In rectangle ABCD, AB = 5 cm and BC = 3 cm. Point F and G are on the line segment CD so that DF = 1 cm and GC = 2 cm. Lines AF and BG intersect at. E. What is the area of AEB ?

Options

A. 10 sq. cm
B. `(15)/(2)` sq. cm
C. `25//2` sq. cm
D. 20 sq. cm

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Rectangle and Points
- We have rectangle ABCD where AB = 5 cm and BC = 3 cm.
- Points F and G are on line segment CD such that DF = 1 cm and GC = 2 cm.
- Therefore, CF = CD - DF = 5 cm - 1 cm = 4 cm.
- And, DG = CD - GC = 5 cm - 2 cm = 3 cm.

### Step 2: Determine the Coordinates
- Assign coordinates to the rectangle:
- A(0, 0)
- B(5, 0)
- C(5, 3)
- D(0, 3)
- Point F on CD is at (1, 3) (since DF = 1 cm).
- Point G on CD is at (3, 3) (since GC = 2 cm).

### Step 3: Find the Equations of Lines AF and BG
- The slope of line AF (from A to F):
- Slope = (y2 - y1) / (x2 - x1) = (3 - 0) / (1 - 0) = 3.
- Equation of line AF: y = 3x.

- The slope of line BG (from B to G):
- Slope = (y2 - y1) / (x2 - x1) = (3 - 0) / (3 - 5) = -3/2.
- Equation of line BG: y - 0 = (-3/2)(x - 5) → y = (-3/2)x + (15/2).

### Step 4: Find the Intersection Point E
- Set the equations of AF and BG equal to find E:
- 3x = (-3/2)x + (15/2).
- Multiply through by 2 to eliminate the fraction: 6x = -3x + 15.
- Combine like terms: 9x = 15 → x = 15/9 = 5/3.

- Substitute x = 5/3 into y = 3x to find y:
- y = 3(5/3) = 5.

- Therefore, point E is at (5/3, 5).

### Step 5: Calculate the Area of Triangle AEB
- The area of triangle AEB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, base AB = 5 cm and height is the y-coordinate of point E, which is 5 cm.

- Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times 5 \times 5 = \frac{25}{2} \text{ cm}^2. \]

### Final Answer
The area of triangle AEB is \(\frac{25}{2} \text{ cm}^2\).

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