ABCD is a trapezium in which ABIIDC and P and Qare points on AD and BC respectively such that `PQabs()DC`. If PD= 12 cm, BQ = 42 cm and QC = 18cm, find AD.

Answer

To solve the problem step by step, we will use the properties of trapeziums and the Basic Proportionality Theorem.

### Step 1: Understand the trapezium and the given points
We have a trapezium ABCD where AB is parallel to DC. Points P and Q are on sides AD and BC respectively, and line segment PQ is parallel to DC.

### Step 2: Identify the lengths given in the problem
- PD = 12 cm
- BQ = 42 cm
- QC = 18 cm

### Step 3: Use the Basic Proportionality Theorem
According to the Basic Proportionality Theorem (also known as Thales' theorem), if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

In triangle BDC:
\[ \frac{OB}{OD} = \frac{BQ}{QC} \]
Let \( OB = x \) and \( OD = y \). Then we have:
\[ \frac{x}{y} = \frac{42}{18} \]
Simplifying this gives:
\[ \frac{x}{y} = \frac{7}{3} \]
This means \( 3x = 7y \) or \( x = \frac{7}{3}y \) (Equation 1).

### Step 4: Apply the theorem in triangle ADB
In triangle ADB, since OP is parallel to AB:
\[ \frac{OD}{OB} = \frac{PD}{AP} \]
Using \( PD = 12 \) cm, we can write:
\[ \frac{y}{x} = \frac{12}{AP} \]
Rearranging gives:
\[ AP = \frac{12x}{y} \quad (Equation 2) \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, we know \( x = \frac{7}{3}y \). Substitute this into Equation 2:
\[ AP = \frac{12 \cdot \frac{7}{3}y}{y} \]
This simplifies to:
\[ AP = \frac{12 \cdot 7}{3} = 28 \text{ cm} \]

### Step 6: Find the length of AD
Now, we can find the length of AD:
\[ AD = AP + PD \]
Substituting the values we found:
\[ AD = 28 \text{ cm} + 12 \text{ cm} = 40 \text{ cm} \]

### Final Answer
Thus, the length of AD is **40 cm**.

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