Question

A rectangle $ABCD$ has sides $AB = 45 \, \text{cm}$ and $BC = 26 \, \text{cm}$. Point $E$ is the midpoint of side $CD$. Find the radius of the incircle of the triangle $\triangle AED$.

Answer 1

Rectangle $ABCD$ with sides:
$AB = 45 \, \text{cm}, \quad BC = 26 \, \text{cm}.$
$E$ is the midpoint of $CD$, so:
$CE = ED = \frac{CD}{2} = \frac{45}{2} = 22.5 \, \text{cm}.$
Coordinates of points:
$A(0, 0), \, B(45, 0), \, D(0, 26), \, C(45, 26), \, E(22.5, 26).$
Step 1: Calculate the lengths of the sides of $\triangle AED$
1. Length of $AE$:
$AE = \sqrt{(22.5 - 0)^2 + (26 - 0)^2} = \sqrt{22.5^2 + 26^2}.$
Simplifying:
$AE = \sqrt{506.25 + 676} = \sqrt{1182.25} \approx 34.39 \, \text{cm}.$
2. Length of $ED$:
$ED = 22.5 \, \text{cm}.$
3. Length of $AD$:
$AD = 26 \, \text{cm}.$
Step 2: Calculate the area of $\triangle AED$
Using Heron's formula, the semi-perimeter ($s$) is:
$s = \frac{AE + ED + AD}{2} = \frac{34.39 + 22.5 + 26}{2} = 41.445 \, \text{cm}.$
The area ($\Delta$) is given by:
$\Delta = \sqrt{s(s - AE)(s - ED)(s - AD)}.$
Substitute the values:
$\Delta = \sqrt{41.445 \cdot (41.445 - 34.39) \cdot (41.445 - 22.5) \cdot (41.445 - 26)}.$
Simplify each term:
$\Delta = \sqrt{41.445 \cdot 7.055 \cdot 18.945 \cdot 15.445}.$
$\Delta \approx \sqrt{85952.84} \approx 293.17 \, \text{cm}^2.$
**Step 3: Radius of the incircle **
The radius of the incircle ($r$) is given by:
$r = \frac{\Delta}{s}.$
Substitute the values:
$r = \frac{293.17}{41.445} \approx 7.07 \, \text{cm}.$

Answer 2

Rectangle $ABCD$ with sides:
$AB = 45 \, \text{cm}, \quad BC = 26 \, \text{cm}.$
$E$ is the midpoint of $CD$, so:
$CE = ED = \frac{CD}{2} = \frac{45}{2} = 22.5 \, \text{cm}.$
Coordinates of points:
$A(0, 0), \, B(45, 0), \, D(0, 26), \, C(45, 26), \, E(22.5, 26).$
Step 1: Calculate the lengths of the sides of $\triangle AED$
1. Length of $AE$:
$AE = \sqrt{(22.5 - 0)^2 + (26 - 0)^2} = \sqrt{22.5^2 + 26^2}.$
Simplifying:
$AE = \sqrt{506.25 + 676} = \sqrt{1182.25} \approx 34.39 \, \text{cm}.$
2. Length of $ED$:
$ED = 22.5 \, \text{cm}.$
3. Length of $AD$:
$AD = 26 \, \text{cm}.$
Step 2: Calculate the area of $\triangle AED$
Using Heron's formula, the semi-perimeter ($s$) is:
$s = \frac{AE + ED + AD}{2} = \frac{34.39 + 22.5 + 26}{2} = 41.445 \, \text{cm}.$
The area ($\Delta$) is given by:
$\Delta = \sqrt{s(s - AE)(s - ED)(s - AD)}.$
Substitute the values:
$\Delta = \sqrt{41.445 \cdot (41.445 - 34.39) \cdot (41.445 - 22.5) \cdot (41.445 - 26)}.$
Simplify each term:
$\Delta = \sqrt{41.445 \cdot 7.055 \cdot 18.945 \cdot 15.445}.$
$\Delta \approx \sqrt{85952.84} \approx 293.17 \, \text{cm}^2.$
**Step 3: Radius of the incircle **
The radius of the incircle ($r$) is given by:
$r = \frac{\Delta}{s}.$
Substitute the values:
$r = \frac{293.17}{41.445} \approx 7.07 \, \text{cm}.$