In `Delta ABC` the straight line parallel to the side `BC `meets `AB and AC `at the points `P and Q` respectively. If `AP = QC`, the length of `AB `is 12 units and the length of `AQ `is 2 units, then the length (in units) of `CQ` is

Options

A. 4
B. 6
C. 8
D. 10

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have triangle ABC with a line parallel to side BC, which intersects sides AB and AC at points P and Q respectively. We know:
- Length of AB = 12 units
- Length of AQ = 2 units
- AP = QC (let's denote this common length as X)

### Step 2: Set Up the Ratios
According to the theorem related to parallel lines in triangles, the segments created by the intersection of a line parallel to one side of a triangle with the other two sides are proportional. Therefore, we can write the following ratio:

\[ \frac{AP}{PB} = \frac{AQ}{QC} \]

Where:
- AP = X
- PB = AB - AP = 12 - X
- AQ = 2
- QC = X

Thus, we can express the ratio as:

\[ \frac{X}{12 - X} = \frac{2}{X} \]

### Step 3: Cross Multiply to Solve for X
Cross-multiplying gives us:

\[ X^2 = 2(12 - X) \]

Expanding this, we have:

\[ X^2 = 24 - 2X \]

Rearranging the equation results in:

\[ X^2 + 2X - 24 = 0 \]

### Step 4: Factor the Quadratic Equation
We need to factor the quadratic equation \(X^2 + 2X - 24 = 0\). We look for two numbers that multiply to -24 and add to 2. These numbers are 6 and -4.

Thus, we can factor the equation as:

\[ (X + 6)(X - 4) = 0 \]

### Step 5: Solve for X
Setting each factor to zero gives us:

1. \(X + 6 = 0 \Rightarrow X = -6\) (not valid since length cannot be negative)
2. \(X - 4 = 0 \Rightarrow X = 4\)

### Step 6: Find the Length of CQ
Since we have established that \(QC = X\), we find:

\[ CQ = 4 \text{ units} \]

### Final Answer
The length of CQ is **4 units**.

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