ABCD is a square . Draw a triangle QBC on side BC considering BC as base and draw a triangle PAC on AC as its base such that `Delta QBC ~ Delta PAC`. Then, `("Area of " Delta QBC)/("Area of " Delta PAC)` is equal to :

Options

A. `(1)/(2)`
B. `(2)/(1)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a square ABCD, and we need to draw two triangles: QBC on side BC and PAC on diagonal AC, such that triangle QBC is similar to triangle PAC. We need to find the ratio of the areas of these triangles.

### Step 2: Define the Square
Let the side length of the square ABCD be \( s \). Thus, the coordinates of the square can be defined as:
- A(0, s)
- B(0, 0)
- C(s, 0)
- D(s, s)

### Step 3: Identify the Bases of the Triangles
- For triangle QBC, the base is BC, which has a length of \( s \).
- For triangle PAC, the base is AC, which is the diagonal of the square.

### Step 4: Calculate the Length of AC
Using the Pythagorean theorem, we find the length of diagonal AC:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2} \]

### Step 5: Calculate the Areas of the Triangles
1. **Area of Triangle QBC**:
\[ \text{Area}_{QBC} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base BC = \( s \). Let the height from point Q to line BC be \( h_1 \). Thus:
\[ \text{Area}_{QBC} = \frac{1}{2} \times s \times h_1 \]

2. **Area of Triangle PAC**:
\[ \text{Area}_{PAC} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base AC = \( s\sqrt{2} \). Let the height from point P to line AC be \( h_2 \). Thus:
\[ \text{Area}_{PAC} = \frac{1}{2} \times s\sqrt{2} \times h_2 \]

### Step 6: Use Similarity of Triangles
Since triangles QBC and PAC are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Thus:
\[ \frac{\text{Area}_{QBC}}{\text{Area}_{PAC}} = \left(\frac{s}{s\sqrt{2}}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Conclusion
The ratio of the area of triangle QBC to the area of triangle PAC is:
\[ \frac{\text{Area}_{QBC}}{\text{Area}_{PAC}} = \frac{1}{2} \]