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?
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