Question:
In a triangle ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of triangle ABC and hence its altitude on AC.
Solution:
Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.
So given,
a = 15 cm
b = 13 cm
c = 14 cm
By using Heron's Formula
The Area of the triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
Semi perimeter of a triangle = 2s
2s = a + b + c
s = (a + b + c)/2
s = (15 + 13 + 14)/2
s = 21 cm
Therefore,
Area of the triangle $=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$
$=\sqrt{21 \times(21-15) \times(21-13) \times(21-14)}$
$=84 \mathrm{~cm}^{2}$
BE is a perpendicular on AC
Now, area of triangle = 84 cm2
1/2 × BE × AC = 84 cm2
BE = 12 cm
The length of BE is 12 cm