Finding the lengths of 2 similar triangles?

Question

The lengths of two similar triangles are x^2 and xy, respectively. What is the ratio of the areas?

Answer 1

given a triangle with sides a, b and c.
and angles A B C opposite to the sides
(a*b *sin C)/2 is the area of this triangle.

If your triangles are similar a1/a2=b1/b2=c1/c2 , where a1,b1,c1 are the sides of triangle1
a2,b2,c2 are the sides of triangle2

So first area is : (a1*b1*sin C1)/2
second area: (a2*b2*sin C2)/2

the ratio is: a1*b1*sin C1 / a2 *b2 *sinC2 . But angle C1=C2 because they are similar. So ratio = a1/a2 *b1/b2.
But being similar a1/a2 = b1/b2 = c1/c2=x^2/(x*y)=x/y .
So the ratio is (x/y)^2

If you haven't studied sin yet you can prove that the heights that correspond to a1 and a2 for instance have the same x/y ratio as the sides (by showing that these heights are themselves sides in similar triangles). Then (a1*h1)/(a2*h2) is the ratio, and the final answer is the same.

Answer 2

x^2/y^2

Answer 3

(x^2/xy)^2=x^2/y^2