Question

The sides of a triangle are 16 cm, 12 cm and 20 cm. Find:

(i) Area of the triangle

(ii) Height of the triangle, corresponding to the largest side

(iii) Height of the triangle, corresponding to the smallest side.

Answer 1

Sides of $\Delta$ are

a =20 cm

b =12 cm

c =16 cm

$S=\frac{a+b+c}{2}$

$=\frac{20+12+16}{2}$

$=\frac{48}{2}=24$

area of $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{24(24-20)(24-12)(24-16)}$

$=\sqrt{24\times 4\times 12\times 8}$

$=\sqrt{12\times 2\times 4\times 12\times 2\times 4}$

$=\sqrt{12\times 12\times 4\times 4\times 2\times 2}$

$=12\times 4\times 2=96c{m}^2$

AD is height of $\Delta$ corresponding to largest side.

$\therefore \frac{1}{2}\times BC\times AD=96$

$\frac{1}{2}\times 20\times AD=96$

$AD=\frac{96\times 2}{20}$

BE is height of $\Delta$ corresponding to smallest side.

$\therefore \frac{1}{2}AC\times BE=96$

$\frac{1}{2}\times 12\times BE=96$

$BE=\frac{96\times 2}{12}$

BE =16 cm

(i)96 $c{m}^2$

(ii) 9.6 cm

(iii)16 cm