Find the area of the triangle whose sides are 13 cm, 14 cm and 15 cm. Also, find the height of the triangle, corresponding to the longest side.

Let a = 13 cm, b = 14 cm and c = 15 cm.

Then,

s = $\dfrac{1}{2}(a + b + c) = \dfrac{1}{2}(13 + 14 + 15) = \dfrac{42}{2}$ = 21 cm.

⇒ (s - a) = (21 - 13) cm = 8 cm.

⇒ (s - b) = (21 - 14) cm = 7 cm.

⇒ (s - c) = (21 - 15) cm = 6 cm.

We know that,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] \Rightarrow \sqrt{21 × 8 × 7 × 6} \\[1em] \Rightarrow \sqrt{7056} \\[1em] \Rightarrow 84 \text{ cm}^2.$

Length of the longest side = 15 cm.

Let the corresponding height be x cm. Then,

$\Rightarrow Area = \dfrac{1}{2} \times \text{ Base } \times \text{ Corresponding height } \\[1em] \Rightarrow 84 = \dfrac{1}{2} \times 15 \times x \\[1em] \Rightarrow x = \dfrac{168}{15} \\[1em] \Rightarrow 11.2 \text{ cm}. \\[1em]$

Hence, area of triangle = 84 cm² and the height of the triangle = 11.2 cm.