ABCD is an isosceles trapezium with AB parallel to DC, AD = BC = 12 cm, `angle A = 60^(@) and DC = 16 cm.` Taking `sqrt3 = 1.732,` find
area of trapezium ABCD.

Answer

To find the area of the isosceles trapezium ABCD, we will follow these steps:

### Step 1: Understand the given information
We have an isosceles trapezium ABCD where:
- AB is parallel to DC
- AD = BC = 12 cm
- Angle A = 60°
- DC = 16 cm

### Step 2: Draw the trapezium
Draw trapezium ABCD with AB parallel to DC. Label the lengths and angles accordingly.

### Step 3: Identify the height and bases
Let:
- AB = a (unknown)
- DC = b = 16 cm
- Height = h (unknown)

### Step 4: Find the base length (AB)
Since angle A = 60°, we can use trigonometric ratios to find the base length (AB).

In triangle ACD:
- AD = 12 cm (hypotenuse)
- Angle A = 60°

Using cosine to find the base (let's denote the base as x):
\[ \cos(60°) = \frac{x}{AD} \]
\[ \cos(60°) = \frac{x}{12} \]
Since \(\cos(60°) = \frac{1}{2}\):
\[ \frac{1}{2} = \frac{x}{12} \]
Multiplying both sides by 12 gives:
\[ x = 6 \text{ cm} \]
Since AD and BC are equal and symmetrical, the length of the base AB will be:
\[ AB = DC + 2x = 16 + 2(6) = 28 \text{ cm} \]

### Step 5: Find the height (h)
Using sine to find the height (h):
\[ \sin(60°) = \frac{h}{AD} \]
\[ \sin(60°) = \frac{h}{12} \]
Since \(\sin(60°) = \frac{\sqrt{3}}{2}\):
\[ \frac{\sqrt{3}}{2} = \frac{h}{12} \]
Multiplying both sides by 12 gives:
\[ h = 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm} \]

### Step 6: Calculate the area of trapezium
The area \(A\) of a trapezium is given by the formula:
\[ A = \frac{1}{2} \times (a + b) \times h \]
Substituting the values we found:
\[ A = \frac{1}{2} \times (28 + 16) \times (6\sqrt{3}) \]
\[ A = \frac{1}{2} \times 44 \times 6\sqrt{3} \]
\[ A = 22 \times 6\sqrt{3} \]
\[ A = 132\sqrt{3} \text{ cm}^2 \]

### Step 7: Substitute the value of \(\sqrt{3}\)
Using \(\sqrt{3} \approx 1.732\):
\[ A \approx 132 \times 1.732 \approx 228.624 \text{ cm}^2 \]

### Final Answer
The area of trapezium ABCD is approximately \(228.624 \text{ cm}^2\).

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