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.
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