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.