Circumscribed Triangles and High-Weightage Board Riders for CBSE Class 10

When a circle is inscribed inside a triangle, it touches all three sides, transforming every side into a tangent and partitioning the perimeter into pairs of equal tangent segments. Solving for unknown side lengths in these circumscribed configurations requires synthesizing Theorem 10.2, Heron's area formula, and the inradius partitioning method.

In CBSE Class 10 Mathematics, Chapter 10 (Circles), the 12cm Circumscribed Triangle Problem (NCERT Exercise 10.2, Question 12) is universally recognized as the single most challenging, prestigious 5-mark question in the board examination.

What You Will Learn

• How an inradius ($r$) connects perimeter to total enclosed triangle area: $\text{Area} = r \times s$
• Complete, step-by-step solution to the NCERT 12cm Circumscribed Triangle Problem
• Equating Heron's Formula with the inradius tripartite partition method
• Solving for tangent lengths when chord lengths are given ($PQ = 8\text{ cm}, r = 5\text{ cm}$)
• Presentation standards for securing full marks in 5-mark Section D questions

1. The Mathematical Engine: Two Ways to Calculate Triangle Area

When a circle of radius $r$ is inscribed inside a triangle $\Delta ABC$ with sides $a, b, c$:

                                      A
                                     /|                                    / |                                    /  |                                    /   O                                    /  r/ 
                                  B---+---+---C
                                    D

Method 1: Heron's Formula (Class 9)

$s = \frac{a + b + c}{2} \quad (\text{Semi-perimeter})$ $\mathbf{\text{Area}(\Delta ABC) = \sqrt{s(s - a)(s - b)(s - c)}}$

Method 2: Partitioning into Three Sub-Triangles

Connect the center $O$ to vertices $A, B,$ and $C$. This splits $\Delta ABC$ into three smaller triangles: $\Delta OBC, \Delta OCA,$ and \Delta OAB$:

• Base of $\Delta OBC$ is $a$, height is inradius $r \implies \text{Area} = \frac{1}{2} \times a \times r$
• Base of $\Delta OCA$ is $b$, height is inradius $r \implies \text{Area} = \frac{1}{2} \times b \times r$
• Base of $\Delta OAB$ is $c$, height is inradius $r \implies \text{Area} = \frac{1}{2} \times c \times r$

Adding all three areas together: $\text{Total Area} = \frac{1}{2} r (a + b + c) = r \times \left(\frac{a + b + c}{2}\right) = \mathbf{r \times s}$

Important: <u>Equating these two independent expressions for the same triangle's area—$\sqrt{s(s-a)(s-b)(s-c)} = r \times s$—is the master key that unlocks the famous 12cm circumscribed triangle problem!</u>

2. The Benchmark NCERT 12cm Triangle Problem (5-Mark Classic)

Problem Statement:

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. Find the lengths of sides $AB$ and $AC$.

                                      A
                                     /                                   x /   \ x
                                   F     E
                                8 /   O   \ 6
                                 /  r=4                                    B---- D ----C
                                  8     6

Step-by-Step Complete Solution:

Step 1: Assign Tangent Variables

Let the circle touch side $AC$ at point $E$ and side $AB$ at point $F$. By Theorem 10.2 (lengths of tangents from an external point are equal):

• From vertex $C$: $CE = CD = 6\text{ cm}$
• From vertex $B$: $BF = BD = 8\text{ cm}$
• From vertex $A$: Let $AF = AE = \mathbf{x\text{ cm}}$

Step 2: Determine Side Lengths of $\Delta ABC$

• Side $a = BC = BD + DC = 8 + 6 = \mathbf{14\text{ cm}}$
• Side $b = AC = AE + EC = \mathbf{x + 6\text{ cm}}$
• Side $c = AB = AF + FB = \mathbf{x + 8\text{ cm}}$

Step 3: Compute the Semi-Perimeter ($s$)

$s = \frac{a + b + c}{2} = \frac{14 + (x + 6) + (x + 8)}{2} = \frac{2x + 28}{2} = \mathbf{x + 14\text{ cm}}$

Step 4: Compute Area Using Heron's Formula

First, evaluate $(s - a), (s - b),$ and $(s - c)$:

• $s - a = (x + 14) - 14 = x$
• $s - b = (x + 14) - (x + 6) = 8$
• $s - c = (x + 14) - (x + 8) = 6$

Substitute into Heron's Formula: $\text{Area}(\Delta ABC) = \sqrt{s(s - a)(s - b)(s - c)}$ $\text{Area} = \sqrt{(x + 14)(x)(8)(6)} = \mathbf{\sqrt{48x(x + 14)}} \quad \text{--- (1)}$

Step 5: Compute Area Using Inradius Partitioning

Join $OA, OB,$ and $OC$. The inradius is $r = 4\text{ cm}$. $\text{Area}(\Delta ABC) = \text{Area}(\Delta OBC) + \text{Area}(\Delta OCA) + \text{Area}(\Delta OAB)$ $\text{Area} = \frac{1}{2} \times 4 \times a + \frac{1}{2} \times 4 \times b + \frac{1}{2} \times 4 \times c$ $\text{Area} = 2(a + b + c) = 2(2s) = 4s = 4(x + 14) \quad \text{--- (2)}$

Step 6: Equate Both Area Expressions and Solve for $x$

$\sqrt{48x(x + 14)} = 4(x + 14)$

Square both sides to eliminate the radical: $48x(x + 14) = [4(x + 14)]^2$ $48x(x + 14) = 16(x + 14)^2$

Divide both sides by $16(x + 14)$ (since $x + 14 \ne 0$): $\frac{48x}{16} = x + 14$ $3x = x + 14$ $2x = 14 \implies \mathbf{x = 7\text{ cm}}$

Step 7: Calculate the Required Side Lengths

• Length of side $AB = c = x + 8 = 7 + 8 = \mathbf{15\text{ cm}}$
• Length of side $AC = b = x + 6 = 7 + 6 = \mathbf{13\text{ cm}}$

Conclusion: <u>The length of side $AB$ is $15\text{ cm}$, and the length of side $AC$ is $13\text{ cm}$</u>.

3. High-Yield Rider: Tangent Length with Given Chord Length

Problem: $PQ$ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length of tangent $TP$.

Solution:

1. Join $OT$ intersecting chord $PQ$ at point $R$.
2. $\Delta TPQ$ is isosceles ($TP = TQ$), and $OT$ is the angle bisector of $\angle PTQ$. Therefore, $OT$ is the perpendicular bisector of chord $PQ$: $PR = RQ = \frac{8}{2} = 4\text{ cm} \quad \text{and} \quad \angle PRO = 90^\circ$
3. In Right Triangle $\Delta OPR$: $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$
4. Compare Similar Triangles: In $\Delta TPR$ and $\Delta OPR$: $\angle TRP = \angle PRO = 90^\circ$, and $\angle TPR = 90^\circ - \angle RPO = \angle POR$. Therefore, by AA similarity: $\Delta TRP \sim \Delta PRO$ $\frac{TP}{OP} = \frac{PR}{OR} \implies \frac{TP}{5} = \frac{4}{3}$ $TP = \frac{5 \times 4}{3} = \mathbf{\frac{20}{3}\text{ cm}}$
5. Therefore, <u>the length of tangent $TP$ is $\frac{20}{3}\text{ cm}$ (or $6.67\text{ cm}$)</u>.

4. Summary and Examination Tips

Exam Tip: In the 12cm circumscribed triangle problem, when you arrive at $48x(x+14) = 16(x+14)^2$, DO NOT expand the quadratic equation into $48x^2 + 672x = \dots$! Simply divide both sides by $(x+14)$ directly. This avoids cumbersome quadratic factoring!

Common Mistake: Forgetting that side $a$ is $14\text{ cm}$ ($8 + 6$), not just $8$ or $6$.