# Just curious

• Jun 15th 2007, 02:45 AM
Janner
Just curious
First of all, I don't know if this is the right sub-forum for this question. I never had the difference between geometry and trigonometry, so keep that in mind before flaming me.

Now my question;
If you know the length of the hypotenusa in a right triagle, and you know the something* of the two other sides, can you calculate what those lengths are? I think it is very possible. What if you only know the something** of the hypotenusa and the two sides?

*If AB = 4 and BC = 2, then the 'something' is 4:2, or 2:1.
**If AB = 4 and BC = 2, then the 'something' is sqrt(20):4:2, or sqrt(20):2:1.

And another thing, I've seen people use images of square roots and such, how do I use those too? And is there any good (and possibly free) maths software out the for geometry, parabolae and things like that? (I'm talking about 'visual' maths)

Thanks in advance for reading this, and another thank you to the people who will help me.

Bye, Jan.
• Jun 15th 2007, 04:40 AM
CaptainBlack
Quote:

Originally Posted by Janner
First of all, I don't know if this is the right sub-forum for this question. I never had the difference between geometry and trigonometry, so keep that in mind before flaming me.

Now my question;
If you know the length of the hypotenusa in a right triagle, and you know the something* of the two other sides, can you calculate what those lengths are? I think it is very possible. What if you only know the something** of the hypotenusa and the two sides?

*If AB = 4 and BC = 2, then the 'something' is 4:2, or 2:1.
**If AB = 4 and BC = 2, then the 'something' is sqrt(20):4:2, or sqrt(20):2:1.

1. If you know the hypotenuse (call it $b$), and you know the ratio $r$ of the
other two sides (call them $a$ and $c$) then you have:

$
a^2+c^2 = b^2
$

and

$
a/c=r
$
,

so we may write $a= c r$ which can be substituted into the first equation to give:

$
c^2 r^2 +c^2 = b^2
$

so:

$
c=\sqrt{b^2/(r^2+1)}
$

then $a$ can be found.

Quote:

And another thing, I've seen people use images of square roots and such, how do I use those too?
The type setting here uses LaTeX, which is places between $$and$$ tags. The tutorial is here.

RonL
• Jun 15th 2007, 04:40 AM
Soroban
Hello, Jan!

I think I understand your question . . .

Quote:

If you know the length of the hypotenusa in a right triagle,
and you know the ratio of the two other sides,
can you calculate what those lengths are? . Yes!

Example

The hypotenuse is 100. .The two sides are in the ratio 3:4.

We are given: . $a:b = 3:4$ and $c =100.$

Then $a$ and $b$ are of the form: . $\begin{array}{c}a \:=\:3k\\ b\:=\:4k\end{array}$ . for some $k > 0$

Using Pythagorus, we have: . $a^2 +b^2\:=\:c^2\quad\Rightarrow\quad(3k)^2 + (4k)^2 \:=\:100^2$

. . $9k^2 + 16k^2\:=\:10,000\quad\Rightarrow\quad 25k^2 \:=\:10,000\quad\Rightarrow\quad k^2 = 400\quad\Rightarrow\quad k = 20$

Hence: . $\begin{array}{c}a \:=\:3(20) \:=\:60 \\b \:=\:4(20) \:=\:80\end{array}$

Therefore, the triangle is: . $(a,\,b,\,c) \;=\;(60,\,80,\,100)$

Example

One side is 10. .The ratio of the other side to the hypotenuse is 2:3.

We are given: . $a = 10$ .and . $b\!:\!c = 2\!:\!3\quad\Rightarrow\quad b = 2k,\;c =3k$

Pythagorus says: . $a^2 + b^2\:=\:c^2\quad\Rightarrow\quad10^2 + (2k)^2\;=\;(3k)^2$

. . $100 + 4k^2 \:=\:9k^2\quad\Rightarrow\quad 5k^2 = 100\quad\Rightarrow\quad k^2 = 20\quad\Rightarrow\quad k = 2\sqrt{5}$

Hence: . $\begin{array}{c}b \:=\:2(2\sqrt{5}) \:=\:4\sqrt{5} \\ c \:=\:3(2\sqrt{5}) \:=\:6\sqrt{5}\end{array}$

Therefore, the triangle is: . $(a,\,b,\,c) \;=\;(10,\,4\sqrt{5},\,6\sqrt{5})$

. . [As you can see, $k$ need not be an integer.]
• Jun 16th 2007, 11:12 PM
Janner
I want to thank you both for the replies. The second by Soroban made more sense to me, because he used examples and so on, but I'm sure CaptainBlack has made a good point too. Still, there is one question left to be answered (atleast I think so) can you calculate the three sides of you have just the ratios? a:b:c = 1:2:3, where c is (ofcourse) the hypotenusa. I don't think it's really possible, atleast not in the way Soroban explained the other question.

Thanks, Jan.
• Jun 17th 2007, 01:21 AM
earboth
Quote:

Originally Posted by Janner
I want to thank you both for the replies. The second by Soroban made more sense to me, because he used examples and so on, but I'm sure CaptainBlack has made a good point too. Still, there is one question left to be answered (atleast I think so) can you calculate the three sides of you have just the ratios? a:b:c = 1:2:3, where c is (ofcourse) the hypotenusa. I don't think it's really possible, atleast not in the way Soroban explained the other question.

Thanks, Jan.

Goeien morgen,

you are right with only the ratio of sides given you'll get similar triangles but you can't calculate the sides of a specific triangle.

By the way: The ratios you gave here will never produce a right triangle:
$a : b : c = 1 : 2 : 3 \Longrightarrow \frac{a}{b} = \frac{1}{2} \text{ and } \frac{b}{c} = \frac{2}{3}$ That means: $a = \frac{1}{2}b$ and $c = \frac{3}{2}b$
Apply Pythagorean theorem:
$a^2 + b^2 = c^2 \longrightarrow \left( \frac{1}{2}b \right)^2 +\left( b \right)^2 \ne \left( \frac{3}{2}b \right)^2$
• Jun 17th 2007, 01:57 AM
Janner
I just picked some random numers, not really focusing on the numbers themself. Thanks for the answer though =)