Area of a quadrilateral on cartesian plane A(0,0), B(4,0), C(3,${\pi \over 8 }$), D(1, ${3\pi\over 8}$)

Question

I'm having trouble on this question.

Could anyone find a solution and answer for this?

What is the area of quadrilateral ABCD whose vertices have polar coordinates

A(0,0), B(4,0), C(3,${\pi \over 8 }$), D(1, ${3\pi\over 8}$)

Thanks!

Answer 1

Hint

Area of a triangle with side lengths $a$ and $b$ and the angle in between $\theta$ is given by $\frac{1}{2}ab \sin \theta$.

Consider the triangle $\Delta ABC$,the side $AB=4$, $AC=3$ and the angle in between ($\angle BAC$) is $\theta=\frac{\pi}{8}$. Thus the area of $\Delta ABC$ is $\frac{1}{2}(4)(3) \sin \frac{\pi}{8}$. Now do the same for the other triangle $\Delta ACD$.

Answer 2

You can develop a formula for polar coordinates like

$$ \frac12 ( r_ 1 r_ 2\sin \theta_1 + r_ 2 r_ 3\sin \theta_2 + r_ 3 r_ 4\sin \theta_3 + r_ 4 r_ 1\sin \theta_4) $$

$ \theta $ is angle between radius vectors, used with sign convention for angles $\theta $ ( CCW >0, CW <0).