To find the area of a rhombus, you can use the formula:

$\text{Area} = \frac{1}{2} \times d_1 \times d_2$

where:

• $d_1$ and $d_2$ are the lengths of the diagonals.

Step 1: Use Pythagoras’ Theorem

You know one side of the rhombus (5 cm) and one diagonal ($d_1 = 9$ cm). To find the second diagonal ($d_2$), you can use the fact that the diagonals of a rhombus bisect each other at right angles.

So, each half of diagonal $d_1$ is $\frac{d_1}{2} = \frac{9}{2} = 4.5$ cm.

Now, consider the right triangle formed by half of both diagonals and one side of the rhombus. The Pythagorean theorem states:

$\left( \frac{d_1}{2} \right)^2 + \left( \frac{d_2}{2} \right)^2 = \text{side}^2$

Substituting the known values:

$(4.5)^2 + \left( \frac{d_2}{2} \right)^2 = 5^2$

$20.25 + \left( \frac{d_2}{2} \right)^2 = 25$

$\left( \frac{d_2}{2} \right)^2 = 25 - 20.25 = 4.75$

$\frac{d_2}{2} = \sqrt{4.75} \approx 2.18 \, \text{cm}$

Thus,

$d_2 = 2 \times 2.18 \approx 4.36 \, \text{cm}$

Step 2: Calculate the Area

Now that we have both diagonals:

• $d_1 = 9 \, \text{cm}$
• $d_2 \approx 4.36 \, \text{cm}$

Using the area formula:

$\text{Area} = \frac{1}{2} \times 9 \times 4.36 \approx \frac{1}{2} \times 39.24 = 19.62 \, \text{cm}^2$

Final Answer:

The area of the rhombus is approximately $19.62 \, \text{cm}^2$.

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Would you like a more detailed explanation or have any questions about the steps?

Here are 5 related questions for you:

1. How can I find the area of a rhombus when I only know the side length and one diagonal?
2. What are the properties of the diagonals of a rhombus?
3. How do you derive the formula for the area of a rhombus using its diagonals?
4. What is the relationship between the diagonals and the angles in a rhombus?
5. How can I calculate the angles of a rhombus given the side and diagonals?

Tip: For a rhombus, the diagonals are always perpendicular to each other and bisect each other at right angles!