Find the Trapezoid Area: Right Triangle with 5cm Parallel Line

To solve this problem, we'll follow these steps:

• Establish similarity between the triangles $\triangle ADE$ and $\triangle ABC$.
• Use similarity to find the height of the trapezoid DECB.
• Calculate the area of the trapezoid using the area formula for a trapezoid.

Let's work through each step:

Step 1: The triangles are similar, $\triangle ADE \sim \triangle ABC$, because DE is parallel to BC, establishing similarity by AA (angle-angle) criteria.

Step 2: In similar triangles, corresponding sides are proportional. Thus,

$\frac{DE}{BC} = \frac{AD}{AB}$

Given $DE = 5$ cm, $BC = 10$ cm, and $AB = 20$ cm, we can set up the proportion:

$\frac{5}{10} = \frac{AD}{20}$

This simplifies to $\frac{1}{2} = \frac{AD}{20}$, giving $AD = 10$ cm since $AD = \frac{20}{2}$.

Step 3: The height of the trapezoid DECB is now $(AB - AD) = 20 - 10 = 10$ cm.

Step 4: Calculate the area of trapezoid DECB using the trapezoid area formula:

$\text{Area} = \frac{1}{2} \times (BC + DE) \times \text{Height}$

Substitute the values:

$\text{Area} = \frac{1}{2} \times (10 + 5) \times 10 = \frac{1}{2} \times 15 \times 10 = \frac{150}{2} = 75 \text{ cm}^2$

Therefore, the area of the trapezoid is 75 cm².