2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think you should find the ratio of areas first, then get the scale factor $\endgroup$ – AgentS Mar 30 '15 at 2:27
  • $\begingroup$ I tried that at first, calculating area of one and the other triangle and dividing those to get the scale factor. It got me confused to hell though. I got 578.68 cm^2 over 21.12 cm^2. Then I tried that scale factor for new sides with old sides lengths... but it is most probably wrong since it didn't work. Maybe if I put the ratio of one area over the other into equation and one of the lengths over x on the other side? It would make for a one looong equation if I were to include Heron's formula though. $\endgroup$ – MathAgain Mar 30 '15 at 2:30
  • 2
    $\begingroup$ @ganeshie8 you were right. I overcomplicated things! I took a ratio of one area over the other, took a square root of that ratio and divided each of the side lengths by that ratio to get to the side lengths. If you enter an answer I'll mark it as solved. Thanks! $\endgroup$ – MathAgain Mar 30 '15 at 3:06
  • $\begingroup$ that looks perfect!! $\endgroup$ – AgentS Mar 30 '15 at 4:05
1
$\begingroup$

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 $\{4.4,10,12\}$

$\endgroup$
  • $\begingroup$ Thanks for the effort Joffan ! @ganeshie8 already put me on the right track in the mean time. I did the same, just switched the sides of the areas for an arbitrary reason of my own :) $\endgroup$ – MathAgain Mar 30 '15 at 3:08
  • $\begingroup$ @MathAgain no problem, when I wrote this you still seemed to be struggling but clearly you've used the difficulty to reach a better understanding, which is great. $\endgroup$ – Joffan Mar 30 '15 at 3:11
  • $\begingroup$ the struggle is always present and real! $\endgroup$ – MathAgain Mar 30 '15 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.