The difference between the semiperimeter and the sides of a $$\Delta$$ABC are 8cm, 7cm and 5cm respectively.Find the area of the triangle.
Consider a, b and c as the sides of a triangle and s as the semi perimeter
So we get
s - a = 8cm
s - b = 7cm
s - c = 5cm
We know that(s - a) + (s - b) + (s - c) = 8 + 7 + 5
On further calculation
3s - (a + b + c) = 20
We know that a + b + c = 2s
3s - 2s = 20
So we gets = 20
By substituting the value of s
a = s - 8 = 20 - 8 = 12cm
b = s - 7 = 20 - 7 = 13cm
c = s - 5 = 20 - 5 = 15cm
We know that
Area=$$\sqrt{s(s-a)(s-b)(s-c)}$$
By substituting the values
Area=$$ \sqrt{20(20-12(20-13)(20-15)}$$
So we get
Area=$$ \sqrt{20\times8\times7\times5}$$
It can be written as
Area=$$\sqrt{4\times5\times4\times2\times7\times5}$$
By multiplication
Area=$$ 20\sqrt{14}$$
Therefore, the area of the triangle is $$20\sqrt{14}cm^2$$