Area △ ABC is equal to the sum of areas △ ABM and △ CBM:

S △ ABC = S △ ABM + S △ CBM;

5MH / 2 + 5MK / 2 = 12;

(5MH + 5MK) / 2 = 12;

5 (MH + MK) = 2 * 12 (proportional);

MH + MK = 24/5;

MH + MK = 4.8.

Answer: MH + MK = 4.8.