SOLUTION: In the diagram given below, three circles having centres at A, B and C touch each other externally. (a) given that the radius of the three circles are 4cm, x cm, and (2x+2) cm res

AB = 4+x
BC = x + (2x+2) = x + 2x+2 = 3x + 2
AC = 4 + (2x+2) = 4 + 2x+2 = 2x + 6

By the Pythagorean theorem:

                 AB² + BC² = AC²

          (4+x)² + (3x+2)² = (2x+6)²

 (4+x)(4+x) + (3x+2)(3x+2) = (2x+6)(2x+6)

  (16+8x+x²) + (9x²+12x+4) = 4x²+24x+36

        16+8x+x²+9x²+12x+4 = 4x²+24x+36

               10x²+20x+20 = 4x²+24x+36

                 6x²-4x-16 = 0

Divide through by 2

                  3x²-2x-8 = 0

That was the first thing you were asked to show.

Factor the left side:

               (3x+4)(x-2) = 0

            3x+4 = 0       x-2 = 0
              3x = -4        x = 2
               x = -4/3

We ignore the negative answer. 
So the value of x is 2.  That was the second 
thing you wanted. 

We calculate the lengths of the sides:
  
AB = 4+x = 4+2 = 6
BC = 3x+2 = 3(2)+2 = 6+2 = 8
AC = 2x+6 = 2(2)+ 6 = 4+6 = 10



Now we want to find the length of the perpendicular from 
B to AC.  We'll draw that perpendicular AD in and label 
it h (in green):



Now the circles just get in the way, so we'll erase them
and just look at the triangle, whose sides are

AB=6, BC=8, AC=10

So we can dispense with the circles and just keep the triangle and
the green altitude AD = h:



A perpendicular from the right angle to the hypotenuse of a right 
triangle divides the right triangle into two smaller similar right
triangles, both of which are similar to the whole right triangle.
So

ᅀADB∼ᅀBDC∼ᅀABC

Using the fact that ᅀADB∼ᅀABC and the fact that AC=10





Cross-multiply:





So the length of the perpendicular AD is 4.8

Edwin