Question

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: $\sqrt{3}=1.73$]

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=10\mathrm{cm},b=16\mathrm{cm}\mathrm{and}c=14\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{10+16+14}{2}=20\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{20(20-10)(20-16)(20-14)} \\ =\sqrt{20\times 10\times 4\times 6} \\ =\sqrt{10\times 2\times 10\times 2\times 2\times 3\times 2} \\ =10\times 2\times 2\sqrt{3} \\ =69.2{\mathrm{cm}}^2 \end{gathered}$$

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = $2\times 69.2{\mathrm{cm}}^2=138.4{\mathrm{cm}}^2$