What we know about the trapezoid ABCD (AB and CD are respectively its large and small bases) is that AB=9 cm,AC=6 cm,AD=8 cm,CD=4 cm. Determine the perimeter of the trapezoid.
It is quite tricky.
I tried using the Pythagorean theorem, but ended with:
$(9 -x)^2 + y^2 = 36$, which led nowhere.
I am confused as to what I can do now.
$\endgroup$ 71 Answer
$\begingroup$Let $x = BC$ and let $\theta = \angle BAC = \angle ACD$. Then by applying Cosine Law to each triangle, we have: \begin{align*} x^2 &= 6^2 + 9^2 - 2 \cdot 6 \cdot 9 \cdot \cos \theta \\ 8^2 &= 6^2 + 4^2 - 2 \cdot 6 \cdot 4 \cdot \cos \theta \end{align*} Divide the first equation by $9$, divide the second equation by $4$: \begin{align*} \frac{x^2}{9} &= 4 + 9 - 2 \cdot 6 \cdot \cos \theta \\ 16 &= 9 + 4 - 2 \cdot 6 \cdot \cos \theta \end{align*} Subtract the two equations: $$ \frac{x^2}{9} - 16 = 0 \iff x^2 = 9 \cdot 16 \iff x = 3 \cdot 4 $$
$\endgroup$
