Question

Find the area of a parallelogram ABCD in which AB = 60 cm, BC = 40 cm and AC = 80 cm.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=40\mathrm{cm},b=80\mathrm{cm}\mathrm{and}c=60\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{40+80+60}{2}=90\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{90(90-40)(90-80)(90-60)} \\ =\sqrt{90\times 50\times 10\times 30} \\ =\sqrt{30\times 3\times 10\times 5\times 10\times 30} \\ =30\times 10\sqrt{15} \\ =300\times 3.87 \\ =1161{\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 1161{\mathrm{cm}}^2=2322{\mathrm{cm}}^2$