-2
$\begingroup$

My sister asked me to help her with her homework for mathematics, however frustratingly I was not able to figure out how to solve it.

The assignment is as follows where it was requested to calculate the length between G and I. How should this assignment be solved?

Diagram of triangles

Thanks!

$\endgroup$
  • 1
    $\begingroup$ Please include your sister's effort as well as your attempts to solve it. $\endgroup$ – John Douma Jan 25 at 17:30
  • $\begingroup$ What has this to do with Pythagorean-triples? $\endgroup$ – José Carlos Santos Jan 25 at 17:31
  • 1
    $\begingroup$ As general advice, always compute what you can compute. Here, $\overline {HE}$ is easy. From that you can get $\overline {GE}$. Now you've got two right triangles left that you haven't used... $\endgroup$ – lulu Jan 25 at 17:34
0
$\begingroup$

It is a messy plane geometry and algebra problem. All the following using Pythagorean theorem. First $|H-E|=4$, then $|E-G|^2=32$. Let $x=|E-I|$ and $h=|G-I|$. The equation for $x$ and $h$ are $x^2+h^2=32$ and $(5-x)^2+h^2=49$. Solving these equations gives $x=.8$ and $h=5.6$, where $h$ is the answer to the problem.

$\endgroup$
  • $\begingroup$ The use of Cartesian coordinates (analytic geometry) may be outside the scope of the course material this exercise was intended to reinforce. $\endgroup$ – hardmath Jan 25 at 18:15
  • $\begingroup$ Thanks herb steinberg, this is indeed the way she was expected to solve it! My sister got some algebra lectures a while ago, so she apparently had to combine it with the current lectures about the pythagorean theorem. Nice ingenious way of solving it, thanks again! $\endgroup$ – A. Huijgen Jan 25 at 19:05
0
$\begingroup$

Hint: Let $$\angle{EFG}=\alpha$$ then $$\cos(\alpha)=\frac{3}{5}$$ and $$\sin(\alpha)=\frac{GI}{7}$$

$\endgroup$
  • $\begingroup$ I'd be concerned that trigonometry may be outside the scope of the course material this exercise was intended to reinforce. $\endgroup$ – hardmath Jan 25 at 17:59
  • $\begingroup$ Thanks Dr. Sonnhard Graubner for your response. So far she hasn't got lectures about trigonometry, but nice to see there are multiple ways to solve it. $\endgroup$ – A. Huijgen Jan 25 at 19:06

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.