a, b and c are the integral sides of a right angled triangle in which c is the hypotenuse measuring `3sqrt5` metres. If side 'a' is increased by `(400)/(3)%` and side 'b' is increased by
`(50)/(3)%`, then their total becomes 14 metres. Find the area of triangle.

Explanation

By Pythagoras theorem,
`a^(2)+b^(2)=(3sqrt5)^2`
`impliesa^(2)+b^(2)=45" "......(1)`
Now length of base `=a+axx(400)/(3)xx(1)/(100)=(7a)/(3)`
and perpendicular `=b+bxx(50)/(3xx100)=(7a)/(6)`
According to the problem, `(7a)/(3)+(7a)/(6)=14`
`implies14a+7b=84impliesb=12-2a`
Substituting this value in equation (1), we get
`a^(2)+(12-2a)^(2)=45`
`impliesa^(2)+144+4a^(2)-48a=45`
`implies5a^(2)-48a+99=0`
`implies5a^(2)-15a-33a+99=0`
`implies5a(a-3)-33(a-3)=0`
`implies(a-3)(5a-33)=0`
`becausea=3ora=(33)/(5)`
But `a=(33)/(5)` is not an integer.
So, we reject `a=(33)/(5)`
therefore a=3
Hence, `a=3m,b=12-2xx3=6`
`because"Area of Delta=(1)/(2)ab=(1)/(2)xx3xx6=9m^(2)`