4q
Square and Rectangle Perimeter Area Problem Explained
This solution explains how to find the square of the difference between the length and breadth of a rectangle, given its perimeter and area relationship with a square.
Define Variables and Perimeter Relationship
- Let the side length of the square be \(s\).
- Let the length and breadth of the rectangle be \(l\) and \(b\) respectively.
- The perimeter is given as \(p\).
- Perimeter of the square: \(P_s = 4s\).
- Perimeter of the rectangle: \(P_r = 2(l+b)\).
- Given \(P_s = P_r = p\).
- Therefore, \(4s = p \implies s = \frac{p}{4}\).
- And \(2(l+b) = p \implies l+b = \frac{p}{2}\).
Area Calculations and Difference
- Area of the square: \(A_s = s^2 = (\frac{p}{4})^2 = \frac{p^2}{16}\).
- Area of the rectangle: \(A_r = l \times b\).
- The areas differ by \(q\). Since the square maximizes area for a given perimeter, \(A_s \ge A_r\).
- So, \(A_s - A_r = q\).
- Substituting the area of the square: \(\frac{p^2}{16} - A_r = q\).
- This gives the area of the rectangle: \(A_r = lb = \frac{p^2}{16} - q\).
Finding the Square of the Length-Breadth Difference
- We need to find the value of \((l-b)^2\).
- Use the algebraic identity: \((l-b)^2 = (l+b)^2 - 4lb\).
- Substitute the known values for \((l+b)\) and \(lb\): \((l-b)^2 = (\frac{p}{2})^2 - 4(\frac{p^2}{16} - q)\)
- Simplify the expression: \((l-b)^2 = \frac{p^2}{4} - \frac{4p^2}{16} + 4q\) \((l-b)^2 = \frac{p^2}{4} - \frac{p^2}{4} + 4q\) \((l-b)^2 = 4q\)
The square of the difference between the length and breadth of the rectangle is \(4q\).
Similar Questions
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A Question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.
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Important Questions from 2-D Mensuration
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The shorter side of a rectangle is 15 cm less than the longer side. The numerical value of its area is equal to 5 times the numerical value of its perimeter. What is the length (in cm) of its longer side?
- In a circle of radius 10.5 cm, if the angle of a sector is $\frac{2\pi}{3}$, then the perimeter of the sector is (in cm):
(Take $\pi = \frac{22}{7}$) - Find the circumference (in m) of the largest circle that can be inscribed in a rectangle whose dimensions are given as 114 m and 63 m.
Take $\pi = \frac{22}{7}$
