Triangles - Area through inradius and circumradius

Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.

The three side lengths are 6, 8 and 10 units

The inradius of a circle can be calculated using the formula: $$rs=Area$$, where s is the semi-perimeter of the triangle 

The Area would be $$\frac{1}{2}\times\ 6\times\ 8=24$$
and the semi-perimeter would be $$\frac{10+6+8}{2}=12$$

Giving the inradius to be $$\frac{24}{12}=2$$ units

Therefore, 2 is the correct answer. 

Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x/(4*R), where R is circumradius.

Also Area = 0.5*x*3 .

Equating both, we have 3 = 17.5*9 / (2*R)

=> R = 26.25.

Consider side of the cube as x.

So diagonal will be of length $$\sqrt{3}$$ * x.

Now if diagonals are side of equilateral triangle we get area = 3*$$\sqrt{3}*x^2$$ /4 .

Also in a triangle

4 * Area * R = Product of sides

4* 3*$$\sqrt{3}*x^2$$ /4 * R = .3*$$\sqrt{3}*x^3$$

R = x