line PQ meets triangle ABC such that P lies on AB and Q lies on AC. If `AP=1cm, PB=3cm, AQ=1.5cmandQC=4.5cm`, then what is the ratio of area of `DeltaAPQ` and quadrilateral PBCQ ?

Options

A. `1:16`
B. `1:15`
C. `1:9`
D. `1:8`

Answer

To find the ratio of the area of triangle \( \Delta APQ \) and quadrilateral \( PBCQ \), we can follow these steps:

### Step 1: Determine the lengths of sides AB and AC
Given:
- \( AP = 1 \, \text{cm} \)
- \( PB = 3 \, \text{cm} \)
- \( AQ = 1.5 \, \text{cm} \)
- \( QC = 4.5 \, \text{cm} \)

Calculate \( AB \) and \( AC \):
\[ AB = AP + PB = 1 + 3 = 4 \, \text{cm} \]
\[ AC = AQ + QC = 1.5 + 4.5 = 6 \, \text{cm} \]

### Step 2: Find the ratio of corresponding sides
We need to find the ratio of the sides \( AP \) to \( AB \) and \( AQ \) to \( AC \):
\[ \text{Ratio of } AP \text{ to } AB = \frac{AP}{AB} = \frac{1}{4} \]
\[ \text{Ratio of } AQ \text{ to } AC = \frac{AQ}{AC} = \frac{1.5}{6} = \frac{1}{4} \]

### Step 3: Establish similarity of triangles
Since the ratios of corresponding sides are equal and angle \( A \) is common, we can conclude that triangles \( \Delta APQ \) and \( \Delta ABC \) are similar by the SAS (Side-Angle-Side) similarity criterion.

### Step 4: Calculate the ratio of the areas of the triangles
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\[ \frac{\text{Area of } \Delta APQ}{\text{Area of } \Delta ABC} = \left(\frac{AP}{AB}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 5: Relate areas of triangle APQ and quadrilateral PBCQ
Let \( \text{Area of } \Delta APQ = x \) and \( \text{Area of } \Delta ABC = 16x \) (since \( \Delta ABC \) is 16 times the area of \( \Delta APQ \)).

The area of quadrilateral \( PBCQ \) can be expressed as:
\[ \text{Area of } PBCQ = \text{Area of } \Delta ABC - \text{Area of } \Delta APQ = 16x - x = 15x \]

### Step 6: Find the ratio of the areas of triangle APQ and quadrilateral PBCQ
Now we can find the ratio:
\[ \frac{\text{Area of } \Delta APQ}{\text{Area of } PBCQ} = \frac{x}{15x} = \frac{1}{15} \]

### Final Answer
Thus, the ratio of the area of triangle \( \Delta APQ \) to the area of quadrilateral \( PBCQ \) is:
\[ \boxed{\frac{1}{15}} \]