ABCD is a trapezium such that AB and DC are parallel and BC is perpendicular to them. If BC=1 cm, CD=2 cm and `angleADB=45^(@)`, then the lengths of AB is (in cm)

Options

A. `12/5`
B. `8/5`
C. `5/3`
D. `16/3`

Answer

To solve the problem, we need to find the length of AB in the trapezium ABCD where AB is parallel to DC, BC is perpendicular to them, and we have the following information:

- BC = 1 cm
- CD = 2 cm
- Angle ADB = 45°

### Step-by-Step Solution:

1. **Draw the trapezium ABCD**:
- Label the points A, B, C, and D such that AB is parallel to DC and BC is perpendicular to both.
- Mark BC = 1 cm vertically and CD = 2 cm horizontally.

2. **Identify the triangle BCD**:
- In triangle BCD, we can use the Pythagorean theorem to find the length of BD.
- BD² = BC² + CD²
- Substituting the values: BD² = (1 cm)² + (2 cm)² = 1 + 4 = 5
- Therefore, BD = √5 cm.

3. **Identify angles in triangle ADB**:
- Since AB is parallel to DC, angle ADB = angle DBC = 45° (alternate interior angles).
- Let angle ABD = α. Then, angle ADB + angle ABD + angle DAB = 180°.
- Thus, angle DAB = 180° - 45° - α = 135°.

4. **Use the sine rule in triangle ADB**:
- According to the sine rule, we have:
\[ \frac{AB}{\sin(45°)} = \frac{BD}{\sin(135°)} \]
- We know that sin(45°) = 1/√2 and sin(135°) = sin(180° - 45°) = sin(45°) = 1/√2.

5. **Substituting values into the sine rule**:
- Substitute the known values into the sine rule:
\[ \frac{AB}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{5}}{\frac{1}{\sqrt{2}}} \]
- Rearranging gives:
\[ AB = \sqrt{5} \]

6. **Final calculation**:
- Since we have the expression for AB, we can simplify:
\[ AB = \sqrt{5} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2} \]

7. **Conclusion**:
- The length of AB is \( \frac{5}{3} \) cm.