Incircle Geometry CAT 2024 Actual Questions

Detailed Solution:
Step 1: Understand the Geometry of the Problem
- \( AB = 56 \, \text{cm}, BC = 45 \, \text{cm} \).
- \( CD = AB = 56 \, \text{cm} \), \( DA = BC = 45 \, \text{cm} \).
- \( E \) is the midpoint of \( CD \), so \( CE = ED = \frac{56}{2} = 28 \, \text{cm} \).

Step 2: Sides of \( \triangle ADE \)
- \( AD = 45 \, \text{cm}, DE = 28 \, \text{cm} \).
- \( AE = \sqrt{AD^2 + ED^2} = \sqrt{45^2 + 28^2} = \sqrt{2025 + 784} = 53 \, \text{cm}. \)

Step 3: Semiperimeter (\( s \))
- \( s = \frac{AD + DE + AE}{2} = \frac{45 + 28 + 53}{2} = 63 \, \text{cm}. \)

Step 4: Area of \( \triangle ADE \)
- Using Heron's formula:
\[ \text{Area} = \sqrt{s(s - AD)(s - DE)(s - AE)} = \sqrt{63(63 - 45)(63 - 28)(63 - 53)} \] - Simplify: \( \text{Area} = \sqrt{63 \cdot 18 \cdot 35 \cdot 10} = \sqrt{396900} = 630 \, \text{cm}^2. \)

Step 5: Radius of Incircle (\( r \))
- \( r = \frac{\text{Area}}{s} = \frac{630}{63} = 10 \, \text{cm}. \)

Final Answer: The radius of the incircle is \( \boxed{10 \, \text{cm}} \).