Question 1097349
For similar shapes (triangles, pyramids, circles, spheres, or any other kind of 2-D or 3-D shape),
if the ratio of corresponding lengths is {{{k}}} .
the ratio of corresponding surfaced areas is {{{k^2}}} , and
the ratio of volumes (if they are 3-D shapes) is {{{k^3}}} .
You can see that is true for squares and cubes,
but you can make any shape you want with a very large number
of much smaller squares or cubes.
 
In this case, the height of triangle DEF is {{{13/5}}} times the height of triangle ABC,
so the area of DEF is {{{(13/5)^2=13^2/5^5}}} times the area of ABC.
That would be, in square cm,
{{{5*(13^2/5^2)=13^2/5=169/5=33.8}}} .
It rounds to {{{highlight(34cm^2)}}} .