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
length of side AB.

Answer

To find the length of side AB in the isosceles trapezium ABCD, we will follow these steps:

### Step 1: Understand the trapezium configuration
We have an isosceles trapezium ABCD where AB is parallel to DC. Given:
- \( AD = BC = 12 \, \text{cm} \)
- \( DC = 16 \, \text{cm} \)
- \( \angle A = 60^\circ \)

### Step 2: Draw perpendiculars from points D and C to line AB
To find the length of AB, we can drop perpendiculars from points D and C to line AB. Let's denote the foot of the perpendicular from D as point P and from C as point Q. This creates two right triangles, ADP and BCQ.

### Step 3: Identify the height and base
In triangles ADP and BCQ:
- The height (DP and CQ) is equal to the length of AD (or BC), which is \( 12 \, \text{cm} \).
- We need to find the base lengths AP and BQ.

### Step 4: Use trigonometric ratios
Using the cosine function in triangle ADP:
\[ \cos(60^\circ) = \frac{AP}{AD} \]
Given \( AD = 12 \, \text{cm} \) and \( \cos(60^\circ) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{AP}{12} \]
Multiplying both sides by 12 gives:
\[ AP = 6 \, \text{cm} \]

### Step 5: Find BQ using the same reasoning
Since ABCD is an isosceles trapezium, the lengths AP and BQ are equal:
\[ BQ = AP = 6 \, \text{cm} \]

### Step 6: Calculate the length of AB
Now, we can find the length of AB:
\[ AB = DC + AP + BQ \]
Substituting the known values:
\[ AB = 16 + 6 + 6 = 28 \, \text{cm} \]

### Final Answer
The length of side AB is \( 28 \, \text{cm} \).

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