Question

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Answer 1

Now in $\Delta ABC$,
$s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21$
$\therefore$ Area = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\times 8\times 7\times 6}=\sqrt{3\times 7\times 7\times 3\times 2\times 2\times 2\times 2}=3\times 7\times 2\times 2=84c{m}^2$
Now area of parallelogram = 84 $c{m}^2$
Base BC =14 cm
$\therefore$ Height = $\frac{Area}{base}=\frac{84}{14}=6cm$