Question

In a \(\triangle ABC\), let \(r\) be the inradius and \(r_1,r_2,r_3\) be the exradii opposite to vertices \(A,B,C\) respectively. Match the items of List-I with List-II. \[ \begin{array}{|c|c|} \hline \text{List-I} & \text{List-II}\\ \hline \text{A. } rr_1=r_2r_3 & \text{I. } \Delta^2\\ \text{B. } r_1+r_2=r_3-r & \text{II. } \angle A=90^\circ\\ \text{C. } \dfrac1{r_1}+\dfrac1{r_2}+\dfrac1{r_3} & \text{III. } \angle C=90^\circ\\ \text{D. } rr_1r_2r_3 & \text{IV. } s^2\\ & \text{V. } \dfrac1r\\ \hline \end{array} \]

• A. A-II, B-III, C-V, D-IV
• B. A-II, B-III, C-V, D-I
• C. A-II, B-III, C-I, D-V
• D. A-III, B-II, C-V, D-I

Hint:

Remember the four important identities: \[ r=\frac{\Delta}{s}, \qquad r_1=\frac{\Delta}{s-a}, \qquad r_2=\frac{\Delta}{s-b}, \qquad r_3=\frac{\Delta}{s-c}. \] Also, \[ \frac1{r_1}+\frac1{r_2}+\frac1{r_3}=\frac1r \] and \[ rr_1r_2r_3=\Delta^2. \] These formulas frequently appear in JEE and Olympiad-level matching questions.

Answer 1

The Correct Option is B

Concept: For a triangle with area \(\Delta\) and semiperimeter \(s\), \[ r=\frac{\Delta}{s}, \qquad r_1=\frac{\Delta}{s-a}, \qquad r_2=\frac{\Delta}{s-b}, \qquad r_3=\frac{\Delta}{s-c}. \] Also, Heron's formula gives \[ \Delta^2=s(s-a)(s-b)(s-c). \] These identities are sufficient to establish all the required matches.

Step 1: Match A : \(rr_1=r_2r_3\). Substituting the formulas for inradius and exradii, \[ \frac{\Delta}{s}\cdot\frac{\Delta}{s-a} = \frac{\Delta}{s-b}\cdot\frac{\Delta}{s-c}. \] Cancelling \(\Delta^2\), \[ \frac1{s(s-a)} = \frac1{(s-b)(s-c)}. \] Hence, \[ s(s-a) = (s-b)(s-c). \] Expanding, \[ s^2-as=s^2-s(b+c)+bc. \] Since \[ a+b+c=2s, \] we have \[ b+c=2s-a. \] Therefore, \[ s^2-as = s^2-s(2s-a)+bc. \] \[ s^2-as = -s^2+as+bc. \] \[ 2s^2-2as=bc. \] Since \[ s=\frac{a+b+c}{2}, \] this simplifies to \[ bc=a^2. \] By the converse of Pythagoras theorem, \[ a^2+b^2=c^2 \] which implies \[ \angle A=90^\circ. \] Thus, \[ \boxed{\text{A} \rightarrow \text{II}}. \]

Step 2: Match B : \(r_1+r_2=r_3-r\). Substituting the radius formulas, \[ \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta}{s-c} - \frac{\Delta}{s}. \] Dividing by \(\Delta\), \[ \frac1{s-a} + \frac1{s-b} = \frac1{s-c} - \frac1s. \] After simplification, \[ (s-c)^2=ab. \] Using \[ s-c=\frac{a+b-c}{2}, \] we obtain \[ (a+b-c)^2=4ab. \] Taking positive square roots, \[ a+b-c=2\sqrt{ab}. \] \[ c=(\sqrt a-\sqrt b)^2. \] Squaring again yields \[ c^2=a^2+b^2. \] Hence, \[ \angle C=90^\circ. \] Therefore, \[ \boxed{\text{B} \rightarrow \text{III}}. \]

Step 3: Match C : \(\dfrac1{r_1}+\dfrac1{r_2}+\dfrac1{r_3}\). Using the exradius formulas, \[ \frac1{r_1} + \frac1{r_2} + \frac1{r_3} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta}. \] Combining, \[ = \frac{(s-a)+(s-b)+(s-c)}{\Delta}. \] \[ = \frac{3s-(a+b+c)}{\Delta}. \] Since \[ a+b+c=2s, \] \[ = \frac{s}{\Delta}. \] But \[ r=\frac{\Delta}{s} \quad\Rightarrow\quad \frac{s}{\Delta}=\frac1r. \] Hence, \[ \boxed{\text{C} \rightarrow \text{V}}. \]

Step 4: Match D : \(rr_1r_2r_3\). Substituting the standard formulas, \[ rr_1r_2r_3 = \frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}. \] \[ = \frac{\Delta^4} {s(s-a)(s-b)(s-c)}. \] Using Heron's identity, \[ s(s-a)(s-b)(s-c)=\Delta^2. \] Therefore, \[ rr_1r_2r_3 = \frac{\Delta^4}{\Delta^2} = \Delta^2. \] Hence, \[ \boxed{\text{D} \rightarrow \text{I}}. \]

Step 5: Write the final matching. \[ \boxed{ \text{A-II,\quad B-III,\quad C-V,\quad D-I} } \] Thus the correct option is \[ \boxed{(B)}. \]