AC, a diagnonal of the rectange ABCD , measure 5 units. The area of the rectangle is 12 sq. units. What is the perimeter of the rectangle?

Options

A. 7
B. 14
C. 17
D. 20

Answer

To solve the problem step by step, we need to find the perimeter of rectangle ABCD given that the diagonal AC measures 5 units and the area of the rectangle is 12 square units.

### Step 1: Define Variables
Let the length of the rectangle be \( L \) and the breadth be \( B \).

### Step 2: Use Area Formula
The area \( A \) of a rectangle is given by the formula:
\[ A = L \times B \]
Given that the area is 12 square units, we can write:
\[ L \times B = 12 \quad \text{(1)} \]

### Step 3: Use Diagonal Formula
The diagonal \( d \) of a rectangle can be calculated using the Pythagorean theorem:
\[ d^2 = L^2 + B^2 \]
Given that the diagonal AC measures 5 units, we have:
\[ 5^2 = L^2 + B^2 \]
This simplifies to:
\[ 25 = L^2 + B^2 \quad \text{(2)} \]

### Step 4: Substitute for B
From equation (1), we can express \( B \) in terms of \( L \):
\[ B = \frac{12}{L} \]
Now substitute \( B \) into equation (2):
\[ 25 = L^2 + \left(\frac{12}{L}\right)^2 \]
This expands to:
\[ 25 = L^2 + \frac{144}{L^2} \]

### Step 5: Multiply Through by \( L^2 \)
To eliminate the fraction, multiply the entire equation by \( L^2 \):
\[ 25L^2 = L^4 + 144 \]
Rearranging gives us:
\[ L^4 - 25L^2 + 144 = 0 \quad \text{(3)} \]

### Step 6: Let \( M = L^2 \)
Let \( M = L^2 \). Then equation (3) becomes:
\[ M^2 - 25M + 144 = 0 \]

### Step 7: Solve the Quadratic Equation
Now we will solve this quadratic equation using the quadratic formula:
\[ M = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -25, c = 144 \):
\[ M = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 144}}{2 \cdot 1} \]
Calculating the discriminant:
\[ M = \frac{25 \pm \sqrt{625 - 576}}{2} \]
\[ M = \frac{25 \pm \sqrt{49}}{2} \]
\[ M = \frac{25 \pm 7}{2} \]
This gives us two possible values for \( M \):
\[ M = \frac{32}{2} = 16 \quad \text{or} \quad M = \frac{18}{2} = 9 \]

### Step 8: Find Length and Breadth
Since \( M = L^2 \), we have:
1. If \( M = 16 \), then \( L = 4 \) (since length cannot be negative).
2. If \( M = 9 \), then \( L = 3 \).

Now we can find \( B \):
- For \( L = 4 \):
\[ B = \frac{12}{4} = 3 \]
- For \( L = 3 \):
\[ B = \frac{12}{3} = 4 \]

Thus, the dimensions of the rectangle are \( L = 4 \) and \( B = 3 \) (or vice versa).

### Step 9: Calculate Perimeter
The perimeter \( P \) of a rectangle is given by:
\[ P = 2(L + B) \]
Substituting the values:
\[ P = 2(4 + 3) = 2 \times 7 = 14 \text{ units} \]

### Final Answer
The perimeter of the rectangle is **14 units**.

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