Exercise 128

Convex quadrilateral $ABCD$ has $AB=7$, $BC=24$, $CD=15$, $AD=20$, and $\angle ABC=90^\circ$. What is the area of the quadrilateral?

Solution

Connect $AC$: since $\angle ABC=90^\circ$, $AC=\sqrt{7^2+24^2}=25$ (a $7$-$24$-$25$ right triangle).
Since $CD=15,AD=20,AC=25$ and $15^2+20^2=25^2$, $\triangle ACD$ is also a right triangle (theorem 19).
Total area $=\frac12(7)(24)+\frac12(15)(20)=84+150=\boxed{234}$.