Question:

A rectangle ABCDABCD has sides AB=45cmAB = 45 \, \text{cm} and BC=26cmBC = 26 \, \text{cm}. Point EE is the midpoint of side CDCD. Find the radius of the incircle of the triangle AED\triangle AED.

Updated On: Nov 24, 2024
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Correct Answer: 7.07

Solution and Explanation

Rectangle ABCDABCD with sides:
AB=45cm,BC=26cm.AB = 45 \, \text{cm}, \quad BC = 26 \, \text{cm}.
EE is the midpoint of CDCD, so:
CE=ED=CD2=452=22.5cm.CE = ED = \frac{CD}{2} = \frac{45}{2} = 22.5 \, \text{cm}.
Coordinates of points:
A(0,0),B(45,0),D(0,26),C(45,26),E(22.5,26).A(0, 0), \, B(45, 0), \, D(0, 26), \, C(45, 26), \, E(22.5, 26).
Step 1: Calculate the lengths of the sides of AED\triangle AED
1. Length of AEAE:
AE=(22.50)2+(260)2=22.52+262.AE = \sqrt{(22.5 - 0)^2 + (26 - 0)^2} = \sqrt{22.5^2 + 26^2}.
Simplifying:
AE=506.25+676=1182.2534.39cm.AE = \sqrt{506.25 + 676} = \sqrt{1182.25} \approx 34.39 \, \text{cm}.
2. Length of EDED:
ED=22.5cm.ED = 22.5 \, \text{cm}.
3. Length of ADAD:
AD=26cm.AD = 26 \, \text{cm}.
Step 2: Calculate the area of AED\triangle AED
Using Heron's formula, the semi-perimeter (ss) is:
s=AE+ED+AD2=34.39+22.5+262=41.445cm.s = \frac{AE + ED + AD}{2} = \frac{34.39 + 22.5 + 26}{2} = 41.445 \, \text{cm}.
The area (Δ\Delta) is given by:
Δ=s(sAE)(sED)(sAD).\Delta = \sqrt{s(s - AE)(s - ED)(s - AD)}.
Substitute the values:
Δ=41.445(41.44534.39)(41.44522.5)(41.44526).\Delta = \sqrt{41.445 \cdot (41.445 - 34.39) \cdot (41.445 - 22.5) \cdot (41.445 - 26)}.
Simplify each term:
Δ=41.4457.05518.94515.445.\Delta = \sqrt{41.445 \cdot 7.055 \cdot 18.945 \cdot 15.445}.
Δ85952.84293.17cm2.\Delta \approx \sqrt{85952.84} \approx 293.17 \, \text{cm}^2.
Step 3: Radius of the incircle  
The radius of the incircle (rr) is given by:
r=Δs.r = \frac{\Delta}{s}.
Substitute the values:
r=293.1741.4457.07cm.r = \frac{293.17}{41.445} \approx 7.07 \, \text{cm}.

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