Can someone help me find a solution for the height of the $3,\;4,\;5$ triangle?

Question

$AB=5,0;\; AC=3,0\;$and $BC=4,0$. What is the height?

Answer 1

Compute the area of triangle in two ways. One way is $\frac{1}{2}(h)(5)$ and the other way is $\frac{1}{2}(3)(4)$.

Answer 2

METHOD $1$.

Suppose the perpendicular drops on a point $P$ (on line $AB$). Then suppose the length $AP$ is $x$ and $BP$ ultimately becomes $5-x$ .

Now use Pythagoras theorem and you will have $4^2-h^2=(5-x)^2$ and $3^2-h^2=x^2$. You have two equations and two variables.

From here you can go further as: $4^2-h^2-(3^2-h^2)=(5-x)^2-(x^2)$ which gives $x=9/5$.

Since $3^2-h^2=x^2$ thus $h= 12/5$.

METHOD $2$.

As known area of triangle is $\frac{(ab)sinC}{2}$ and is also equal to $\frac{hc}{2}$.

THus $\frac{(ab)sinC}{2}$=$\frac{hc}{2}$. Putting $a=4,b=3,c=5$ and $sinC=1$ we get that $h=12/5$.

METHOD $3$.

Since you know all the sides, you can use Heron's formula to get area as $AREA=\sqrt{s(s-a)(s-b)(s-c)}$. Then equate it to $\frac{hc}{2}$ and you will get $h=12/5$ again.

Answer 3

Let $D$ be the point where $h$ strikes $AB$ .Then let $AD=y\implies DB=5-y$

Now $(5-y)^2+h^2=16;y^2+h^2=9$

Subtracting we get $9-10y=-9\implies y=\dfrac{9}{5}$

Hence $AD=\dfrac{9}{5}\implies h=\dfrac{12}{5}$

Answer 4

You can find the area of the triangle by using Hero's formula $\sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$ and then substituting area and base in $Area=\frac{1}{2}*base*height$