Length of three altitudes of a triangle are 3 cm, 4 cm and 5 cm. Find

SOLUTION

Let BC =a, AB = c, AC =b

Also AD=  =3cm , CE= h₃ = 5 cm , BF = h₂ = 4 cm

⇒

So sides of triangle 20x, 15x, 12x

Now by Heron’s formula

Area of triangle =  , where s is the semi- perimeter

Here s = ( 20x + 15x + 12x )/ 2 = 47x/2

Area of triangle is also given by  = (1/2)(20x)(3) = 30x

So perimeter of triangle = 20x + 15x + 12x = 47x