I've been working through this task in an old textbook and can't figure out where I'm wrong. I suspect my whole approach is wrong.
Task says: Given the side lengths of a triangle that are equal to 5.5 cm, 12.5 cm, and 15 cm. Similar triangle has an area of 21.12 cm^2. What are the lengths of this triangle?
So, my approach is basically to try and do this via ratios.
$\frac{12.5}{5.5}=\frac{x}{y}$
$\frac{x}{y}=\frac{25}{2}\frac{2}{11}=\frac{25}{11}$
everything times $11y$
$11x=25y$
that is, $y = \frac{11x}{25}$
and then, $x\frac{25}{11x}=\frac{25}{11}$ and everything times $11x$ gives $25x=25$ or $x=1$
so, now I have $x=1, y = \frac{11}{25}$ and via further substition I got the third side to be equal to 1.2
So, my final relations to the task were $$a = 5.5 cm = 0.44 cm$$ $$b = 12.5 cm = 1 cm$$ $$c = 15 cm = 1.2 cm$$
However, when I add these new side lengths to an area (via Heron's formula) the number is right, by 100x less. Off by a factor of two.
I can't understand where I'm wrong. I SUSPECT that I am doing the task the wrong way, and I think the hint is that I was also given an area for the triangle I need to find the sides for.
Use Heron's formula for area, $T = \sqrt{s(s-a)(s-b)(s-c)}$
where $s= \frac{a+b+c}{2}$ is half of the triangle's perimeter, to get the area for the initial triangle.
$$\begin{align}s &= \frac{5.5+12.5+15}{2}=16.5 \\[1em] T &= \sqrt{16.5\times 11 \times 4 \times 1.5} \\[0.5em] &= 33 \end{align}$$
Then the ratio of lengths is $\sqrt{21.12/33} = \sqrt{0.64} = 0.8$
And the similar triangle has side lengths $\{4.4,10,12\}$