Given scalene trapezoid with the following characteristics:
- $c = DC = 2$ cm (shorter base)
- $b = BC = 8$ cm (leg)
- $d = AD = 5$ cm (leg)
- $\cos \angle ABC = 1/7$
Find $AB = a$ (longer base) and $S$ (area of the trapezoid).
What I know for sure
$S = 1/2 \times ((a + c) \times h)$
$P = a + b + c + d$
$\alpha + \delta = \beta + \gamma = 180°$
$h = d \sin \alpha = b \sin \beta$
My attempt
Given $\cos \angle ABC = 1/7$, I believe I have to use the cosine theorem.
$BD = \sqrt{a^2 + b^2 + 2ab \cos(B)}$
$AC = \sqrt{a^2 + b^2 – 2ab \cos(B)}$
We don't know $BD$, $AC$ and $a$ and I'm kinda stuck.
By looking at the formulas above, we could probably find the height of the trapezoid using $h = d \sin \alpha = b \sin \beta$. We have $b$ and $\cos \beta$. However, we need $\sin$, not the $\cos$, which stomps me here.
Best Answer
If perp from $C$ to $AB$ meets at $G$, $ \displaystyle BG = b \cos \beta$
Applying Pythagoras, $h = CG = \sqrt{b^2 - BG^2}$
If perp from $D$ to $AB$ meets at $H$ then,
$AH = \sqrt{d^2 - h^2}$
$ \displaystyle AB = AH + GH + BG~$ and $GH = CD = c$
The only problem is that once you find $h$, you realize $h \gt d$ which is not possible. So the given measurements are wrong.