Question

In parallelogram ABCD, AB = $10$ cm, AD = $24$ cm and BD = $26$ cm, find the area of parallelogram(in cm${}^2$)

Answer 1

Area of the parallelogram ABCD $=$ Area of triangle ABD $+$ Area of triangle BCD
Since both the triangles are same, Area of the parallelogram ABCD $=2\times$( Area of triangle ABD)

Area of triangle with sides a, b, and c and s
$=\frac{a+b+c}{2}$ is $\sqrt{s(s-a)(s-b)(s-c)}$.
For triangle with sides 10 cm, 24 cm and 26 cm, $s=\frac{10+24+26}{2}=30cm$
Area of the triangle with sides 10 cm, 24 cm and 26 cm $=\sqrt{30(30-10)(30-24)(30-26)}=\sqrt{30\times 20\times 6\times 4}=\sqrt{6\times 5\times 4\times 5\times 6\times 4}=6\times 5\times 4=120{cm}^2$

Hence, Area of the parallelogram ABCD $=2\times$( Area of triangle ABD) $=2\times 120=240{cm}^2$