posted by K .
1) The base angles of an isosceles triangle are (50-x)^6 and (30x-12)^6 what is the vertex angle?
2) The area of a trapezoid is 52cm^2 if it's bases are 8 cm apart find the sum of their heights
3) What is the area of a circle inscribed inside an equilateral triangle whose area is 4√3 cm^2
Please show solution
1) the base angles are equal, so
(50-x)^6 = (30x-12)^6
50-x = 30x-12
Find x, evaluate the angles, and subtract them from 180
2) I assume you mean lengths. If they are b and B,
8(b+B)/2 = 52
3)The medians of a triangle meet 2/3 of the way from a vertex to the opposite side. This triangle's medians are also its altitude. The side length is 4, so its altitude is 2√3. Thus, the radius of the circle is 2/3 of that.