A triangle is drawn to circumscribe a circle of radius such that the segments and into which is divided by the point of contact are of lengths and respectively (see Fig. 10.14). Find the sides and .

Understand the Question Tangents drawn from an external point to a circle are equal in length. This allows us to express all three side lengths of in terms of an unknown tangent segment . We can calculate the area of in two independent ways: 1. Using Heron's formula: 2. Splitting into three smaller triangles (, , ) connected to the incentre , each having height equal to the inradius . Equating both area expressions gives an equation in , which we solve to find and . Step 1 · Express Side Lengths and Semi-perimeter in Terms of Let the incircle touch sides and at points and respectively. Since tangents drawn from an external point to a circle are equal: From point : From point : From point : let The side lengths of are: Semi-perimeter : Step 2 · Find Area of Using Heron's Formula Using Heron's formula with , , : Step 3 · Find Area of as Sum of Three Triangles Join , , and . The perpendicular heights from incenter to each side are equal to the radius : Summing the three areas: Step 4 · Equate Areas and Solve for Equating the two area expressions: Squaring both sides: This gives: Since side length cannot be negative, . Step 5 · Calculate Sides and Substitute into the side expressions: Answer > and Common Mistakes Tangent Identification Error: Forgetting that and , which leads to incorrect side length representations. Algebraic Cancellation: Dividing both sides by without justifying that , or missing the negative root (which must be formally rejected since side length ). Radius as Height: Forgetting that the inradius is perpendicular to all three sides, so it acts as the exact altitude for , , and .

Circles | Exercise 10.2

Question 12

A triangle $ABC$ is drawn to circumscribe a circle of radius $4 \text{ cm}$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8 \text{ cm}$ and $6 \text{ cm}$ respectively (see Fig. 10.14). Find the sides $AB$ and $AC$ .

Solution

Understand the Question

Tangents drawn from an external point to a circle are equal in length. This allows us to express all three side lengths of $\triangle ABC$ in terms of an unknown tangent segment $x = AE = AF$ .
We can calculate the area of $\triangle ABC$ $△ A B C$ in two independent ways:
1. Using Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
2. Splitting $\triangle ABC$ into three smaller triangles ( $\triangle OBC$ , $\triangle OCA$ , $\triangle OAB$ ) connected to the incentre $O$ , each having height equal to the inradius $r = 4 \text{ cm}$ .
Equating both area expressions gives an equation in $x$ , which we solve to find $AB$ and $AC$ .

Step 1 · Express Side Lengths and Semi-perimeter in Terms of $x$

Let the incircle touch sides $AB$ and $AC$ at points $E$ and $F$ respectively. Diagram 1

Since tangents drawn from an external point to a circle are equal:

From point $C$ : $CF = CD = 6 \text{ cm}$
From point $B$ : $BE = BD = 8 \text{ cm}$
From point $A$ : let $AF = AE = x$

The side lengths of $\triangle ABC$ are:

\begin{aligned} BC &= BD + DC = 8 + 6 = 14 \text{ cm} \\[0.6em] AB &= AE + EB = x + 8 \\[0.6em] AC &= AF + FC = x + 6 \end{aligned}

Semi-perimeter $s$ :

\begin{aligned} 2s &= AB + BC + AC \\[0.6em] 2s &= (x + 8) + 14 + (x + 6) \\[0.6em] 2s &= 2x + 28 \\[0.6em] s &= x + 14 \end{aligned}

Step 2 · Find Area of $\triangle ABC$ Using Heron's Formula

Using Heron's formula with $a = 14$ , $b = x + 6$ , $c = x + 8$ :

\begin{aligned} s - a &= (x + 14) - 14 = x \\[0.6em] s - b &= (x + 14) - (x + 6) = 8 \\[0.6em] s - c &= (x + 14) - (x + 8) = 6 \end{aligned}

\begin{aligned} \text{Area}(\triangle ABC) &= \sqrt{s(s-a)(s-b)(s-c)} \\[0.6em] &= \sqrt{(x+14)(x)(8)(6)} \\[0.6em] &= \sqrt{48x(x+14)} \end{aligned}

Step 3 · Find Area of $\triangle ABC$ as Sum of Three Triangles

Join $OA$ , $OB$ , and $OC$ . The perpendicular heights from incenter $O$ to each side are equal to the radius $r = 4 \text{ cm}$ :

\begin{aligned} \text{Area}(\triangle OBC) &= \dfrac{1}{2} \times BC \times OD = \dfrac{1}{2} \times 14 \times 4 = 28 \text{ cm}^2 \\[0.6em] \text{Area}(\triangle OCA) &= \dfrac{1}{2} \times AC \times OF = \dfrac{1}{2} \times (x+6) \times 4 = 2(x+6) = 2x + 12 \text{ cm}^2 \\[0.6em] \text{Area}(\triangle OAB) &= \dfrac{1}{2} \times AB \times OE = \dfrac{1}{2} \times (x+8) \times 4 = 2(x+8) = 2x + 16 \text{ cm}^2 \end{aligned}

Summing the three areas:

\begin{aligned} \text{Area}(\triangle ABC) &= 28 + (2x + 12) + (2x + 16) \\[0.6em] &= 4x + 56 \\[0.6em] &= 4(x + 14) \end{aligned}

Step 4 · Equate Areas and Solve for $x$

Equating the two area expressions: $\sqrt{48x(x+14)} = 4(x+14)$

Squaring both sides:

\begin{aligned} 48x(x+14) &= 16(x+14)^2 \\[0.6em] 3x(x+14) &= (x+14)^2 \\[0.6em] 3x(x+14) - (x+14)^2 &= 0 \\[0.6em] (x+14)[3x - (x+14)] &= 0 \\[0.6em] (x+14)(2x - 14) &= 0 \end{aligned}

This gives: $x + 14 = 0 \implies x = -14$ $2x - 14 = 0 \implies x = 7$

Since side length cannot be negative, $x = 7 \text{ cm}$ .

Step 5 · Calculate Sides $AB$ and $AC$

Substitute $x = 7$ into the side expressions:

\begin{aligned} AB &= x + 8 = 7 + 8 = 15 \text{ cm} \\[0.6em] AC &= x + 6 = 7 + 6 = 13 \text{ cm} \end{aligned}

Answer

$AB = 15 \text{ cm}$ and $AC = 13 \text{ cm}$

Common Mistakes

Tangent Identification Error: Forgetting that $CF = CD = 6\text{ cm}$ and $BE = BD = 8\text{ cm}$ , which leads to incorrect side length representations.
Algebraic Cancellation: Dividing both sides by $(x+14)$ without justifying that $x+14 \neq 0$ , or missing the negative root $x = -14$ (which must be formally rejected since side length $> 0$ ).
Radius as Height: Forgetting that the inradius $r = 4\text{ cm}$ is perpendicular to all three sides, so it acts as the exact altitude for $\triangle OBC$ , $\triangle OCA$ , and $\triangle OAB$ .

Question 12

More questions in Exercise 10.2