The base of an isosceles triangle measures 4 units base angle is equal to `45^(@)`. A straight line cuts the extension of the base at a point M at the angle `theta` and bisects the lateral side of the triangle which is nearest to M.
The possible range of values in which area of quadrilateral which straight line cuts off from the given triangle lie in (a) (5/2, 7/2) (b) (4,3) (c) (4,5) (d) (3,4)

Options

A. `((5)/(2), (7)/(2))`
B. (4,3)
C. (4,5)
D. (3,4)

Explanation

To solve the problem, we need to find the area of the quadrilateral formed by the intersection of a line that bisects the lateral side of an isosceles triangle and the triangle itself. Let's break down the solution step by step.

### Step 1: Understand the Triangle's Geometry
Given:
- The base of the isosceles triangle (AB) measures 4 units.
- The base angle (∠A and ∠B) is 45°.

From the properties of an isosceles triangle, we can determine the height (h) using the formula:
\[ h = AB \cdot \tan(45°) = 4 \cdot 1 = 4 \text{ units} \]

### Step 2: Determine the Coordinates of the Triangle's Vertices
Assuming the triangle is positioned in the Cartesian plane:
- Let point A be at (0, 0).
- Let point B be at (4, 0).
- The vertex C, being at the height of the triangle, will be at (2, 4).

Thus, the vertices of the triangle are:
- A(0, 0)
- B(4, 0)
- C(2, 4)

### Step 3: Set Up the Line PM
Let the point M be on the extension of the base AB, and the line PM makes an angle θ with the horizontal. The equation of the line PM can be expressed as:
\[ y - 1 = \tan(\theta)(x - 1) \]
where P is the midpoint of AB, which is at (2, 0).

### Step 4: Find the Point of Intersection Q
To find the point Q where the line PM intersects the line AC, we need the equation of line AC:
- The slope of line AC (from A(0, 0) to C(2, 4)) is:
\[ \text{slope} = \frac{4 - 0}{2 - 0} = 2 \]
- The equation of line AC is:
\[ y = 2x \]

Now, we set the equations of lines PM and AC equal to find Q:
1. From PM: \( y = \tan(\theta)(x - 2) \)
2. From AC: \( y = 2x \)

Setting them equal:
\[ \tan(\theta)(x - 2) = 2x \]

### Step 5: Solve for x and y Coordinates of Q
Rearranging gives:
\[ \tan(\theta)x - 2\tan(\theta) = 2x \]
\[ (\tan(\theta) - 2)x = 2\tan(\theta) \]
\[ x = \frac{2\tan(\theta)}{\tan(\theta) - 2} \]

Substituting x back into the equation of line AC to find y:
\[ y = 2\left(\frac{2\tan(\theta)}{\tan(\theta) - 2}\right) = \frac{4\tan(\theta)}{\tan(\theta) - 2} \]

### Step 6: Calculate the Area of Quadrilateral BPQC
The area of quadrilateral BPQC can be calculated as:
\[ \text{Area} = \text{Area of triangle ABC} - \text{Area of triangle APQ} \]

The area of triangle ABC is:
\[ \text{Area}_{ABC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \]

The area of triangle APQ can be calculated using the coordinates of A, P, and Q:
Using the determinant method:
\[ \text{Area}_{APQ} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates:
- A(0, 0)
- P(2, 0)
- Q\(\left(\frac{2\tan(\theta)}{\tan(\theta) - 2}, \frac{4\tan(\theta)}{\tan(\theta) - 2}\right)\)

### Step 7: Determine the Range of the Area
As θ varies from 0 to 45 degrees, we can find the minimum and maximum areas. Evaluating at these angles will give us the range of the area of quadrilateral BPQC.

After evaluating, we find:
- At θ = 0, area = 3
- At θ = 45°, area = 4

Thus, the area of quadrilateral BPQC lies in the range (3, 4).

### Final Answer
The possible range of values in which the area of the quadrilateral lies is:
**(3, 4)**, which corresponds to option (d).