In `DeltaABC`, a=12 cm, `angleB=30^(@)` and `angleC=90^(@)`, then area of `DeltaABC=?`

Options

A. `6sqrt(3) m^(2)`
B. `24sqrt(3) m^(2)`
C. `36sqrt(3) m^(2)`
D. None of these

Answer

To find the area of triangle ABC where side \( a = 12 \, \text{cm} \), \( \angle B = 30^\circ \), and \( \angle C = 90^\circ \), we can follow these steps:

### Step 1: Identify the triangle and its properties
In triangle ABC:
- \( \angle C = 90^\circ \) (right angle)
- \( \angle B = 30^\circ \)
- Therefore, \( \angle A = 180^\circ - (30^\circ + 90^\circ) = 60^\circ \)

### Step 2: Use the formula for the area of a triangle
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 consider side \( BC \) as the base and height \( AC \) as the height.

### Step 3: Find the length of side \( BC \)
Since \( \angle B = 30^\circ \) and \( a = 12 \, \text{cm} \) (which is opposite to angle A), we can use the sine function to find \( BC \):
\[ \sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB} \]
Here, \( AB = a = 12 \, \text{cm} \), therefore:
\[ \sin(30^\circ) = \frac{AC}{12} \]
We know that \( \sin(30^\circ) = \frac{1}{2} \), so:
\[ \frac{1}{2} = \frac{AC}{12} \]
Multiplying both sides by 12 gives:
\[ AC = 12 \times \frac{1}{2} = 6 \, \text{cm} \]

### Step 4: Find the length of side \( AC \)
Now we can use the tangent function to find \( AC \):
\[ \tan(30^\circ) = \frac{AC}{BC} \]
We know \( BC = 12 \, \text{cm} \) (the side opposite angle A):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{AC}{12} \]
Thus, we have:
\[ AC = 12 \times \frac{1}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \, \text{cm} \]

### Step 5: Calculate the area
Now we can substitute the values of base \( BC \) and height \( AC \) into the area formula:
\[ A = \frac{1}{2} \times BC \times AC = \frac{1}{2} \times 12 \times 4\sqrt{3} \]
Calculating this gives:
\[ A = \frac{1}{2} \times 48\sqrt{3} = 24\sqrt{3} \, \text{cm}^2 \]

### Final Answer
The area of triangle ABC is:
\[ \boxed{24\sqrt{3} \, \text{cm}^2} \]