Which is the type of the triangle where: (3^1/2)*a-(3^1/2)*b=c a+b=(3^1/2)*c a,b,c=the length of each side of the triangle

We'll try to determine the length of b and the length of c, considering the length of a side.

We'll multiply the second relation with the value (3^1/2) and after that we'll add the equivalent obtained relation to the first one.

(3^1/2)*a + (3^1/2)*b + (3^1/2)*a - (3^1/2)*b = 3*c + c

We'll group the same terms:

2*(3^1/2)*a = 4*c

(3^1/2)*a = 2 *c

c= [(3^1/2)*a]/2

With the c value written in function of "a" value, we'll go in the second relation and substitute it:

a + b = [(3^1/2)*(3^1/2)*a]/2

a + b = 3*a/2

We'll have the same denominator on the left side of the equality:

2*a + 2*b = 3*a

2*b = 3*a - 2*a

2*b = a

b = a/2

If the triangle is a right one, then, using the Pythagorean theorem, we'll have the following relation between the sides of triangle:

a^2 = b^2 + c^2

Now, we have to plug in the values of "b" and "c", in the relation above:

a^2 = a^2/4 + 3*a^2/4

a^2 = 4*a^2/4

a^2 = a^2

We've shown that the equality is true, so the triangle is right, where "a" is hypotenuse and "b","c" are cathetus.