What is the length of the base of an isosceles triangle if the center of the inscribed circle divides the altitude to the base into the ratio of 12: 5 (from the vertex to the base), and the length of a leg is 60 cm?

  50 cm

Step-by-step explanation:

Consider isosceles triangle AEF in the attachment. Point B is the center of the incircle, and it divides altitude AC into segments having the ratio 12:5.

Triangle ABD is similar to triangle AEC by AA similarity. (Angle A is the same for both right triangles. Then the ratio of hypotenuse to short leg will be the same for each. In triangle ABD, that ratio is 12:5, as given by the problem statement. Since we know AE = 60 cm, also from the problem statement, we know that ...

  AB/BD = AE/EC

  12/5 = 60 cm/EC

so ...

  EC = (60 cm)·(5/12) = 25 cm

Base length EF is twice that, or 50 cm.

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step-by-step explanation: