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?

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- Aug 6th 2009, 06:44 PMConfused25length 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?

- Aug 6th 2009, 08:13 PMANDS!
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 $\displaystyle B+A+H=60CM$, 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:

$\displaystyle B+12+H=60CM$

Now we also know that $\displaystyle B^{2}+A^{2}=H^{2}$ 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? - Aug 6th 2009, 09:06 PMWilmer
- Aug 6th 2009, 09:14 PMANDS!
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.

My bad. - Aug 6th 2009, 09:30 PMWilmer
- Aug 6th 2009, 09:43 PMWilmer
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. - Aug 7th 2009, 06:19 AMSoroban
Hello, Confused25!

Quote:

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 $\displaystyle ABC,\;C = 90^o$

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

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

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

. . $\displaystyle 120c \:=\:3312 \quad\Rightarrow\quad\boxed{ c \:=\:27.6}$

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

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

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

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

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

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

- Aug 7th 2009, 06:34 AMmalaygoel
- Aug 7th 2009, 08:39 AMWilmer
I think Mr Soroban's having a bad day (Crying)

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