An equilateral triangle has a side length of 6. What is the height of the triangle?
Accepted Solution
A:
Since the height of an equilateral triangle in terms of its side s is sβ3/2, the height of the triangle is 6β3/2 = 3β3 and so the area is (1/2)(6)(3β3) = 9β3.Β
If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3β3 - h.Β
Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have:Β
(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle)Β ==> x/6 = (3β3 - h)/(3β3)Β ==> x = (6β3 - 2h)/β3Β
Thus, the area of the upper triangle is:Β
A = (1/2)[(6β3 - 2h)/β3](3β3 - h) = [(6β3 - 2h)(3β3 - h)]/(2β3).Β (Made a dumb mistake about the height here for some reason)Β
Since we require that the area of this triangle is to be half of the total area (9β3/2), we need to solve:Β