Area of a Trapezoid: Suggesting options for terms when the formula result is known

To solve this problem, we'll use the trapezoid area formula:

Given:

• Area, $S = 20 \, \text{cm}^2$
• Sum of bases, $b_1 + b_2 = 10 \, \text{cm}$

The formula for the area of a trapezoid is:

$S = \frac{(b_1 + b_2)}{2} \times h$

Substituting the known values into the formula, we have:

$20 = \frac{10}{2} \times h$

Simplifying the equation:

$20 = 5 \times h$

Solving for $h$, we divide both sides by 5:

$h = \frac{20}{5} = 4 \, \text{cm}$

Therefore, the height of the trapezoid is $4 \, \text{cm}$.

To solve this problem, we'll use the following plan:

• Identify the given values: The area is $30 \, \text{cm}^2$, and the height is $5 \, \text{cm}$.
• Use the area formula for a trapezoid, which is $\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$.
• Recognize the relationship: Since AB is half the length of DC, let the length of AB be $x$ and DC be $2x$.
• Substitute the known values and relationships into the formula to find $x$.

Let's proceed with the calculation:

The formula for the area of a trapezoid is:

$\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$

Substitute the known values:

$30 = \frac{1}{2} \times (x + 2x) \times 5$

Combine the bases:

$30 = \frac{1}{2} \times 3x \times 5$

Simplify: $30 = \frac{15x}{2}$

Multiply both sides by 2 to clear the fraction:

$60 = 15x$

Divide both sides by 15:

$x = 4$

Therefore, the bases are:

• AB (the shorter base) is $4 \, \text{cm}$.
• DC (the longer base) is $2x = 2 \times 4 = 8 \, \text{cm}$.

The lengths of the bases of the trapezoid are $\textbf{4 cm and 8 cm.}$

The formula for the area of a trapezoid is given by:

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

where $A$ is the area, $b_1$ and $b_2$ are the lengths of the two bases, and $h$ is the height.

We are given:

• $A = 20 \, \text{cm}^2$
• $b_1 + b_2 = 10 \, \text{cm}$

We substitute these values into the formula:

$20 = \frac{1}{2} \times 10 \times h$

To find $h$, we simplify the equation:

$20 = 5h$

Dividing each side by 5, we get:

$h = \frac{20}{5}$

Hence, the height of the trapezoid is:

$h = 4 \, \text{cm}$

Therefore, the height of the trapezoid is 4 cm.

To solve for the bases of the trapezoid, follow these steps:

• Let the length of $AB$ be $x$ cm, and since $AB$ is half of $DC$, let $DC$ be $2x$ cm.
• Using the formula for the area of a trapezoid:
$\text{Area} = \frac{1}{2} \times (AB + DC) \times \text{height}$ \li>Substitute the known values:$\frac{1}{2} \times (x + 2x) \times 5 = 30$ \item>Simplify and solve for $x$: $\frac{1}{2} \times 3x \times 5 = 30$ $7.5x = 30$ $x = \frac{30}{7.5} = 4$ \item>Thus, $AB = x = 4 \, \text{cm}$ and $DC = 2x = 8 \, \text{cm}$.

Therefore, the lengths of the trapezoid's bases are $AB = 4 \, \text{cm}$ and $DC = 8 \, \text{cm}$.