In a right angled triangle `DeltaPQR`, PR is the hypotenuse of length 20 cm, `anglePRQ = 30^(@)`, the area of the triangle is

Options

A. `50sqrt(3)cm^(2)`
B. `100sqrt(3) cm^(2)`
C. `25 sqrt(3) cm^(2)`
D. `(100)/(sqrt(3)) cm^(2)`

Answer

To find the area of the right-angled triangle \( \Delta PQR \) where \( PR \) is the hypotenuse of length 20 cm and \( \angle PRQ = 30^\circ \), we can follow these steps:

### Step 1: Identify the sides of the triangle
In a right-angled triangle, we can use trigonometric ratios to find the lengths of the other two sides (PQ and QR) using the hypotenuse (PR).

### Step 2: Find the length of side PQ using sine
Using the sine function:
\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]
Here, \( \theta = 30^\circ \), the opposite side is \( PQ \), and the hypotenuse is \( PR = 20 \, \text{cm} \).
\[ \sin(30^\circ) = \frac{PQ}{20} \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{PQ}{20} \]
Multiplying both sides by 20:
\[ PQ = 10 \, \text{cm} \]

### Step 3: Find the length of side QR using cosine
Using the cosine function:
\[ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \]
Here, the adjacent side is \( QR \):
\[ \cos(30^\circ) = \frac{QR}{20} \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{QR}{20} \]
Multiplying both sides by 20:
\[ QR = 10\sqrt{3} \, \text{cm} \]

### Step 4: Calculate the area of triangle PQR
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this triangle, we can take \( PQ \) as the height and \( QR \) as the base:
\[ A = \frac{1}{2} \times PQ \times QR \]
Substituting the values we found:
\[ A = \frac{1}{2} \times 10 \times 10\sqrt{3} \]
Calculating this gives:
\[ A = \frac{1}{2} \times 100\sqrt{3} = 50\sqrt{3} \, \text{cm}^2 \]

### Final Answer
The area of triangle \( \Delta PQR \) is \( 50\sqrt{3} \, \text{cm}^2 \).

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