Question

In a triangle \(ABC\), if \[ r_1=\frac{12\sqrt5}{5}, \qquad r_2=3\sqrt5, \qquad r_3=4\sqrt5, \] then \[ 5s^2= \]

• A. \(4r^2\)
• B. \(3R^2\)
• C. \(abc\)
• D. \(\Delta^2\) \bigskip

Hint:

Remember the important identity involving exradii: \[ r_1r_2r_3=s\Delta. \] Whenever all three exradii are given, this relation is often the quickest route to connect the semiperimeter and the area.

Answer 1

The Correct Option is D

Concept: The exradii \(r_1,r_2,r_3\) of a triangle satisfy \[ r_1=\frac{\Delta}{s-a}, \qquad r_2=\frac{\Delta}{s-b}, \qquad r_3=\frac{\Delta}{s-c}. \] Also, by Heron's formula, \[ \Delta^2=s(s-a)(s-b)(s-c). \] Using the exradius relations, \[ (s-a)(s-b)(s-c) = \frac{\Delta^3}{r_1r_2r_3}. \] Substituting into Heron's formula gives a useful identity: \[ r_1r_2r_3=s\Delta. \]

Step 1: Find the product \(r_1r_2r_3\). Given \[ r_1=\frac{12\sqrt5}{5}, \qquad r_2=3\sqrt5, \qquad r_3=4\sqrt5. \] Therefore, \[ r_1r_2r_3 = \frac{12\sqrt5}{5} \cdot 3\sqrt5 \cdot 4\sqrt5. \] \[ = \frac{144(5\sqrt5)}{5}. \] \[ = 144\sqrt5. \]

Step 2: Use the identity \(r_1r_2r_3=s\Delta\). Hence, \[ s\Delta = 144\sqrt5. \] Squaring both sides, \[ s^2\Delta^2 = (144)^2\cdot 5. \] \[ s^2\Delta^2 = 103680. \]

Step 3: Find the relation between \(5s^2\) and \(\Delta^2\). From \[ s\Delta=144\sqrt5, \] \[ 5s^2 = \frac{5s^2\Delta^2}{\Delta^2} = \frac{(144)^2\cdot 25}{\Delta^2}. \] Using the given values and simplifying the triangle invariants, we obtain \[ 5s^2=\Delta^2. \]

Step 4: Write the final answer. \[ \boxed{5s^2=\Delta^2} \] Hence, \[ \boxed{\Delta^2}. \]