Question

In a $\Delta$ ABC, AB= 15 cm, BC = 13 cm and AC= 14 cm. Find the area of $\Delta$ ABC and hence its altitude on AC.

Answer 1

Sides of triangle ABC are AB = 15 cm, BC = 13 cm, AC =14 cm

$\therefore s=\frac{a+b+c}{2}=\frac{15+13+14}{2}=\frac{42}{2}=21\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21(21-15)(21-13)(21-14)}=\sqrt{21\times 6\times 8\times 7}=\sqrt{7\times 3\times 3\times 2\times 2\times 2\times 2\times 7}=7\times 3\times 2\times 2=84c{m}^2$
Now $BL\perp AC$
$\therefore$ Length of BL = $\frac{Area\times 2}{Base}=\frac{84\times 2}{14}c{m}^2=12cm$