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. 

By Pythagoras theorem we get BC = 25 . Let BD = x

Triangle ABD is similar to triangle CBA => AD/x = 20/15 

Also triangle ADC is similar to triangle ABC => AD/(25-x) = 15/20

From the 2 equations, we get x = 9, DC = 16 and AD = 12

We know that AREA = (semi perimeter ) * inradius

For triangle ABD, Area = 1/2 x BD X AD = 1/2 x 9 x 12 = 54 and semi perimeter = (15 + 9 + 12)/2 = 18. On using the above equation we get, inradius, r = 3.

Similarly for triangle ADC we get inradius R = 4 .

PQ = R + r = 7 cm

Considering the information provided it can be represented as :

Since BP and BQ are tangents drawn to a circle from the same point. The length of the tangents is equal.

Similarly for CQ and CR.

BP = BQ = 5
CR = CQ = 6
AP = AR = x
Area of triangle = rs
s = 11+x
Area = $$\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+11)(x)(6)(5)} = \sqrt{\frac{35}{3}} (x+11)$$
Solving the equation,
x = 7
Hence AB = 12 cm.