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Perimeter and Area - Perimeter and Area of Squares and Rectangles

Grade 7CBSE

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

🔑Concepts

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Perimeter is the total distance around the edge of a closed 2D shape. For a square, since all four sides are equal, the perimeter is 4×side4 \times \text{side}. For a rectangle, it is 2×(length+breadth)2 \times (\text{length} + \text{breadth}).

A rectangle showing length and breadth dimensions.
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Area measures the surface covered by a 2D shape. It is expressed in square units like cm2cm^{2} or m2m^{2}. The area of a square is side×side\text{side} \times \text{side}, and the area of a rectangle is length×breadth\text{length} \times \text{breadth}.

A square with side 's' showing the area region.
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When a path is built around a rectangular field, the outer dimensions change. If a path of width ww is built outside, the new length becomes L+2wL + 2w and the new breadth becomes B+2wB + 2w.

A rectangle with a path surrounding it.
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Unit Conversion: 1 m2=10,000 cm21\text{ m}^{2} = 10,000\text{ cm}^{2} and 1 hectare=10,000 m21\text{ hectare} = 10,000\text{ m}^{2}. This is crucial when calculating costs for surfacing or fencing.

📐Formulae

Perimeter of a Square = 4×s4 \times s (where ss is the side)

Area of a Square = s×s=s2s \times s = s^{2}

Perimeter of a Rectangle = 2×(l+b)2 \times (l + b) (where ll is length and bb is breadth)

Area of a Rectangle = l×bl \times b

Side of a Square = Perimeter4\frac{Perimeter}{4}

Side of a Square = Area\sqrt{Area}

💡Examples

Problem 1:

A square park has a side length of 15m15 m. Find the total cost of fencing the park at a rate of 2020 per meter.

Solution:

Step 1: Identify the side of the square, s=15ms = 15 m. Step 2: Calculate the perimeter (boundary) for fencing. Perimeter=4×s=4×15=60mPerimeter = 4 \times s = 4 \times 15 = 60 m Step 3: Calculate the total cost. Cost=Perimeter×Rate=60×20=1200Cost = Perimeter \times Rate = 60 \times 20 = 1200 Final Answer: The total cost of fencing is 12001200.

Explanation:

To find the cost of fencing, we first need the total length of the boundary, which is the perimeter. Once the perimeter is found in meters, we multiply it by the cost per meter.

Problem 2:

The area of a rectangular hall is 96m296 m^{2}. If the length of the hall is 12m12 m, find its breadth and its perimeter.

Solution:

Step 1: Use the area formula to find the breadth (bb). Area=l×b  ⟹  96=12×bArea = l \times b \implies 96 = 12 \times b b=9612=8mb = \frac{96}{12} = 8 m Step 2: Use the length (12m12 m) and breadth (8m8 m) to find the perimeter. Perimeter=2×(l+b)=2×(12+8)Perimeter = 2 \times (l + b) = 2 \times (12 + 8) Perimeter=2×20=40mPerimeter = 2 \times 20 = 40 m Final Answer: The breadth is 8m8 m and the perimeter is 40m40 m.

Explanation:

We start by rearranging the area formula to solve for the unknown breadth. Once both dimensions are known, we apply the perimeter formula for a rectangle.

Problem 3:

A wire in the shape of a square with side 10 cm10\text{ cm} is bent into a rectangle of length 12 cm12\text{ cm}. Find its breadth. Which shape encloses more area?

Diagram showing a square and a rectangle made from the same wire.

Solution:

Perimeter of Square = 4×10=40 cm4 \times 10 = 40\text{ cm}. Since the same wire is used for the rectangle, Perimeter of Rectangle = 40 cm40\text{ cm}. 2×(12+b)=402 \times (12 + b) = 40 12+b=2012 + b = 20 b=8 cmb = 8\text{ cm} Area of Square = 10×10=100 cm210 \times 10 = 100\text{ cm}^{2}. Area of Rectangle = 12×8=96 cm212 \times 8 = 96\text{ cm}^{2}. The square encloses more area.

Explanation:

Because the length of the wire remains constant, the perimeters are equal. We then compare the areas using the derived dimensions.

Problem 4:

A rectangular garden is 30 m30\text{ m} long and 25 m25\text{ m} wide. A path 2.5 m2.5\text{ m} wide is constructed outside the garden. Find the area of the path.

Diagram of a garden with an outer path of 2.5m width.

Solution:

Inner length = 30 m30\text{ m}, Inner breadth = 25 m25\text{ m}. Inner Area = 30×25=750 m230 \times 25 = 750\text{ m}^{2}. Outer length = 30+(2×2.5)=35 m30 + (2 \times 2.5) = 35\text{ m}. Outer breadth = 25+(2×2.5)=30 m25 + (2 \times 2.5) = 30\text{ m}. Outer Area = 35×30=1050 m235 \times 30 = 1050\text{ m}^{2}. Area of Path = Outer Area - Inner Area 1050−750300\begin{array}{r} 1050 \\ - 750 \\ \hline 300 \end{array} Area of Path = 300 m2300\text{ m}^{2}.

Explanation:

To find the area of the path, we subtract the area of the inner rectangle from the area of the outer rectangle formed by adding the path's width to all sides.