Question

The radius of the in circle of a triangle is $ 4 $ cm and the segment into which one side divided by the point of contact are $ 6\,cm $ and $ 8\,cm $ . Determine the value of x.

Answer 1

Hint: In this problem we first find the area of triangle ABC by using Heron’s formulas as three sides of the triangle are known and also using the basic area formula to find the area of three sub triangles which formed inside the given triangle ABC. Then equate the area of triangle ABC with the sum of three areas of triangles to find the required value of ‘x’.

Complete step-by-step answer:
Area of triangle = $ \dfrac{1}{2} \times base \times altitude $ , Heron's formula: $ S(semi\,perimeter) = \dfrac{{a + b + c}}{2},\,\, $ and $ Area = \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $ where a, b, and c are three sides of triangle.
From fig we clearly see that BL and BN are tangents from external point to circle.
 $ BL = BN $ = (x) (tangents from external point are equal in length)

Also, CL and CM are tangents from external point to circle
 $ \therefore CL = CM $ = (8cm) (tangents from external point are equal in length)
Also, AM and AN are tangents from the external point to circle.
 $ \therefore AM = AN $ = (6cm) (tangents from external point are equal in length)
Therefore, from above we see that length of three sides of triangle ABC will be given as:
AB = AN + BN = $ \left( {6 + x} \right)\,cm $
BC = BL + LC = $ \left( {x + \,8} \right)\,cm $
AC = AM + MC = $ 6 + 8 = 14cm $
As three sides of the triangle are known to us therefore we use Heron’s formula to find its area. For this we first calculate semi perimeter.
 $ S(semi\,perimeter) = \dfrac{{a + b + c}}{2},\,where\,a,\,b,\,c\,\,are\,\,three\,\,sides\,\,of\,\,triangle\, $
 $
   \Rightarrow S = \dfrac{{\left( {x + 8} \right) + \left( {x + 6} \right) + 14}}{2} \\
   \Rightarrow S = \dfrac{{2x + 28}}{2} \\
   \Rightarrow S = x + 14 \\
  $
Therefore, area of triangle is given as
Area = $ \sqrt {S\left( {S - A} \right)\left( {S - B} \right)\left( {S - C} \right)} $
\[
  Area = \sqrt {\left( {x + 14} \right)\left( {x + 14 - x - 8} \right)\left( {x + 14 - x - 6} \right)\left( {x + 14 - 14} \right)} \\
  Area = \sqrt {\left( {x + 14} \right)\left( 6 \right)\left( 8 \right)\left( x \right)} \\
  Area = \sqrt {48x\left( {x + 14} \right)} ....................(i) \\
 \]
Also, in triangle ABC we see that the area of triangle ABC is equal to the sum of the area of three triangles.
 $ ar\left( {\Delta ABC} \right) = ar\left( {\Delta AIC} \right) + ar\left( {\Delta BIC} \right) + ar\left( {\Delta AIB} \right) $
We also know that area of triangle is also given as: $ \dfrac{1}{2} \times base \times altitude $
 $ \therefore ar(\Delta ABC) = \dfrac{1}{2} \times AC \times IM + \dfrac{1}{2} \times BC \times IL + \dfrac{1}{2} \times AB \times IN $
Substituting values from triangle in above formula we have
 $ ar(\Delta ABC) = \dfrac{1}{2} \times \left( {14} \right) \times 4 + \dfrac{1}{2} \times \left( {x + 8} \right) \times 4 + \dfrac{1}{2} \times \left( {x + 6} \right) \times 4 $
On simplifying right hand side of above equation
 $ ar(\Delta ABC) = 28 + 2x + 16 + 2x + 12 $
 $
   \Rightarrow ar(\Delta ABC) = 4x + 56 \\
   \Rightarrow ar(\Delta \left( {ABC} \right) = 4\left( {x + 14} \right)..................(ii) \\
  $
From (i) and (ii) we have
 $ 4\left( {x + 14} \right) = \sqrt {48x\left( {x + 14} \right)} $
To solve under root squaring both side
 $
  16{\left( {x + 14} \right)^2} = 48x\left( {x + 14} \right) \\
   \Rightarrow \left( {x + 14} \right) = 3x \\
   \Rightarrow 2x = 14 \\
   \Rightarrow x = 7 \\
  $
Hence, from above we see the required value of x is $ 7 $ .

Note: We know that if a given figure is divided into different parts. Then the sum of the area of smaller parts will be equal to the area of the figure whose parts have been made.