Given rectangle ABCD with the following dimensions:Nbsp;
- AB = 56 cm
- BC = 45 cm
- CD = 56 cm (opposite sides of a rectangle are equal)
- DA = 45 cm (opposite sides of a rectangle are equal)
- E is the midpoint of CD, so CE = ED = $\frac{56}{2} = 28$ cm.
We need to find the radius of the incircle of $\triangle ADE$. The formula for the incircle radius $r$ of a triangle is:
$r = \frac{A}{s}$
where A is the triangle's area and s is its semi-perimeter.
Calculate the Semi-perimeter s:
The sides of $\triangle ADE$ are DA = 45 cm, DE = 28 cm, andNbsp;\(AE = \sqrt{AB^2 + DE^2}\)
\(= \sqrt{56^2 + 28^2}\)
\(= \sqrt{3136 + 784}\)
\(= \sqrt{3920} \approx 62.61\ cm.\)
The semi-perimeter s is:
$s = \frac{DA + DE + AE}{2} = \frac{45 + 28 + 62.61}{2} = 67.805$ cm.
Calculate the Area A:
Since $\triangle ADE$ is a right-angled triangle (with the right angle at D), its area can be calculated as:
$A = \frac{1}{2} \times base \times height = \frac{1}{2} \times DE \times DA$
$A = \frac{1}{2} \times 28 \times 45 = 630 \, cm^2$.
Calculate the Radius r:
Now, use the formula $r = \frac{A}{s}$:
$r = \frac{630}{67.805} \approx 9.29$ cm.
The radius of the incircle is approximately 9.29 cm.