# Thread: length of three sides

1. ## length of three sides

The perimeter of a right triangle is 60 cm and the altitude perpendicular to the hypotenuse is 12 cm. What are the lengths of the three sides?

2. I'm assuming altitude means height. If that is so, what do we know:

We know the perimiter of the triangle is 60cm, which means we know that $B+A+H=60CM$, where B is the base, A is the altitude, and H is the hypotenuse of our triangle.

We also know that the Altitude (A) is 12CM, so:

$B+12+H=60CM$

Now we also know that $B^{2}+A^{2}=H^{2}$ by the Pytha Theorem yes?

Using these two formulas, we can do some substitutions and solve for these worrisome sides!

Can you take it from here?

3. Originally Posted by ANDS!
We also know that the Altitude (A) is 12CM, so:

$B+12+H=60CM$
No. 12 is not the length of a side; it is the length of a line inside the triangle,
from the right angle corner to the hypotenuse, perpendicular to the hypotenuse.

So this line forms 2 right triangles inside the given right triangle: side 12
is common to both.

4. Well good show. Even easier. We've two right triangles formed by splitting an angle of 90 degrees into two. From there its simply a matter of using properties of a 45-45-90 triangle to figure out the sides.

5. Originally Posted by ANDS!
Well good show. Even easier. We've two right triangles formed by splitting an angle of 90 degrees into two. From there its simply a matter of using properties of a 45-45-90 triangle to figure out the sides.
No. The 90 degree angle is not split 45-45.

6. OK; works like this (not as easy as it looks!):

Let a = one leg, b = other leg, c = hypotenuse

Since perimeter = 60, then c = 60 - a - b; so:
a^2 + b^2 = (60 - a - b)^2 ; simplify to get:
b = (60a - 1800) / (a - 60) [1]

Using areas: ab = 12(60 - a - b) ; simplify to get:
b = (720 - 12a) / (a + 12) [2]

From [1] and [2]:
(60a - 1800) / (a - 60) = (720 - 12a) / (a + 12) ; simplify to get:
a^2 - 35a + 300 = 0
(a - 20)(a - 15) = 0
a = 20 or 15 : that's the lengths of the 2 legs.
So hypotenuse = 60 - 15 - 20 = 25.

So you end up with a 15-20-25 right triangle.

7. Hello, Confused25!

The perimeter of a right triangle is 60 cm and the altitude to the hypotenuse is 12 cm.
What are the lengths of the three sides?
Code:
            C
*
*|  *
b * |     *   a
*  |12      *
*   |           *
*    |              *
A *-----+-----------------* B
: - - - - - c - - - - - :

Given: right triangle $ABC,\;C = 90^o$

We have: . $\begin{array}{cc}a + b + c \:=\:60 & {\color{blue}[1]} \\ a^2 +b^2 \:=\:c^2 & {\color{blue}[2]} \\ ab \:=\:12^2 & {\color{blue}[3]} \end{array}$

From [1], we have: . $a + b \:=\:60-c$

$\text{Square both sides: }\;(a+b)^2 \:=\:(60-c)^2 \quad\Rightarrow\quad \underbrace{a^2 + b^2}_{c^2} + 2\underbrace{(ab)}_{144} \:=\:3600 - 120c + c^2$
. . $120c \:=\:3312 \quad\Rightarrow\quad\boxed{ c \:=\:27.6}$

Substitute into [1]: . $a + b + 27.6 \:=\:60 \quad\Rightarrow\quad a + b \:=\:32.4 \quad\Rightarrow\quad b \:=\:32.4 - a$ .[4]

Substitute into [2]: . $a^2 + b^2 \:=\:27.6^2$ .[5]

Substitute [4] into [5]: . $a^2 + (32.4-a)^2 \:=\:27.6^2 \quad\Rightarrow\quad a^2 - 32.4a + 144 \:=\:0$

Quadratic Formula: . $a \;=\;\frac{32.4 \pm \sqrt{473.76}}{2} \;=\; \frac{32.4 \pm 12\sqrt{3.29}}{2}$

Hence: . $\boxed{a \;=\;16.2 \pm6\sqrt{3.29}}$

Substitute into [4]: . $b \;=\;32.4 - (16.2 \pm6\sqrt{3.29}) \quad\Rightarrow\quad \boxed{b \;=\;16.2 \mp\sqrt{3.29}}$

8. Originally Posted by Soroban
[size=3]

We have: . $\begin{array}{cc}a + b + c \:=\:60 & {\color{blue}[1]} \\ a^2 +b^2 \:=\:c^2 & {\color{blue}[2]} \\ ab \:=\:12^2 & {\color{blue}[3]} \end{array}$

I think your third equation is incorrect
It should be:
$ab=12c$

9. I think Mr Soroban's having a bad day

Right triangle 3-4-5 has height to hypotenuse = 1.2 is something
we memorize in grade 9 or so.

Anyhow, answer is a = 15, b = 20, c = 25.

As a general case, given p = perimeter and h = height to hypotenuse:

b,a = [-v +,- SQRT(v^2 - 4uw)] / (2u), where:
u = 2(p + h)
v = -p(p + 2h)
w = p^2 h

OK, Mr.Confused; now try this one:
perimeter = 1400, height to hypotenuse = 168 ; calculate a and b