Final answer:
To solve for the legs of the isosceles triangle ABC with 30-degree base angles and a median to the base, we equate the sum of the perimeters of sub-triangles ACD and BCD to the perimeter of ABC plus 20 cm. Solving the equation, we find that the length of each leg of the triangle ABC is 20 cm.
Step-by-step explanation:
The question involves calculating the lengths of the legs of an isosceles triangle given the perimeters of the triangle and its sub-triangles formed by a median. First, let's denote the lengths of the legs of the isosceles triangle ABC as AB = BC = l and the base as AC = b. Since CD is the median, we have AD = DC = b/2. The base angles of the triangle are 30 degrees.
The perimeter of triangle ABC is PABC = 2l + b. According to the problem, the sum of the perimeters of triangles ACD and BCD is 20 cm more than the perimeter of triangle ABC. Let PACD denote the perimeter of triangle ACD and PBCD denote the perimeter of triangle BCD. The sum of the perimeters of these two triangles is PACD + PBCD = 2l + 2b, because the median length is added twice - once for each triangle.
Now we can form the equation PACD + PBCD = PABC + 20. Plugging in the known expressions for perimeters, we get 2l + 2b = 2l + b + 20, which simplifies to b = 20. Now we must use the 30-degree angle to find l. In an isosceles triangle with a 30-degree base angle, the height forms a 30-60-90 triangle. Since the short side opposite the 30-degree angle is half the hypotenuse in such a triangle, l is equal to b, which means l = 20 cm.