Question

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
 $
  {\text{A}}{\text{. 13 and 15}} \\
  {\text{B}}{\text{. 15 and 13}} \\
  {\text{C}}{\text{. 16 and 12}} \\
  {\text{D}}{\text{. None of these}} \\
 $

Answer 1

Hint: In order to find the lengths of sides AB and AC, we construct an appropriate figure in such a way that all the given data is represented and we make use of concepts such as the tangents coming to a circle from the same point are equal, semi-perimeter of a triangle and area of a triangle in terms of its semi-perimeter. Using this we establish a relation between areas and solve for answers.

Complete step-by-step answer:
Given data,
Radius of circle = 4 cm
BD = 8 cm and DC = 6 cm

Let us make a figure which represents the data given in the question,

Let the center of the circle be O touching the sides of the triangle AB abs AC at E and F respectively. And let the length of the side AE be x cm.
Now we know the length of tangents from a single point to the circle are always equal. Therefore from the figure, there are two tangents each coming from point A, B and C respectively. I.e.,
CD = CF = 6 cm
BD = BE = 8 cm
AE = AF = x cm

Now the sides of the triangle,
AB = AE + EB = x + 8 cm
BC = BD + DC = 8 + 6 = 14 cm
CA = CF + FA = 6 + x cm

Now we know the semi perimeter of a triangle is given by the formula,
Semi perimeter, $ {\text{s = }}\dfrac{{{\text{a + b + c}}}}{2} $ , where a, b and c are the sides of the triangle AB, BC and AC respectively.
 $
   \Rightarrow {\text{s = }}\dfrac{{{\text{AB + BC + AC}}}}{2} \\
   \Rightarrow {\text{s = }}\dfrac{{\left( {{\text{x + 8}}} \right){\text{ + 14 + }}\left( {{\text{x}} + 6} \right)}}{2} \\
   \Rightarrow {\text{s = x + 14 cm}} \\
 $

Now the formula of area of the triangle in terms of the semi perimeter is given by
Area of the ∆ ABC = $ \sqrt {{\text{s}}\left( {{\text{s - a}}} \right)\left( {{\text{s - b}}} \right)\left( {{\text{s - c}}} \right)} $
 $
   \Rightarrow \sqrt {\left( {14{\text{ + x}}} \right)\left( {14{\text{ + x - x - 8}}} \right)\left( {14{\text{ + x - 14}}} \right)\left( {14{\text{ + x - x - 6}}} \right)} \\
   \Rightarrow \sqrt {\left( {14{\text{ + x}}} \right)\left( {\text{x}} \right)\left( 8 \right)\left( {\text{6}} \right)} \\
   \Rightarrow 4\sqrt {{\text{3x}}\left( {14{\text{ + x}}} \right)} \\
 $
Area of the ∆ ABC = $ 4\sqrt {{\text{3x}}\left( {14{\text{ + x}}} \right)} $ sq cm
Now from the figure we can say,
Area of the ∆ ABC = Area of the ∆ OAC + Area of the ∆ OBC + Area of the ∆ OAB

We know the area of a triangle is given by, A = $ \dfrac{1}{2} \times {\text{b}} \times {\text{h}} $
From the figure,
Given radius of circle is 4 cm, hence OF = OD = OE = 4 cm
Area of the ∆ OAC,
= $ \dfrac{1}{2} \times {\text{OF}} \times {\text{AC}} $
= $ \dfrac{1}{2} \times 4 \times \left( {6{\text{ + x}}} \right) $
= 12 + 2x sq cm

Area of the ∆ OBC,
= $ \dfrac{1}{2} \times {\text{OD}} \times {\text{BC}} $
= $ \dfrac{1}{2} \times 4 \times 14 $
= 28 sq cm

Area of the ∆ OAB,
= $ \dfrac{1}{2} \times {\text{OE}} \times {\text{AB}} $
= $ \dfrac{1}{2} \times 4 \times \left( {{\text{8 + x}}} \right) $
= 16 + 2x sq cm

Now Area of the ∆ ABC = Area of the ∆ OAC + Area of the ∆ OBC + Area of the ∆ OAB
⟹ $ 4\sqrt {{\text{3x}}\left( {14{\text{ + x}}} \right)} $ = 16 + 2x + 12 + 2x + 28
⟹ $ \sqrt {{\text{3x}}\left( {14{\text{ + x}}} \right)} $ = 14 + x
Squaring on both sides we get,
\[
   \Rightarrow {\text{3x}}\left( {14{\text{ + x}}} \right) = {\left( {14{\text{ + x}}} \right)^2} \\
   \Rightarrow {\text{3x = 14 + x - - - - - }}\left( {14{\text{ + x = 0}} \Rightarrow {\text{x = - 14 is not possble}}} \right) \\
   \Rightarrow 3{\text{x - x = 14}} \\
   \Rightarrow {\text{x = 7 cm}} \\
\]

Therefore the lengths of the sides AB and AC are,
AB = x + 8 = 7 + 8 = 15 cm
AC = 6 + x = 6 + 7 = 13 cm

Option B is the correct answer.

Note: In order to solve this type of questions the key is to express the area of the triangle ABC in terms of smaller triangles formed with respect to the center of the circle O. This helps us in forming a relation between the variable x, which when solved gave us the answer.Having a good knowledge of the formulae of area of a triangle, semi perimeter and area of the triangle in terms of its semi perimeter is essential.Area is always expressed in square units.