Finding all dimensions of an isosceles triangle equally divided by 3?

Question

If given an isosceles triangle with a total area of 30, and angles 30, 75, 75 (with the shortest side down), then divided by 3 horizontally (such that no piece is symmetrical, but each piece contains 10 for the area), what is the dimensions of all sides of what would now be a smaller isoscelese triangle, and 2 trapezoids? And how do you go about solving this?

Answer 1

First the dimensions of the base and the height

(1/2)bh = 30

bh = 60

h / (0.5b) = tan 75

h = 0.5 tan 75 b

h = (0.5)(3.732)b

60 / b = 1.866b

60 = 1.866b²

32.1539 = b²

5.67 = b

10.58 = h

Consider the total area of the top two sections is (2/3) of the entire triangle.

Therefore the linear scale factor is 4/9

The lower base of the middle trapezoid is (4/9)(5.57).

This base is (4/9)(10.58) from the vertex.

The area of the top triangle is 1/2 that of the triangle in the above solution.

Therefore the linear scale factor is 1/4.

The upper base of the middle trapezoid is (1/4)(4/9)(5.57).

This base is (1/2)(4/9)(10.58) from the vertex.

You now have the dimensions of the bases and heights of the three parts.

Answer 2

I solved the first part (almost)!

This is a long problem!

I gave you some guidance toward finding the rest of the pieces.