In triangle ABC, a median line is drawn parallel to the side AC. She divided the triangle into a quadrilateral and a triangle. Their perimeters are respectively 12:11. Find AC and perimeter of triangle ABC
Let’s take advantage of the fact that this middle line divides the sides of the triangle AB and BC in half and is equal to half of the AC side.
Let us denote the point at which the middle line crosses the AB side through M, the point at which the middle line crosses the BC side through N.
| AM | = | MV | = (1/2) * | AB |;
| ВN | = | NC | = (1/2) * | ВС |;
| MN | = (1/2) * | AC |.
According to the condition of the problem, the perimeters of the quadrilateral and triangle, into which the middle line divides the triangle ABC, are 12 and 11.
Therefore, we can write the following relationship:
(1/2) * | AB | + (1/2) * | BC | + (1/2) * | AC | = 11;
(1/2) * | AB | + (1/2) * | BC | + (1/2) * | AC | + | AC | = 12.
Subtracting the first ratio from the second, we get:
(1/2) * | AB | + (1/2) * | BC | + (1/2) * | AC | + | AC | – (1/2) * | AB | – (1/2) * | ВС | – (1/2) * | AC | = 12 – 11;
| AU | = 1.
Multiplying both sides of the first ratio by 2, we get:
2 * (1/2) * | AB | + 2 * (1/2) * | BC | + 2 * (1/2) * | AC | = 2 * 11;
| AB | + | Sun | + | AC | = 22.
Answer: | AC | = 1, the perimeter of triangle ABC is 22.