# The lengths of the sides of a triangle are in a ratio of 3: 4: 5 and its perimeter is 144 cm.

Question:

The lengths of the sides of a triangle are in a ratio of 3: 4: 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side?

Solution:

Given the perimeter of a triangle is 160m and the sides are in a ratio of 3: 4: 5

Let the sides a, b, c of a triangle be 3x, 4x, 5x respectively

So, the perimeter = 2s = a + b + c

144 = a + b + c

144 = 3x + 4x + 5x

Therefore, x = 12 cm

So, the respective sides are

a = 36 cm

b = 48 cm

c = 60 cm

Now, semi perimeter

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

$=\frac{36+48+60}{2}$

= 72 cm

By using Heron's Formula

The area of a triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

$=\sqrt{72 \times(72-36) \times(72-48) \times(72-60)}$

$=864 \mathrm{~cm}^{2}$

Thus, the area of a triangle is $864 \mathrm{~cm}^{2}$

The altitude will be smallest provided the side corresponding to this altitude is longest.

The longest side = 60 cm

Area of the triangle = 12 × h × 60

12 × h × 60 = 864 cm2

h = 28.8 cm

Hence the length of the smallest altitude is 28.8 cm