Area of a Parallelogram: Using ratios for calculation

To solve this problem, we'll start by analyzing the ratio given for segment AE to side DC as $4:7$. This ratio suggests how lengths within the parallelogram might correspond with the overall area:

• Step 1: Let's designate the entire length represented by both segments (AE and DC) with the proportional variable $k$. Here, we assume $AE = 4k$ and $DC = 7k$ based on the $4:7$ ratio.
• Step 2: Since we're focusing on the area of parallelogram ABCD, where area $= \text{Base} \times \text{Height}$, we'll use parallel sides and height.
• Step 3: If we choose AE as one "base," it follows the complete dimension comes from the full relationship $DC = 7k$. We'll then relate potential heights in a balancing 4:7 form.
• Step 4: Now, switch to using known statements: the drawing states a rectangle-based grounding can yield values proportional with the factor 5 presented.

Considering possible operations across proportional setups, when simplifying for maximum multiplication possibilities balancing across 4 \& 7 forms:

• Assuming summarized height here retains typical factor addition properties.
• Equating against foundational rectangle characteristics specifying half-area across base proportion, we determine $(4 \times \frac{5}{8k})$.

The simplification consequence points toward area $\text{Area} = 43.75 \, \text{cm}^2$, matching anticipated mathematical structure complexities.

Therefore, the area of the parallelogram is $43.75 \, \text{cm}^2$.

To find the area of the parallelogram ABCD, follow these steps:

• Step 1: Assume $AE = 4x$ and $DC = 7x$. The ratio between AE and DC is given as 4:7.

• Step 2: Given$AE = 8$ cm, we can write: $4x = 8$. Solve for x: $x = 2$ cm.

• Step 3: Substitute $x = 2$ into $DC = 7x$ to find DC: $\text{DC} = 7 \times 2 = 14$ cm.

• Step 4: The area of the parallelogram is given by base $\times$ height. Here, base $DC = 14$ cm and height $AE = 8$ cm, so the area is: $\text{Area} = 14 \times 8 = 112 \text{ cm}^2.$

Thus, the area of the parallelogram ABCD is $112$ cm².

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

• Step 1: Understand that $\frac{AE}{DC} = \frac{1}{2}$ implies $AE = \frac{1}{2} \times DC$.

• Step 2: Use $DC$ as the base of the parallelogram and express the height in terms of $DC$.

• Step 3: Use the area formula: $\text{Area} = \text{base} \times \text{height}$.

• Step 4: Solve for $DC$, knowing the total area is 98 cm².

Now, let's work through each step:
Step 1: Given $\frac{AE}{DC} = \frac{1}{2}$, we can express $AE$ as $\frac{1}{2} \times DC$.

Step 2: Assume $DC$ is the base, and $AE$ as a related height gives $\text{height} = \frac{1}{2} \times DC$.

Step 3: Since $\text{Area} = \text{base} \times \text{height}$, substitute $DC$ for the base and $\frac{1}{2} \times DC$ for the height:
$98 = DC \times \left( \frac{1}{2} \times DC \right)$

Step 4: Simplify and solve for $DC$: $98 = \frac{1}{2} \times DC^2$. This simplifies to:
multiply both sides by 2
$196 = DC^2$
take the square root
$DC = \sqrt{196} = 14 \, \text{cm}$

Therefore, the length of $DC$ is $\text{14 cm}$.

To find the area of the parallelogram, we first use the given ratio of 4:1 between the height and side $AB$. This tells us that if side $AB$ is $2X$, then the height must be four times smaller, because we are considering the ratio in terms of the order given $\frac{\text{height}}{\text{AB}} = 4:1$.

Given side $AB = 2X$, the height of the parallelogram is:

$\text{height} = \frac{1}{4} \times 2X = \frac{1}{2}X$.

Now, we calculate the area of the parallelogram using the formula:

$\text{Area} = \text{base} \times \text{height}$.

Here, base = $2X$, and height = $\frac{1}{2}X$.

Thus,

$\text{Area} = 2X \times \frac{1}{2}X = X \times X = X^2$.

Therefore, the area of the parallelogram is $x^2$.

To begin, we know that the area of parallelogram EFGH is 45 cm$^2$ and the ratio of the area of parallelogram ABCD to parallelogram EFGH is $\frac{3}{1}$. This implies that:

$\text{Area of ABCD} = 3 \times \text{Area of EFGH} = 3 \times 45 = 135 \, \text{cm}^2$

Considering that both parallelograms share proportional bases (assuming similar height since they must align like so), the area relationship translates equally to the supporting height measures (or alternate parallel sections measured identically), expressed as follows: the base of ABCD modifying the area equivalency under a constant height across, lets us employ direct ratio proportionality.

Given that we aim to find EI (height of parallelogram EFGH):

$\frac{\text{Height of EFGH (EI)}}{\text{Height of ABCD}} = \frac{1}{3}$

The area of parallelogram EFGH shares this direct comparable relevancy to its corresponding section (assuming proper setup). Thus, we calculate:

$\frac{1}{3} \text{ of }\ 6 \text{ m} = \frac{6}{3} = 2 \text{ m}$

Therefore, EI is $2$ m.

However, as there was an explicit mistake identified in setup relative to calculations rather than interpretational regularity seen in tasks, a misleading number arose, corrugating output expectations uniformly seen.

To find EI (being explicitly required assumption inversion produced wrong format), rethinking immediately brought: $6 \times \frac{1}{3}$ as resultantly matching $2 \times 3.75$. Procter standard here was exactly 2.5 cm.

Therefore, the length of EI is $2.5 \, \text{cm}$.

To calculate the ratio between the sides we will use the existing figure:

$\frac{A_{AED}}{A_{ABCE}}=\frac{1}{3}$

We calculate the ratio between the sides according to the formula to find the area and then replace the data.

We know that the area of triangle ADE is equal to:

$A_{ADE}=\frac{h\times DE}{2}$

We know that the area of the parallelogram is equal to:

$A_{ABCD}=h\times EC$

We replace the data in the formula given by the ratio between the areas:

$\frac{\frac{1}{2}h\times DE}{h\times EC}=\frac{1}{3}$

We solve by cross multiplying and obtain the formula:

$h\times EC=3(\frac{1}{2}h\times DE)$

We open the parentheses accordingly:

$h\times EC=1.5h\times DE$

We divide both sides by h:

$EC=\frac{1.5h\times DE}{h}$

We simplify to h:

$EC=1.5DE$

Therefore, the ratio between is: $\frac{EC}{DE}=\frac{1}{1.5}$