Area of a Trapezoid: Using variables

To solve this problem, we'll utilize the information given about trapezoid $ABCD$:

• $AB = AD$ which implies $x = x = AB$.
• $DC = 2 \times AB$ implies $DC = 2x$.
• The formula for the area of a trapezoid is given by: $\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$.
• The area of the trapezoid is stated as three times longer than side $AB$, giving us $\text{Area} = 3 \times x$.

The bases of trapezoid $ABCD$ are $AB = x$ and $DC = 2x$. Assume the height of trapezoid $ABCD$ is $h$.

Using the area formula, we have:
$\frac{1}{2} \times (x + 2x) \times h = 3x$

This simplifies to:
$\frac{3x}{2} \times h = 3x$

To find $h$, divide both sides by $\frac{3x}{2}$ this yields:
$h = 2$

Next, verify that when $h = 2$, the area calculation matches:
Substitute $h = 2$ back into the expression for area:
$\frac{1}{2} \times 3x \times 2 = 3x$, which holds true as $3x = 3x$.

Thus, the calculations confirm the length of side $AB$ is $2$.

To solve this problem, we'll find the length of side AB given the area of the trapezoid. Follow these steps:

• Step 1: Set up the area formula for a trapezoid:
  The area $A$ of a trapezoid is given by the formula $A = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$.
• Step 2: Substitute the given information:
  Here, $\text{Base}_1 = AB = a$ and $\text{Base}_2 = DC = a + 3$ cm. The height $h = \frac{1}{2}$ cm. The area is given as $A = 2a$ cm².
• Step 3: Substitute into the formula:
  $2a = \frac{1}{2} \times (a + (a + 3)) \times \frac{1}{2}$
• Step 4: Simplify and solve for $a$:
  $2a = \frac{1}{2} \times (2a + 3) \times \frac{1}{2}$ $2a = \frac{(2a + 3)}{4}$
  Multiply through by 4 to clear the fraction: $8a = 2a + 3$
  Subtract $2a$ from both sides: $6a = 3$
  Divide both sides by 6: $a = \frac{3}{6} = \frac{1}{2}$

Therefore, the length of side AB is $\frac{1}{2}$ cm, and the correct choice is (3).

Let's calculate the area of trapezoid $ABCD$:

The formula for the area of a trapezoid is:

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

In this trapezoid, we have:

• $\text{Base}_1 = AB = 5$ cm
• $\text{Base}_2 = DC = 3$ cm
• $\text{Height} = h$

Substituting these into the formula, we get:

$\text{Area} = \frac{1}{2} \times (5 + 3) \times h$

Simplify the calculation:

$\text{Area} = \frac{1}{2} \times 8 \times h = 4h$

Thus, the area of the trapezoid is $4h$ square centimeters.

To express the area of the trapezoid in terms of $X$, follow these steps:

• Step 1: Identify the given values for the trapezoid's dimensions. The top base $b_1$ is $X + 14$, the bottom base $b_2$ is $3X + 7$, and the height $h$ is $2X$.
• Step 2: Use the trapezoid area formula $A = \frac{1}{2} \times (b_1 + b_2) \times h$.
• Step 3: Compute the sum of the bases: $(X + 14) + (3X + 7) = X + 14 + 3X + 7 = 4X + 21$.
• Step 4: Calculate the area using the formula: $A = \frac{1}{2} \times (4X + 21) \times (2X)$.
• Step 5: Simplify: $A = \frac{1}{2} \times (4X + 21) \times 2X = (4X + 21) \times X$.
• Step 6: Simplify further by distributing: $A = 4X^2 + 21X$.

Thus, the area of the trapezoid expressed in terms of $X$ is $4X^2 + 21X$.

To solve this problem, we will use the formula for the area of a trapezoid:

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

Here, $\text{Area} = 9x$ cm², $\text{Base}_1 = 2x$ cm, and $\text{Base}_2 = 2.5x$ cm.

Substitute the known values into the formula:

$9x = \frac{1}{2} \times (2x + 2.5x) \times \text{AE}$

$9x = \frac{1}{2} \times 4.5x \times \text{AE}$

Multiply through by 2 to clear the fraction:

$18x = 4.5x \times \text{AE}$

Solve for AE by dividing both sides by $4.5x$:

$\text{AE} = \frac{18x}{4.5x} = 4$

Thus, the height AE is $\textbf{4 cm}$.

Therefore, the solution to the problem is $\textbf{4}$ cm.

To determine the area of the trapezoid, we will use the formula for the area of a trapezoid:

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

From the problem, the two bases are $2X$ and $3Y$. The height is $X$.

Substituting into the formula, we have:

$A = \frac{1}{2} \times (2X + 3Y) \times X$

Simplifying the expression inside the parenthesis gives:

$A = \frac{1}{2} \times (2X + 3Y) \times X = \frac{1}{2} \times (2X \times X + 3Y \times X)$

Distributing $X$ through the terms inside the parenthesis gives:

$A = \frac{1}{2} \times (2X^2 + 3XY)$

Continuing the simplification:

$A = \frac{1}{2} \times 2X^2 + \frac{1}{2} \times 3XY$

Which simplifies to:

$A = X^2 + 1.5XY$

Therefore, the area of the trapezoid is $X^2 + 1.5XY$ cm².

Through comparison, this expression matches the given choice: $x^2+1.5xy$ cm², which corresponds to choice $3$.

Thus, the correct area of the trapezoid is $x^2 + 1.5xy$ cm².

To solve this problem, we need to apply the area formula for a trapezoid:

• The area $A$ of a trapezoid is given by $A = \frac{1}{2} \times (b_1 + b_2) \times h$ where $b_1$ and $b_2$ are the base lengths, and $h$ is the height.

The problem provides:

• Area $A = 78$ cm²
• Base 1, $b_1 = X + 10$
• Base 2, $b_2 = 12X - 10$
• Height $h = 6$ cm

Substitute these into the area formula:

$78 = \frac{1}{2} \times ((X + 10) + (12X - 10)) \times 6$

Simplify the expression:

$78 = \frac{1}{2} \times (13X) \times 6$

Multiply through by 2 to clear the fraction:

$156 = 13X \times 6$

Simplify further:

$156 = 78X$

Solving for $X$ gives:

$X = \frac{156}{78}$

$X = 2$

Therefore, the solution to the problem is $X = 2$.