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Measuring Space: Perimeter and Area - Squaring a Rectangle

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The concept of 'Squaring a Rectangle' refers to finding a square that has the exact same area as a given rectangle. If a rectangle has length ll and breadth bb, its area is A=l×bA = l \times b. To find the side ss of the equivalent square, we solve the equation s2=l×bs^2 = l \times b, which gives s=l×bs = \sqrt{l \times b}.

A rectangle and a square showing equal area concept.
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While the area remains constant during this conversion, the perimeter changes. A square is the most efficient rectangular shape, meaning for a fixed area, the square will always have a smaller perimeter than any non-square rectangle.

Comparison of perimeters between a rectangle and a square of equal area.
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The geometric mean: The side of the square s=l×bs = \sqrt{l \times b} is mathematically known as the geometric mean of the rectangle's dimensions ll and bb.

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Applications: This principle is used in architecture and land surveying to standardize plot sizes or to compare the material efficiency of different boundary shapes.

📐Formulae

Perimeter of Rectangle=2(l+b)\text{Perimeter of Rectangle} = 2(l + b) pieces

Area of Rectangle=l×b\text{Area of Rectangle} = l \times b

Perimeter of Square=4s\text{Perimeter of Square} = 4s

Area of Square=s2\text{Area of Square} = s^2

Side of a square with area equal to a rectangle(s)=l×b\text{Side of a square with area equal to a rectangle} (s) = \sqrt{l \times b}

💡Examples

Problem 1:

A rectangle has a length of 16 cm16\text{ cm} and a breadth of 9 cm9\text{ cm}. Calculate its area and find the side of a square that has the same area as this rectangle.

Solution:

Given: l=16 cml = 16\text{ cm}, b=9 cmb = 9\text{ cm}. Area of Rectangle=l×b=16×9=144 cm2\text{Area of Rectangle} = l \times b = 16 \times 9 = 144\text{ cm}^2 To find the side ss of the square with the same area: s2=144s^2 = 144 s=144s = \sqrt{144} s=12 cms = 12\text{ cm}

Explanation:

First, the area of the rectangle is found by multiplying length and breadth. Since the square must have the same area, we set the square's area formula s2s^2 equal to the rectangle's area and solve for ss by taking the square root.

Problem 2:

Calculate the difference in perimeter between a rectangle of dimensions 25 m25\text{ m} by 4 m4\text{ m} and a square having the same area.

Solution:

Area of Rectangle=25×4=100 m2\text{Area of Rectangle} = 25 \times 4 = 100\text{ m}^2 Perimeter of Rectangle=2(25+4)=2(29)=58 m\text{Perimeter of Rectangle} = 2(25 + 4) = 2(29) = 58\text{ m} For the square: Area of Square=100 m2\text{Area of Square} = 100\text{ m}^2 s=100=10 ms = \sqrt{100} = 10\text{ m} Perimeter of Square=4×10=40 m\text{Perimeter of Square} = 4 \times 10 = 40\text{ m} Difference in perimeter: 58−4018\begin{array}{r} 58 \\ - 40 \\ \hline 18 \end{array} Difference = 18 m18\text{ m}.

Explanation:

We calculate the area of the rectangle to find the side of the equivalent square. Then, we calculate the perimeters of both shapes and find the difference by subtraction.

Problem 3:

A rectangular plot of land measures 36 m36\text{ m} by 4 m4\text{ m}. A builder wants to construct a square-shaped shed with the same area. Find the side of the shed and determine how much fencing is saved by choosing the square shape instead of the rectangular shape.

Comparison of a 36x4 rectangle and a 12x12 square.

Solution:

  1. Find Area of rectangle: Area=l×b=36×4=144 m2Area = l \times b = 36 \times 4 = 144\text{ m}^2
  2. Find side of the square: s=Area=144=12 ms = \sqrt{Area} = \sqrt{144} = 12\text{ m}
  3. Calculate Perimeter of rectangle: Prect=2(l+b)=2(36+4)=2(40)=80 mP_{rect} = 2(l + b) = 2(36 + 4) = 2(40) = 80\text{ m}
  4. Calculate Perimeter of square: Psq=4s=4×12=48 mP_{sq} = 4s = 4 \times 12 = 48\text{ m}
  5. Calculate fencing saved: Savings=80−48=32 m\text{Savings} = 80 - 48 = 32\text{ m}

Explanation:

First, the area is calculated using the rectangular dimensions. Since the square must have the same area, we take the square root to find its side. Comparing the perimeters shows the efficiency of the square.

Problem 4:

A square courtyard has a side of 10 m10\text{ m}. If this is converted into a rectangle with a breadth of 5 m5\text{ m} while keeping the area constant, find the new length and the increase in the boundary length (perimeter).

Comparison of a 10x10 square and a 20x5 rectangle.

Solution:

  1. Find Area of the square: Area=s2=102=100 m2Area = s^2 = 10^2 = 100\text{ m}^2
  2. Find length of the rectangle: l=Areab=1005=20 ml = \frac{Area}{b} = \frac{100}{5} = 20\text{ m}
  3. Calculate Perimeter of square: Psq=4s=4×10=40 mP_{sq} = 4s = 4 \times 10 = 40\text{ m}
  4. Calculate Perimeter of rectangle: Prect=2(l+b)=2(20+5)=2(25)=50 mP_{rect} = 2(l + b) = 2(20 + 5) = 2(25) = 50\text{ m}
  5. Increase in boundary length: Increase=50−40=10 m\text{Increase} = 50 - 40 = 10\text{ m}

Explanation:

We start with the square's area and divide by the new breadth to find the length of the equivalent rectangle. The perimeter increases because the shape becomes less compact.