Calculate Trapezoid Area: Right Triangle with BC=5cm and AC=13cm

We are tasked with finding the area of trapezoid DECB in the right triangle ABC where DE is the midsegment parallel to BC. Given $BC = 5$ cm, $AC = 13$ cm, let us calculate the area step-by-step.

• First, use the Pythagorean theorem to find the length of $AB$. Since triangle ABC is a right triangle at $B$, we have:

$AB = \sqrt{AC^2 - BC^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

• The midsegment DE of triangle ABC, being parallel to the base BC, is half the length of BC:

$DE = \frac{1}{2} \times BC = \frac{1}{2} \times 5 = 2.5$ cm.

• Now, to find the area of trapezoid DECB, which has bases DE and BC, the height is the same as $AB$ (the vertical side of triangle ABC):

$A = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (2.5 + 5) \times 12$.

Calculate the expression:

• $A = \frac{1}{2} \times 7.5 \times 12 = \frac{1}{2} \times 90 = 45$ cm².
• Since DECB is half of the total triangle, the area is half of 45.

So, the area of trapezoid DECB is $45 \div 2 = 22.5$ cm².

The solution to the problem is $\boxed{22.5}$.