1
$\begingroup$

Let $\square ABCD$ be an isosceles trapezoid where $AD$ is a base. Let point $P$ be a point inside the trapezoid such that $PA$, $PB$, $PC$, $PD$ are angle bisectors of $A$, $B$, $C$, $D$. If $|PA| = 3$ and $\angle APD = 120^\circ$, then what is the area of the trapezoid?

My try: I got $|AD| = {3}\sqrt3$ using the cosine law. I then connected the midpoint of $BC$ to $A$ and $D$, respectively, to get a equilateral triangle. I got the length of $AD$ and the height of the trapezoid, but can't seem to get the length of $BC$.

enter image description here

$\endgroup$
  • $\begingroup$ Where can we find an image of the problem? $\endgroup$ – Dr. Sonnhard Graubner Apr 22 '18 at 7:54
  • $\begingroup$ There was no image given, but I think te picture i inserted is correct $\endgroup$ – SuperMage1 Apr 22 '18 at 8:04
2
$\begingroup$

A reasonably-accurate diagram really helps.

enter image description here

$\endgroup$
0
$\begingroup$

Hint: $$A=\frac{1}{2}\cdot 9\sin(120^{\circ})+2\cdot 3 \cdot \frac{\tan(30^{\circ})}{2}+\frac{1}{2}(3\tan(30^{ \circ}))^2\sin(60^{\circ})$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.