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 in circle of $\triangle$ADE is

Let us first visualize what we have:

Rectangle ABCD where AB = 56 cm and BC = 45 cm

E is the midpoint of CD (meaning E cuts CD exactly in half)

We need the incircle radius of triangle ADE

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In rectangle ABCD, opposite sides are equal and all angles are 90°.

AD = BC = 45 cm (opposite sides of rectangle are equal)

Since E is the midpoint of CD, we have DE = CD/2 = AB/2 = 56/2 = 28 cm

Angle D = 90° (all corners of a rectangle are right angles)

So triangle ADE is a right triangle with:

Right angle at D

Legs: AD = 45 cm and DE = 28 cm

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Using the Pythagorean theorem ($a^2 + b^2 = c^2$ for right triangles):

$AE^2 = AD^2 + DE^2$

$AE^2 = 45^2 + 28^2 = 2025 + 784 = 2809$

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

Quick check: $53^2 = 2809$

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For any triangle, the incircle radius formula is:

$r = \dfrac{\text{Area}}{\text{Semi-perimeter}}$

Since △ADE is a right triangle:

$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 28 \times 45 = 630 \text{ cm}^2$

Semi-perimeter = Half of the perimeter

$s = \dfrac{45 + 28 + 53}{2} = \dfrac{126}{2} = 63 \text{ cm}$

$r = \dfrac{630}{63} = 10 \text{ cm}$

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For right triangles, there's a shortcut formula:

$r = \dfrac{a + b - c}{2}$

where a and b are the legs and c is the hypotenuse.

$r = \dfrac{45 + 28 - 53}{2} = \dfrac{20}{2} = 10 \text{ cm}$

This gives the same answer and is much faster!

In a right triangle, the incircle touches all three sides, and the distances from each vertex to the points where the incircle touches the sides follow a specific pattern that leads to this elegant formula.

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Answer: 10 cm