2020/008) A right angled triangle has perimeter 40 m and area 60 $m^2$ Find the lengths of the sides of the triangle.

Let the 3 sides be a,b & c, where c is the hypotenuse and without loss of generality $a>b$.
We have as it is right angled triangle
$a^2+b^2 = c^2\cdots(1)$
As perimeter is 40 we have a+b+c = 40 and hence
$a+b= 40 -c\cdots(2)$
Area is 60 so we have
$\frac{ab}{2} = 60$ or $ab=120\cdots(3)$
From (2) and (3) $(a+b)^2 - 2ab = (40-c)^2 - 240 $
Or $a^2+b^2 = c^2 - 80c + 1600 - 240 = c^2 - 80c + 1360 $
Or $80c-1360 = c^2-(a^2+b^2) = 0$ sing (2)
Or c = 17
Puttng c =17 in (2) we get
$a+b= 40-17 = 23$
So $(a+b)2^ = 23^2$
Using above and (3) we get $(a-b)^2 = (a+b)^2 - 4ab = 23^2 - 4 * 120 = 49$
Or a-b = 7\cdots(4)
From (2) and (4) a = 15 and b = 8
So sides of triangle are 15m,8m,17m


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