In △ MNK: MN = NK – lateral sides, MK = 10 – base, MA, NB and KC – medians intersecting at point S, NS = 6.
1. It is known that the median points of intersection are divided in a ratio of 2: 1 starting from the top, that is:
NS / BS = 2/1.
Let’s find the length BS:
6 / BS = 2/1;
BS = 6 * 1/2 (proportional);
BS = 6/2 = 3.
2. Find the length NB:
NB = NS + BS;
NB = 6 + 3 = 9.
3. Since NB is the median drawn to the base MK of the isosceles △ MNK, then NB is both the median, and the height, and the bisector.
The area of a triangle is found by the formula:
S = ah / 2,
where a is the length of the side of the triangle, h is the length of the height of the triangle drawn to side a.
Then the area △ MNK is equal to:
S = MK * NB / 2 = 10 * 9/2 = 90/2 = 45.
Answer: S = 45.
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