ABCD is a trapezium in which `AB ||CD, AB =20cm, BC =10` cm, `CD =10` cm and `AD =10` cm . Find `/_ADC`

Options

A. `80^(@)`
B. `100^(@)`
C. `120^(@)`
D. `140^(@)`

Explanation

To find the angle \( \angle ADC \) in trapezium \( ABCD \) where \( AB \parallel CD \), \( AB = 20 \) cm, \( BC = 10 \) cm, \( CD = 10 \) cm, and \( AD = 10 \) cm, we can follow these steps:

### Step 1: Draw the trapezium
Draw trapezium \( ABCD \) such that \( AB \) is parallel to \( CD \). Label the lengths:
- \( AB = 20 \) cm
- \( CD = 10 \) cm
- \( BC = 10 \) cm
- \( AD = 10 \) cm

### Step 2: Draw a line parallel to \( CD \)
Draw a line \( DE \) parallel to \( CD \) such that \( D \) is directly below \( A \) and \( E \) is directly below \( B \). Since \( AB \) and \( CD \) are parallel, \( DE \) will also be parallel to both.

### Step 3: Identify the lengths
Since \( AB \) is longer than \( CD \), we can find the length of \( AE \) (the horizontal distance from \( A \) to \( E \)):
- \( AE = AB - CD = 20 \, \text{cm} - 10 \, \text{cm} = 10 \, \text{cm} \)

### Step 4: Form triangle \( ADE \)
Now, we have triangle \( ADE \) where:
- \( AD = 10 \) cm (given)
- \( DE = 10 \) cm (since \( DE \) is parallel to \( CD \) and equal in length)
- \( AE = 10 \) cm (calculated)

### Step 5: Determine the type of triangle
Since all sides of triangle \( ADE \) are equal (10 cm each), triangle \( ADE \) is an equilateral triangle.

### Step 6: Calculate the angles of triangle \( ADE \)
In an equilateral triangle, all angles are equal to \( 60^\circ \). Therefore:
- \( \angle ADE = 60^\circ \)
- \( \angle A = 60^\circ \)
- \( \angle DAE = 60^\circ \)

### Step 7: Find \( \angle ADC \)
Since \( AB \parallel CD \), the angles \( \angle DAB \) and \( \angle ADC \) are supplementary (they form a linear pair). Thus:
\[ \angle ADC + \angle DAB = 180^\circ \]
Substituting \( \angle DAB = 60^\circ \):
\[ \angle ADC + 60^\circ = 180^\circ \]
\[ \angle ADC = 180^\circ - 60^\circ = 120^\circ \]

### Final Answer
Thus, \( \angle ADC = 120^\circ \).

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