How do I find the height of a triangle when it is tilted downwards at one end?

Question

In the first pic, it is shown that the height of the triangle is $1.5$ m. In the second pic, the point $C$ is moved to point B. How do I find height $h$ so that the perpendicular height of the triangle stays the same at $1.5$ m?

Answer 1

based on your given values, this is impossible if the triangle has just been shifted because if you take AC in figure 1 to be the same as AB in figure 2, the hight h would have to be 0 in order to satisfy Pythagorean theorem. the length between point B ant the endpoint of line h that isn't point A would have to be shorter than 60

Answer 2

Labeling the corner at the base of the altitude as $F$, then $EFB$ and $ADB$ are similar. This gives $$\frac{h}{60} = \frac{1.5}{\left(\frac{\sqrt{60^2+h^2}}{2}\right)}$$ so $$\frac{180}{h} = \sqrt{60^2+h^2}$$ squaring $$\frac{32400}{h^2} = 3600+h^2$$ so $$h^4+3600h^2-32400=0$$ giving (via quadratic formula) $$h^2=180(\sqrt{101}-10)$$ (the other solution is negative so won't help here) $$h=6\sqrt{5\left(\sqrt{101}-10\right)}\approx 2.996266$$ This puts the length of $EB$ at about $30.074$.

Answer 3

Begin by setting some variables. Let $x$ be the length of the congruent sides on the isosceles triangle and $h$ be the height we seek. Consider triangle $ADE,$ which is right. By Pythagoras, we have that $$h^2 + (60 - x)^2 = x^2$$ $$h^2 + 3600 - 120x + x^2 = x^2$$ $$h^2 + 3600 = 120x.$$

Not very pretty right? Well now let us consider the similar triangles $BXE$ and $BDA.$ We can write $$\frac{1.5}{x} = \frac{h}{\sqrt{h^2 + 3600}}.$$

But we know what $\sqrt{h^2 + 3600}$ is! It's from the previous calculation. So after substituting and manipulating the equations, we have the quartic equation in terms of $h,$ $$h^4 + 3600h^2 - 32,400 = 0.$$

So with quadratic equation, we find that the exact value of $h$ is $\boxed{6\sqrt{5(\sqrt{101} - 10)}}$ and this approximates to $\boxed{2.996}$ meters.

Answer 4

Note that the triangle $\triangle ABE$ is similar to $\triangle HFJ$, whose height is $\dfrac{h}{2}$ and whose hypotenuse is $1.5$. The side $JF$ corresponds to side $EB$ and the side $HJ$ corresponds to side $AE$.

Using the Pythagorean theorem we compute the side $HJ$ to have length $\tfrac{1}{2}\sqrt{9-h^2}$.

Because of similar triangles we have the ratio

\begin{equation} \frac{HJ}{JF}=\frac{AE}{EB} \end{equation}

Which reduces to

\begin{equation} \frac{\sqrt{9-h^2}}{h}=\frac{h}{60} \end{equation}

Simplifying gives us the fourth degree equation in $h$:

\begin{equation} h^4+3600h^2-32400=0 \end{equation}

which has the form of a quadratic equation and results in

\begin{equation} h^2=180\sqrt{101}-1800 \end{equation}

from which we conclude that

\begin{equation} h=6\sqrt{5\sqrt{101}-50} \end{equation}