Question 10

Rhombus: A rhombus is a quadrilateral with all four sides equal. Its diagonals bisect each other at right angles. The area of a rhombus can be calculated using its diagonals:

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

where $d_1$ and $d_2$ are the lengths of the two diagonals. Here, the vertices are given in order A, B, C, D — so AC and BD are the diagonals.

Distance Formula to find diagonal lengths:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

We will use the distance formula to find the lengths of the diagonals.

Step 1 — Find length of first diagonal

Let the vertices be A$(3, 0)$, B$(4, 5)$, C$(-1, 4)$, and D$(-2, -1)$. The first diagonal is AC. We use the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2}$

$= \sqrt{(-4)^2 + (4)^2}$

$= \sqrt{16 + 16}$

$= \sqrt{32}$

$= \sqrt{16 \times 2}$

$\boxed{4\sqrt{2} \text{ units}}$

Step 2 — Find length of second diagonal

The second diagonal is BD. We use the distance formula again.

$BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2}$

$= \sqrt{(-6)^2 + (-6)^2}$

$= \sqrt{36 + 36}$

$= \sqrt{72}$

$= \sqrt{36 \times 2}$

$\boxed{6\sqrt{2} \text{ units}}$

Step 3 — Calculate the area of the rhombus

The area of a rhombus is $\frac{1}{2}$ (product of its diagonals). Let $d_1 = \mathbf{4\sqrt{2}}$ and $d_2 = \mathbf{6\sqrt{2}}$.

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

$= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$

$= \frac{1}{2} \times 24 \times (\sqrt{2} \times \sqrt{2})$

$= \frac{1}{2} \times 24 \times 2$

$= 24$

$\boxed{24 \text{ square units}}$

Answer

The area of the rhombus is 24 square units.