Question

solve this..........

Answer 1

In triangle ABC
AB = 20 cm (a)
BC = 34 cm (b)
& AC = 42 cm (c)

semi perimeter, s
$\frac{20 + 34 + 42}{2 } = \frac{96}{2} = 48cm$

Using heron's formula
Area of triangle ABC =
$\sqrt{s(s - a)(s - b)(s - c) } \\ = > \sqrt{48(48 - 20)(48 - 34)(48 - 42)} \\ = > \sqrt{48 \times 28 \times 14 \times 6} \\ = > \sqrt{4 \times 4 \times 3 \times 2 \times 14 \times 14 \times 2 \times 3} \\ = > 4 \times 3 \times 2 \times 14 \\ = > 336 {cm}^{2}$

Now, area of parallelogram ABCD =
$336 {cm}^{2} \times 2 = 672 {cm}^{2}$

Thus, the area of parallelogram ABCD is 672 cm sq.

Mark me as brainliest if it's correct ;-)