Calculating Trapezoid's three equal sides with just knowing: base and height? It is really an architectural problem. For example when you have a bay window and you want to find out the three equal sides of bay window by just having window opening size and how far out is protruding Bay window
That what I'm trying to calculate: the given are the base and the height.
I want to calculate three equal sides of the trapezoid.
Any help greatly appreciated.
Thank you,
 A: The trapezoid can be labelled as follows, where $b$ is the base, $h$ is the height, and $k$ is the length of one of the three equal sides.

When the altitude $h$ is dropped, there is a gap between where it touches the base and where $k$ touches it. Let the length of this gap be $g$. Then
$$g=\sqrt{k^2-h^2}$$
and
$$b=k+2g$$
so
$$b=k+2\sqrt{k^2-h^2}$$
And we want to solve this for $k$.
$$b-k=2\sqrt{k^2-h^2}$$
$$(b-k)^2=4(k^2-h^2)$$
$$b^2-2kb+k^2=4k^2-4h^2$$
$$=3k^2+2bk-4h^2-b^2$$
Then solve using the quadratic formula:
$$k=\frac{-2b+\sqrt{4b^2-4(3)(-4h^2-b^2)}}{6}$$
$$k=\frac{-2b+\sqrt{4b^2+12(4h^2+b^2)}}{6}$$
$$k=\frac{-2b+\sqrt{16b^2+48h^2}}{6}$$
$$k=\frac{-2b+4\sqrt{b^2+3h^2}}{6}$$
$$k=\frac{-b+2\sqrt{b^2+3h^2}}{3}$$
I believe that this is the answer.
A: Let the base be $b$, the height $h$, the length to be found $a$, and the angle between the base and the sloping sides be $\theta$. (Similar triangles implies both such angles are the same). Then we have the two equations
$$ h = a\sin{\theta}, \\
b = a+2a\cos{\theta}. $$
We now eliminate $\theta$: we have
$$a^2\sin^2{\theta} = h^2, \qquad a^2\cos^2{\theta} = \frac{(b-a)^2}{4}, $$
and then adding and using the Pythagorean identity $\cos^2{\theta}+\sin^2{\theta}=1$ gives
$$ 4a^2 = 4h^2+(b-a)^2, $$
(we could get to here by using just Pythagoras and the length from the corner to the foot of the perpendicular), so
$$ 3a^2 +2ba = 4h^2+b^2 $$
Completing the square gives
$$ (a+b/3)^2 + \frac{b^2}{9} = \frac{4h^2+b^2}{3}, $$
so
$$ a = \frac{-b \pm 2\sqrt{3h^2+b^2}}{3} $$
Of course we want a positive root since $a$ is a length, so we have to take the $+$.
A: Let's call $L=230$ the opening size and $h=66$ the protruding depth.
I'll call $x$ the sides and $y$ the complement, i.e. $x+2y=L$.
Applying pythagoras' theorem in one of the rectangle triangle we have $y^2+h^2=x^2$
Now we solve in $y$ : $y^2+h^2=L^2-4Ly+4y^2\iff3y^2-4Ly+L^2-h^2=0$
$\Delta=16L^2-12(L^2-h^2)=4L^2(1+\frac{3h^2}{L^2})$
$y=\frac{4L-2L\sqrt{1+\frac{3h^2}{L^2}}}{6}$

$\displaystyle x=L-2y=\frac L3(2\sqrt{1+\frac{3h^2}{L^2}}-1)$

