A Trig Question !!

The leg of a right triangle is equal to 1/5 the sum of other sides. The triangle has a perimeter of 1. What is the triangle's area? Thank you.

Let the sides be  a, b and c      where a and b are the legs and c is the hypotenuse

Let  a  =  ( b + c) / 5     so    5a   = b + c

And since the perimeter   =  1, we have that

a +  b + c  =  1                

a +  5a   = 1

6a  =  1    →   a  =  1/6 units

And we have that  5(1/6)=  b + c   →  5/6  = b + c  →  c = (5/6) - b

So.....using the Pythagorean Theorem, we have that

a^2 + b^2  =  c^2

(1/6)^2  + b^2   = [ (5/6)  - b ] ^2

(1/36) + b^2   =  b^2  - (10/6)b + 25/36    simplify

(1/36)    = 25/36 - (5/3)b      rearrange as

(5/3)b   =   25/36  -  1/36

(5/3)b  =  24/36

(5/3)b  =  2/3

5b =  2

b = 2/5 

So.......the area  = (1/2)(product of the legs) =  (1/2)(a)(b)   =  (1/2)(1/6)(2/5)   =  1/30 units^2

let the base = b
let the hypotenuse = c
a=(b+c) / 5, or 
5a =(b+c), but we have:.................(1)
a+b+c=1, we sub as
a+5a = 1, and we have:
a = 1/6, now will sub this into (1) above:
5/6 - b = c, Now to solve for b, will use Pytho. theorem:
a^2 + b^2 =c^2...........................(2)
(1/6)^2 + b^2 =(5/6 - b)^2 simplifying this we get:
(10/6)b = 24/36
b = 2/5, we now have:
Area = ab/2
(1/6)(2/5)(1/2) =1/30 sq.units.

Another, albeit, intuitive way!

In a 5-12-13 triangle, 5=(12+13)/5. Its perimeter is 30, area =(5 x 12)/2 =30 sq.units.
The right triangle with a perimeter of 1 is similar to the 5-12-13 triangle. Its sides are then
1/30 of the 5-12-13, so its area is proportional by (1/30)^2. Its area is therefore =area(5-12-13) / 30^2
=30 /30^2 =1/30 sq.units.