Question

in the rhombus pqrs below, pq = 20 mm and pt = 16 mm.
find the area of the rhombus.
be sure to include the correct unit in your answer.

Explanation:

Identify given values and properties

We are given a rhombus \(PQRS\) with:

• Side length \(PQ = 20\text{ mm}\)
• Half-diagonal segment \(PT = 16\text{ mm}\)
• Diagonals \(PR\) and \(QS\) intersect perpendicularly at point \(T\).

Calculate the other half-diagonal segment

Since the diagonals of a rhombus are perpendicular, triangle \(\triangle PTQ\) is a right-angled triangle at \(T\).
Using the Pythagorean theorem:

$$ PT^2 + QT^2 = PQ^2 $$

$$ 16^2 + QT^2 = 20^2 $$

$$ 256 + QT^2 = 400 $$

$$ QT^2 = 144 \implies QT = 12\text{ mm} $$

Determine the full lengths of the diagonals

The diagonals of a rhombus bisect each other:

• Diagonal \(d_1 = PR = 2 \times PT = 2 \times 16 = 32\text{ mm}\)
• Diagonal \(d_2 = QS = 2 \times QT = 2 \times 12 = 24\text{ mm}\)

Calculate the area of the rhombus

The formula for the area of a rhombus is:

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

$$ \text{Area} = \frac{1}{2} \times 32 \times 24 = 16 \times 24 = 384\text{ mm}^2 $$

Answer:

In the rhombus \(PQRS\) below, \(PQ = 20\text{ mm}\) and \(PT = 16\text{ mm}\).
Find the area of the rhombus.

Area: <blank>\(384\text{ mm}^2\)</blank>