Solving Equations Using All Methods: Using additional geometric shapes

To solve this problem, we need to use the formula for the area of a trapezoid and the relationships given in the problem.

Step 1: Identify the given information
From the diagram and problem statement, we have:

• Upper base (top of trapezoid) = $2x$
• Lower base (bottom of trapezoid) = $4x$
• Height of trapezoid = $x$
• Area of trapezoid = $12$ square cm

Step 2: Verify the relationships
Let's confirm the stated relationships:

• Lower base is 2 times upper base: $4x = 2 \times 2x = 4x$ ✓
• Lower base is 4 times height: $4x = 4 \times x = 4x$ ✓

Step 3: Apply the trapezoid area formula
The area of a trapezoid is given by:
$A = \frac{1}{2}(b_1 + b_2) \times h$
where $b_1$ and $b_2$ are the two parallel bases and $h$ is the height.

Step 4: Substitute the values
Substituting our expressions into the formula:
$12 = \frac{1}{2}(2x + 4x) \times x$

Step 5: Simplify and solve for x
$12 = \frac{1}{2}(6x) \times x$
$12 = \frac{6x^2}{2}$
$12 = 3x^2$
$x^2 = \frac{12}{3}$
$x^2 = 4$
$x = 2$ (taking the positive root since x represents a length)

Step 6: Verify the solution
When $x = 2$:

• Upper base = $2x = 4$ cm
• Lower base = $4x = 8$ cm
• Height = $x = 2$ cm
• Area = $\frac{1}{2}(4 + 8) \times 2 = \frac{1}{2}(12) \times 2 = 12$ square cm ✓

Therefore, the value of x equals $x = 2$.

To find the side length of a square when the area is given, follow these steps:

• Step 1: We are given the area $A = 49 \, \text{cm}^2$.
• Step 2: Use the formula for the area of a square, which is $A = x^2$, where $x$ is the side length.
• Step 3: Solve the equation $x^2 = 49$.
• Step 4: To find $x$, take the square root of both sides: $x = \sqrt{49}$.
• Step 5: Calculate $\sqrt{49} = 7$.

Therefore, the side length of the square is $x = 7 \, \text{cm}$.

From the given answer choices, choice 2: $x=7$ is correct.

To find the value of $\angle C$, follow these steps:

Step 1: Set up the equations.
We know:
- $\angle A = \alpha$
- $\angle B = 3\alpha$

Using the given condition $\angle A + \angle B = 2\angle C$:
$\alpha + 3\alpha = 2\angle C \implies 4\alpha = 2\angle C \implies \angle C = 2\alpha$

Step 2: Use the triangle angle sum property.
From the triangle angle sum, we have:
$\angle A + \angle B + \angle C = 180^\circ$ Substituting the expressions for the angles:
$\alpha + 3\alpha + 2\alpha = 180^\circ$ $6\alpha = 180^\circ$ Solving for $\alpha$:
$\alpha = \frac{180^\circ}{6} = 30^\circ$

Step 3: Calculate $\angle C$.
Since $\angle C = 2\alpha$:
$\angle C = 2 \times 30^\circ = 60^\circ$ Therefore, the size of angle $\angle C$ is $\boxed{60^\circ}$.

To solve this problem, let's calculate $∢A$ with the steps outlined below:

• Step 1: Write the equations for each angle based on the given conditions: $∢B = 2A$$∢C = 3B = 3(2A) = 6A$

• Step 2: Use the sum of angles in a triangle: $∢A + ∢B + ∢C = 180^\circ$ Substitute the expressions: $A + 2A + 6A = 180$

• Step 3: Simplify the equation: $9A = 180$ Divide both sides by 9 to solve for $A$: $A = \frac{180}{9} = 20$

Therefore, the solution to the problem is $∢A = 20^\circ$.

To solve this problem, we'll use the properties of a triangle and given ratio:

• Step 1: Let $∢B = x$ and $∢C = 3x$ as per the given ratio $\frac{∢B}{∢C} = \frac{1}{3}$.
• Step 2: Use the triangle sum property: $∢A + ∢B + ∢C = 180^\circ$.
• Step 3: Substitute known values: $70^\circ + x + 3x = 180^\circ$.
• Step 4: Simplify: $4x + 70^\circ = 180^\circ$.
• Step 5: Solve for $x$: $4x = 110^\circ$.
• Step 6: Determine $x$: $x = 27.5^\circ$.
• Step 7: Calculate $∢C$: $∢C = 3x = 3 \times 27.5^\circ = 82.5^\circ$.

Therefore, the measure of angle $∢C$ is $82.5^\circ$.

The area of the rectangle is equal to length multiplied by width.

Let's set up the data in the formula:

$27=3x\times x$

$27=3x^2$

$\frac{27}{3}=\frac{3x^2}{3}$

$9=x^2$

$x=\sqrt{9}=3$

Given the problem, we are tasked to find the value of $x$ for a deltoid where the length is twice the width and the area is given. Let's proceed as follows:

• Step 1: In this deltoid problem, the diagonals correspond to length $2x$ and width $x$. The formula for the area of a deltoid in terms of its diagonals is $A = \frac{1}{2} \times d_1 \times d_2$.
• Step 2: Substitute the values. Thus, the area $16 = \frac{1}{2} \times (2x) \times x$.
• Step 3: Simplify the equation: $16 = \frac{1}{2} \times 2x^2 = x^2$.
• Step 4: Solve for $x$: We find $x^2 = 16$, so $x = \sqrt{16}$.
• Step 5: Conclude $x = 4$.

Therefore, the solution to the problem is $x = 4$.

The area of a rectangle is equal to its length multiplied by its width.

Let's write out the known data:

$22x=\frac{1}{2}x\times(x+8)$

$22x=\frac{1}{2}x^2+\frac{1}{2}x8$

$22x=\frac{1}{2}x^2+4x$

$0=\frac{1}{2}x^2+4x-22x$

$0=\frac{1}{2}x^2-18x$

$0=\frac{1}{2}x(x-36)$

For the equation to be balanced, $x$ needs to be equal to 36.

To solve this problem, we shall adhere to the following steps:

• Step 1: Utilize the area formula for triangles.
• Step 2: Simplify the equation to find the variable $x$.
• Step 3: Verify the result against the multiple-choice options.

Now, let us execute these steps:

Step 1: Start by applying the triangle area formula $A = \frac{1}{2} \times \text{base} \times \text{height}$.
The given area is $10 \, \text{cm}^2$, the base is $x$, and the height is $5x$. Thus, the formula becomes:

$10 = \frac{1}{2} \times x \times 5x$

Step 2: Simplify the equation:
$10 = \frac{1}{2} \times 5x^2$ $10 = \frac{5}{2}x^2$

Multiply both sides by $2$ to eliminate the fraction:

$20 = 5x^2$

Divide both sides by $5$:

$4 = x^2$

Take the square root of both sides:

$x = 2$

So, the value of $x$ is $\boxed{2}$.

Step 3: Upon reviewing the given multiple-choice options, the answer $x = 2$ corresponds to one of the listed choices, ensuring our calculations align with the expected solution.

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