ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of ∆ADE is

The correct answer is 10.

Here is the step-by-step calculation to find the radius of the incircle of triangle ADE:

Step 1: Identify the dimensions of the rectangle and the position of point E
In rectangle ABCD, opposite sides are equal in length. Therefore:

$CD=AB=56\text{cm}$

$AD=BC=45\text{cm}$
Since E is the midpoint of side CD, we can calculate the length of DE as:

$DE=\frac{CD}{2}=\frac{56}{2}=28\text{cm}$

Step 2: Find the length of the hypotenuse AE of triangle ADE
Since ABCD is a rectangle, the angle at corner D is a right angle ($\angle ADE=90\circ$). This makes triangle ADE a right-angled triangle at D. Using the Pythagorean theorem, we find the hypotenuse AE:

$A{E}^2=A{D}^2+D{E}^2$
Substituting the given lengths:

$A{E}^2={45}^2+{28}^2$

$A{E}^2=2025+784=2809$
Taking the square root of both sides:

$AE=\sqrt{2809}=53\text{cm}$

Step 3: Calculate the radius of the incircle
For any right-angled triangle with legs of length a and b and a hypotenuse of length c, the radius r of the incircle (inradius) is given by the formula:

$r=\frac{a+b-c}{2}$
Here, we set $a=AD=45\text{cm}$, $b=DE=28\text{cm}$, and $c=AE=53\text{cm}$:

$r=\frac{45+28-53}{2}$

$r=\frac{73-53}{2}$

$r=\frac{20}{2}=10\text{cm}$

Thus, the radius of the incircle of triangle ADE is 10 cm.