Calculate the area of a triangle whose sides are 13 cm, 5 cm and 12 cm. Hence, calculate the altitude corresponding to the longest side of this triangle. Leave your answer as a fraction.

The sides of the triangle are 13 cm, 5 cm and 12 cm.

Area of triangle = $\sqrt{s(s - a)(s - b)(s- c)}$

where, s = $\dfrac{a + b + c}{2}$

Substituting a = 13 cm, b = 5 cm, c = 12 cm:

s = $\dfrac{13 + 5 + 12}{2}$

= $\dfrac{30}{2}$

= 15

Area of the triangle:

$= \sqrt{15(15 - 13)(15 - 5)(15 - 12)}\\[1em] = \sqrt{15 \times 2 \times 10 \times 3}\\[1em] = \sqrt{900}\\[1em] = 30$

Let the height be h and base = 13 cm.

Area of triangle = $\dfrac{1}{2}$ x base x height

⇒ 30 = $\dfrac{1}{2}$ x 13 x h

⇒ h = $\dfrac{30 \times 2}{13}$

⇒ h = $\dfrac{60}{13}$

⇒ h = $\dfrac{60}{13}$

⇒ h = $4\dfrac{8}{13}$

Hence, the altitude corresponding to the longest side of the triangle is $4\dfrac{8}{13}$ cm.