Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Yes. Length = 20 m, Breadth = 20 m
Step 1: Define the variables.
Let the length of the rectangular park be \(L\) metres and the breadth be \(B\) metres.
Step 2: Use the perimeter condition.
Perimeter of a rectangle is given by \(2(L + B)\).
We are told the perimeter is 80 m, so:
\[2(L + B) = 80\]
Divide both sides by 2:
\[L + B = 40\]
Express \(L\) in terms of \(B\):
\[L = 40 - B\]
Step 3: Use the area condition.
Area of a rectangle is \(L \times B\).
We are told the area is 400 m², so:
\[LB = 400\]
Substitute \(L = 40 - B\):
\[(40 - B)B = 400\]
Step 4: Form the quadratic equation.
Expand the left-hand side:
\[40B - B^2 = 400\]
Rearrange to bring all terms to one side:
\[-B^2 + 40B - 400 = 0\]
Multiply the whole equation by \(-1\) to make the coefficient of \(B^2\) positive:
\[B^2 - 40B + 400 = 0\]
Step 5: Use discriminant to check possibility.
For \(ax^2 + bx + c = 0\), discriminant is \(D = b^2 - 4ac\).
Here, \(a = 1\), \(b = -40\), \(c = 400\).
Compute:
\[D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0\]
Since \(D = 0\), there are equal real roots, so a real solution exists and the design is possible.
Step 6: Find \(B\).
When \(D = 0\), the root is:
\[B = \dfrac{-b}{2a} = \dfrac{-(-40)}{2 \cdot 1} = \dfrac{40}{2} = 20\]
So, \(B = 20\) m.
Step 7: Find \(L\).
Recall \(L = 40 - B\):
\[L = 40 - 20 = 20\,\text{m}\]
Step 8: Verify the conditions.
Perimeter: \(2(L + B) = 2(20 + 20) = 2 \times 40 = 80\,\text{m}\) ✓
Area: \(L \times B = 20 \times 20 = 400\,\text{m}^2\) ✓
Conclusion: Yes, such a park is possible. It must be a square with length 20 m and breadth 20 m.