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Find length and breadth of a rectangular park whose perimeter is $100$ m and area is $600$ $m^2$
Find length and breadth of a rectangular park whose perimeter is $100$ m and area is $600$ $m^2$
This Question has 1 answers.
Let the length and breadth of the rectangular park be $ l $ and $ b $ respectively.
### Step 1: Use the Perimeter Equation: The perimeter of a rectangle is given by:
$ 2(l + b) = 100 $
Dividing by 2:
$ l + b = 50 $ ----(1)
### Step 2: Use the Area Equation: The area of a rectangle is given by:
$ l \times b = 600 $ ----(2)
### Step 3: Solve for $ l $ and $ b $: From equation (1), express $ l $ in terms of $ b $:
$ l = 50 - b $
Substituting in equation (2):
$ (50 - b) \times b = 600 $
Expanding:
$ 50b - b^2 = 600 $
Rearranging:
$ b^2 - 50b + 600 = 0 $
### Step 4: Solve the Quadratic Equation: Using the quadratic formula where:
$ a = 1, \quad b = -50, \quad c = 600 $
The roots are given by:
$ b = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(600)}}{2(1)} $
$ b = \frac{50 \pm \sqrt{2500 - 2400}}{2} $
$ b = \frac{50 \pm \sqrt{100}}{2} $
$ b = \frac{50 \pm 10}{2} $
Solving for $ b $:
$ b = \frac{50 + 10}{2} = \frac{60}{2} = 30 $
or
$ b = \frac{50 - 10}{2} = \frac{40}{2} = 20 $
Thus, the possible dimensions are:
- If $ b = 30 $, then $ l = 20 $.
- If $ b = 20 $, then $ l = 30 $.
### Final Answer: The length and breadth of the rectangular park are $ 30 $ m and $ 20 $ m (or vice versa).
### Step 1: Use the Perimeter Equation: The perimeter of a rectangle is given by:
$ 2(l + b) = 100 $
Dividing by 2:
$ l + b = 50 $ ----(1)
### Step 2: Use the Area Equation: The area of a rectangle is given by:
$ l \times b = 600 $ ----(2)
### Step 3: Solve for $ l $ and $ b $: From equation (1), express $ l $ in terms of $ b $:
$ l = 50 - b $
Substituting in equation (2):
$ (50 - b) \times b = 600 $
Expanding:
$ 50b - b^2 = 600 $
Rearranging:
$ b^2 - 50b + 600 = 0 $
### Step 4: Solve the Quadratic Equation: Using the quadratic formula where:
$ a = 1, \quad b = -50, \quad c = 600 $
The roots are given by:
$ b = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(600)}}{2(1)} $
$ b = \frac{50 \pm \sqrt{2500 - 2400}}{2} $
$ b = \frac{50 \pm \sqrt{100}}{2} $
$ b = \frac{50 \pm 10}{2} $
Solving for $ b $:
$ b = \frac{50 + 10}{2} = \frac{60}{2} = 30 $
or
$ b = \frac{50 - 10}{2} = \frac{40}{2} = 20 $
Thus, the possible dimensions are:
- If $ b = 30 $, then $ l = 20 $.
- If $ b = 20 $, then $ l = 30 $.
### Final Answer: The length and breadth of the rectangular park are $ 30 $ m and $ 20 $ m (or vice versa).
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