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?
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?
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?
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.
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 $\displaystyle ABC,\;C = 90^o$
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]: .$\displaystyle b \;=\;32.4 - (16.2 \pm6\sqrt{3.29}) \quad\Rightarrow\quad \boxed{b \;=\;16.2 \mp\sqrt{3.29}}$
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