IMO A

Let the heights h1 and h2 be 3x and 2x respectively. And sides be a1 and a2.

Find the side (a) of the triangle first:
height = root(3)*a/2 ==> a = height * 2/root(3)

a1 = h1 * 2/root(3) = 3x * 2/root(3) = 6x / root(3)
a2 = h2 * 2/root(3) = 2x * 2/root(3) = 4x / root(3)

Now substitute a and h values in Area formula,
Area = root(3) * a^2/4

Let Area of larger and smaller triangles be A1 and A2.

A1 = root(3) * a1^2/4 = root(3) * (6x/root(3))^2 / 4 = 9x^2 / root(3)
A2 = root(3) * a1^2/4 = root(3) * (4x/root(3))^2 / 4 = 4x^2 / root(3)

A1 / A2 = [9x^2 / root(3)] / [4x^2 / root(3)] = 9/4.


Although this method looks pretty long, believe me that it took only a min to solve.

treker wrote:If the ratio of the heights of two equilateral triangles is 3/2 , what is the ratio of the area of the larger triangle to the smaller triangle?

A. 9/4
b. 3/2
c. (3sqrt(3))/2
d. sqrt(3)/2
e.sqrt(3)/4

Please explain your answers.

Thanks!

Instead of lots of math, let's think this out.

All linear measurements in similar triangles are proportional. So, if the ratio of the heights is 3/2, the ratio of the bases will also be 3/2.

Since area of a triangle is (1/2)(base)(height), the ratio of area is simply (ratio of bases)*(ratio of heights) = (3/2)(3/2) = 9/4... choose A.

Awesome shortcut! Thanks stuart!