Ex.10.2 Q12 Circles Solution - NCERT Maths Class 10

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Question

A triangle \( ABC\) is drawn to circumscribe a circle of radius \(4\, \rm{cm}\) such that the segments \( BD\) and \( DC\) into which \( BC\) is divided by the point of contact \( D\) are of lengths \(8 \,\rm{cm}\) and \(6\, \rm{cm}\) respectively (see Figure). Find the sides \( AB\) and \( AC\).

Text Solution

 

What is the known?

(i) \(\Delta {ABC}\) is drawn to circumscribe a circle of radius \(4\, \rm{cm.}\)

\[\begin{align}{BD}&=8\;\rm{cm}\\{CD}&=6\;\rm{cm}\end{align}\]

What is Unknown?

Sides \({AB}\) and \({AC}\)

Reasoning:

Finding the area of \(\Delta {ABC}\) in two ways and equating them will result in unknown length. Hence other sides can be calculated.

Let,

\({AE = AF = x}\) (The lengths of tangents drawn from an external point to a circle are equal.)

\({CD = CE = 6 \rm cm}\) (Tangents from point \( C\))

\({BD = BF = 8 \rm{cm}}\) (Tangents from point \( B\))

Area of the triangle \(\begin{align}= \sqrt {{ s ( s - a ) ( s - b ) ( s - c )} }  \end{align}\)

where \(\begin{align} s = \frac { 1 } { 2 } ( a + b + c )\end{align}\)

\(a,\; b\) and \(c\) are sides of a triangle

\[\begin{align} a &= AB= x + 8 \\ b &= BC = 8 + 6 \\ & = 14 \\c& = CA = 6 + x \end{align}\]

\[\begin{align} s & = \frac { 1 } { 2 } ( x + 8 + 14 + 6 + x ) \\ s & = \frac { 1 } { 2 } ( 2 x + 28 ) \\ s & = x + 14 \end{align}\]

Area of \(\Delta {ABC}\)

\[\begin{align} & = \sqrt { (x + 14) ( x + 14 - x - 8 ) ( x + 14 - 14 ) ( x + 14 - x + 6 ) } \\ & = \sqrt { ( x + 14 ) ( 6 ) ( x ) ( 8 ) } \\ & = \sqrt { 48 x ( x + 14 ) } \\ & = \sqrt { 48 \left( x ^ { 2 } + 14 x \right) }\;\text {sq units .................... (Equation 1) } \end{align}\]

Area of \(\Delta ABC=\) Area of \(\Delta {AOC}\;+\;\)Area of \(\Delta {AOB}\;+\;\) Area of \(\Delta {BOC}\)

\[\begin{align} & = \frac { 1 } { 2 } ( x + 6 ) 4 + \frac { 1 } { 2 } ( x + 8 ) 4 + \frac { 1 } { 2 } ( 14 ) 4 \\&\qquad \qquad \left(\therefore \operatorname{area~of} \Delta=\frac{1}{2} \times b \times h\right)\\ & = 2 x + 12 + 2 x + 16 + 28 \\ & = 4 x + 56 \\ & = 4 ( x + 14 ) \text{ sq units.................... (Equation 2) } \end{align}\]

Equating (1) and (2) \[\sqrt { 48 \left( x ^ { 2 } + 14 x \right) } = 4 ( x + 14 )\]

Squaring both sides

\[\begin{align} 48 \left( \mathrm { x } ^ { 2 } + 14 x \right) & = 4 ^ { 2 } ( x + 14 ) ^ { 2 } \\ 48 \left( \mathrm { x } ^ { 2 } + 14 x \right) & = 16 \left( x ^ { 2 } + 28 x + 196 \right)\\&\left[ \text{ Using }(a+b)^{2}=a^{2}+2 a b+b^{2}\right] \\ \frac { 48 } { 16 } \left( \mathrm { x } ^ { 2 } + 14 x \right) & = x ^ { 2 } + 28 x + 196\\ 3 x^{2}+42 x &=x^{2}+28 x+196 \\ 3 x ^ { 2 } - x ^ { 2 } + 42 x - 28 x - 196 & = 0 \\ 2 x ^ { 2 } + 14 x - 196 & = 0\\ &\text{(divide this equation by  2 )} \\ x ^ { 2 } + 7 x - 98 & = 0 \end{align}\]

Solving by factorization method

\[\begin{align} x ^ { 2 } + 14 x - 7 x - 98 & = 0 \\ x ( x + 14 ) - 7 ( x + 14 ) & = 0 \\ ( x + 14 ) ( x - 7 ) & = 0 \end{align}\]

\[\begin{align} &{ x + 14 = 0 } \qquad\text{and} \qquad x - 7 = 0 \\ &{ x = - 14 } \qquad \qquad \qquad \;\; { x = 7 } \end{align}\]

Since \(x\) represents length, it cannot be negative.

\(\therefore x = 7\)

\[\begin{align}AB = a = x + 8 = 7 + 8 = 15 \,\rm{cm} \\ AC = c = 6 + x = 6 + 7 = 13\,\rm{cm} \end{align}\]

Hence the sides

\(AB = 15\, \rm{cm}\\AC = 13 \,\rm{cm}\)