Area Of Quadrilateral Given All Its Sides?

Consider $\triangle CAB$ . We have the lengths of two of its sides, but there are infinity triangles with this 2 sides, so there are infinity possible areas for this triangle (with some restrictions).

Now, consider $\triangle CDB$ . For each of the $\triangle CAB$ possible areas, $\overline { CB }$ has a different length; thus, there exists a $\triangle CDB$ for each $\triangle CAB$ . As $(ABCD)=(\triangle CAB)+(\triangle CDB)$ , there are infinity possible values for the area of the quadrilateral. ( $(\triangle CAB)$ denotes the area of $\triangle CAB$ )

Note that the reasoning is based on the Congruence Theorem SSS, which states that 3 sides determine one single triangle, but 2 sides determine infinity triangles. We need, at least, another segment length or a quadrilateral angle.

For example:

If ( $\triangle CAB)= 704$ , as one of the $(\triangle CAB)$ bases is $\overline { AB }$ , then $704=\frac { AB\cdot { h }_{ AB } }{ 2 }$ .

Thus, ${ h }_{ AB }=32$ . As $\sin { (\measuredangle A) } =\frac { { h }_{ AB } }{ AC }$ , therefore, $\measuredangle A\approx 70.25°$ . Applying Law of Cosines:

${ CB }^{ 2 }={ AB }^{ 2 }+{ AC }^{ 2 }+2\cdot AC\cdot AB\cdot \cos { \measuredangle A }$ . Replacing the known values and solving, $AC\approx 64$ .