Question

Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=26\mathrm{cm},b=30\mathrm{cm}\mathrm{and}c=28\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{26+30+28}{2}=42\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{42(42-26)(42-30)(42-28)} \\ =\sqrt{42\times 16\times 12\times 14} \\ =\sqrt{14\times 3\times 4\times 4\times 2\times 2\times 3\times 14} \\ =14\times 4\times 2\times 3 \\ =336{\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 336=672{\mathrm{cm}}^2$