#### Answer

6,8

#### Work Step by Step

Let us assume the legs of a right triangle is 2n, 2+2
The given condition is the numerical value of the area of triangle is three times of that the longer leg.
Area of triangle = 3(2n+2)
The area of the right triangle with legs of the length a and b given by
A = $\frac{1}{2}$ ab
3(2n+2)=$\frac{1}{2}$ 2n (2n+2)
6n+6 = n(2n+2)
6n+6 = 2$n^{2}$ + 2n
2$n^{2}$ - 4n - 6 =0
$n^{2}$ - 2n - 3 =0
$n^{2}$ - 3n + n - 3 =0
n(n-3)+1(n-3) = 0
(n+1)(n-3) = 0
n=-1,3
Therefore the n value cannot be negative,
the value of n = 3
Therefore the legs of triangle 2n , 2n+2
= 2*3, 2*3+2
=6,8
6 and 8 are the length of the required triangle.