SSC CGL 7 Sep 2016 Paper

Area of parallelogram ABCD
= 2 × Area of Δ ABC
AB = 21 cm. = c
BC = 20 cm. = a
AC = 29 cm. = b
∴ Semi-perimeter of Δ ABC
= s = $\frac{\mathit{a}+\mathit{b}+\mathit{c}}{2}$
= $\left(\frac{20+29+21}{2}\right)$ cm
= $\frac{70}{2}$
= 35 cm
∴ Area of Δ ABC
= $\sqrt{8\left(\mathit{s}-\mathit{a}\right)\left(\mathit{s}-\mathit{b}\right)\left(\mathit{s}-\mathit{c}\right)}$
=
= $\sqrt{35\times 15\times 6\times 14}$
=

= 7 × 2 × 3 × 5 = 210 sq.cm.
∴ Area of ABCD
= (2 × 210) sq. cm.
= 420 sq. cm.