20. A cyclic quadrilateral has all four vertices on the circumference of a circle. Brahmagupta (598-670AD) gave the following formula for the area, $A$, of a cyclic quadrilateral whose edges have lengths $a, b, c, d$ : $A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is half of the perimeter of the quadrilateral.
What is the area of the cyclic quadrilateral with sides of length $4 \mathrm{~cm}$, $5 \mathrm{~cm}, 7 \mathrm{~cm}$ and $10 \mathrm{~cm}$ ?
A $6 \mathrm{~cm}^{2}$
B $13 \mathrm{~cm}^{2}$
C $26 \mathrm{~cm}^{2}$
D $30 \mathrm{~cm}^{2}$
E $36 \mathrm{~cm}^{2}$
20. E The length of $s$, half the perimeter of the cyclic quadrilateral, is $\frac{1}{2}(4+5+7+10) \mathrm{cm}=13 \mathrm{~cm}$. So the required area, in $\mathrm{cm}^{2}$, is $\sqrt{(13-4)(13-5)(13-7)(13-10)}=\sqrt{9 \times 8 \times 6 \times 3}=\sqrt{9 \times 144}=3 \times 12=36$.