The length of a rectangle is twice its breadth. If the length is decreased by half of the 10 cm and the breadth is increased by half of the 10 cm,the area of the rectangle is increased by 5 sq cm more than 70 sq cm. Find the length of the rectangle.

Options

A. a. 30 cm
B. b. 40 cm
C. c. 21 cm
D. d. 45 cm

Answer

To find the length of the rectangle, we will follow these steps:

### Step 1: Define the Variables
Let the breadth of the rectangle be \( B \). Since the length \( L \) is twice the breadth, we can express the length as:
\[ L = 2B \]

### Step 2: Adjust the Length and Breadth
According to the problem, the length is decreased by half of 10 cm (which is 5 cm) and the breadth is increased by half of 10 cm (which is 5 cm). Therefore, the new dimensions are:
- New Length: \( L' = L - 5 = 2B - 5 \)
- New Breadth: \( B' = B + 5 \)

### Step 3: Calculate the Original Area
The original area \( A \) of the rectangle is given by:
\[ A = L \times B = (2B) \times B = 2B^2 \]

### Step 4: Calculate the New Area
The new area \( A' \) after the adjustments is:
\[ A' = L' \times B' = (2B - 5)(B + 5) \]

### Step 5: Expand the New Area Expression
Expanding the expression for the new area:
\[ A' = (2B - 5)(B + 5) = 2B^2 + 10B - 5B - 25 = 2B^2 + 5B - 25 \]

### Step 6: Set Up the Equation
According to the problem, the new area is 5 square cm more than 70 square cm:
\[ A' = 70 + 5 = 75 \]
So, we can set up the equation:
\[ 2B^2 + 5B - 25 = 75 \]

### Step 7: Simplify the Equation
Rearranging the equation gives:
\[ 2B^2 + 5B - 25 - 75 = 0 \]
\[ 2B^2 + 5B - 100 = 0 \]

### Step 8: Solve the Quadratic Equation
Now, we will solve the quadratic equation \( 2B^2 + 5B - 100 = 0 \) using the quadratic formula:
\[ B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 5, c = -100 \).

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \times 2 \times (-100) = 25 + 800 = 825 \]

Now substituting into the quadratic formula:
\[ B = \frac{-5 \pm \sqrt{825}}{4} \]

### Step 9: Calculate the Value of B
Calculating \( \sqrt{825} \):
\[ \sqrt{825} = 5\sqrt{33} \]
Thus,
\[ B = \frac{-5 \pm 5\sqrt{33}}{4} \]

Since breadth cannot be negative, we take the positive root:
\[ B = \frac{-5 + 5\sqrt{33}}{4} \]

### Step 10: Find the Length
Now, substituting \( B \) back to find \( L \):
\[ L = 2B = 2 \times \frac{-5 + 5\sqrt{33}}{4} = \frac{-10 + 10\sqrt{33}}{4} = \frac{-5 + 5\sqrt{33}}{2} \]

### Final Answer
Thus, the length of the rectangle is:
\[ L = \frac{-5 + 5\sqrt{33}}{2} \]