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Mensuration - Perimeter and Area of Squares and Rectangles

Grade 7ICSE

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

🔑Concepts

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The perimeter of a rectangle is the total distance around its edge, calculated by summing twice the length and twice the breadth: P=2(l+b)P = 2(l + b). Its area is the region enclosed, given by A=l×bA = l \times b.

A rectangle with labels for length and breadth.
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A square is a special rectangle where all sides (ss) are equal. The perimeter is 4s4s and the area is s2s^2. The diagonal (dd) divides it into two right-angled triangles, where d=s2d = s\sqrt{2} and Area=12d2Area = \frac{1}{2}d^2.

A square with equal sides and a diagonal.
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When a path is built around or inside a rectangular region, the area of the path is the difference between the area of the outer rectangle and the area of the inner rectangle.

Concentric rectangles showing a path surrounding an inner area.
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Units of Measurement: Perimeter is measured in linear units like cmcm or mm. Area is measured in square units like cm2cm^2 or m2m^2. For large land areas, 1 hectare=10,000 m21 \text{ hectare} = 10,000 \text{ m}^2.

📐Formulae

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

textAreaofRectangle=ltimesb\\text{Area of Rectangle} = l \\times b

textDiagonalofRectangle=sqrtl2+b2\\text{Diagonal of Rectangle} = \\sqrt{l^2 + b^2}

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

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

textSideofSquare=fractextPerimeter4\\text{Side of Square} = \\frac{\\text{Perimeter}}{4}

textSideofSquare=sqrttextArea\\text{Side of Square} = \\sqrt{\\text{Area}}

textAreaofSquareusingDiagonal=frac12d2\\text{Area of Square using Diagonal} = \\frac{1}{2}d^2

💡Examples

Problem 1:

A rectangular playground is 25textm25 \\text{ m} long and 18textm18 \\text{ m} wide. Find the cost of fencing it at the rate of ₹15₹ 15 per meter.

Solution:

  1. Identify dimensions: Length l=25textml = 25 \\text{ m} and Breadth b=18textmb = 18 \\text{ m}.
  2. Calculate Perimeter (PP): P=2(l+b)=2(25+18)=2(43)=86textmP = 2(l + b) = 2(25 + 18) = 2(43) = 86 \\text{ m}.
  3. Calculate Cost: textTotalCost=textPerimetertimestextRate=86times15\\text{Total Cost} = \\text{Perimeter} \\times \\text{Rate} = 86 \\times 15.
  4. Final Calculation: 86times15=129086 \\times 15 = 1290.

Explanation:

To find the cost of fencing, we must find the total length of the boundary (Perimeter). We then multiply this distance by the cost per meter provided in the problem.

Problem 2:

The area of a square plot is 144textcm2144 \\text{ cm}^2. Find its perimeter and the length of its diagonal.

Solution:

  1. Given: textAreaA=144textcm2\\text{Area } A = 144 \\text{ cm}^2.
  2. Find the Side (ss): s=sqrttextArea=sqrt144=12textcms = \\sqrt{\\text{Area}} = \\sqrt{144} = 12 \\text{ cm}.
  3. Calculate Perimeter (PP): P=4s=4times12=48textcmP = 4s = 4 \\times 12 = 48 \\text{ cm}.
  4. Calculate Diagonal (dd): d=ssqrt2=12sqrt2textcmd = s\\sqrt{2} = 12\\sqrt{2} \\text{ cm}.

Explanation:

Starting with the area, we determine the length of one side by taking the square root. Once the side is known, we can easily find the perimeter by multiplying by 4 and the diagonal by multiplying the side by sqrt2\\sqrt{2}.

Problem 3:

A path of width 2 m2 \text{ m} is built outside and around a rectangular park of length 30 m30 \text{ m} and breadth 20 m20 \text{ m}. Find the area of the path.

Diagram showing a 30x20 park with a 2m wide path around it.

Solution:

  1. Length of inner park l=30 ml = 30 \text{ m}
  2. Breadth of inner park b=20 mb = 20 \text{ m}
  3. Area of inner park = l×b=30×20=600 m2l \times b = 30 \times 20 = 600 \text{ m}^2
  4. The path is 2 m2 \text{ m} wide on all sides. Outer Length L=30+2+2=34 mL = 30 + 2 + 2 = 34 \text{ m} Outer Breadth B=20+2+2=24 mB = 20 + 2 + 2 = 24 \text{ m}
  5. Area of outer rectangle = L×B=34×24=816 m2L \times B = 34 \times 24 = 816 \text{ m}^2
  6. Area of path = Outer Area - Inner Area 816−600216\begin{array}{r} 816 \\ - 600 \\ \hline 216 \end{array} Area of path = 216 m2216 \text{ m}^2

Explanation:

To find the area of the path, we calculate the area of the larger rectangle (including the path) and subtract the area of the smaller rectangle (the park). Adding the width twice to each dimension accounts for the path on both ends.

Problem 4:

Calculate the area of a square whose diagonal is 10 cm10 \text{ cm}.

Square with a diagonal labeled as 10 cm.

Solution:

  1. Given Diagonal d=10 cmd = 10 \text{ cm}
  2. Formula for Area of a square using diagonal is Area=12d2Area = \frac{1}{2} d^2
  3. Area=12×(10)2Area = \frac{1}{2} \times (10)^2
  4. Area=12×100=50 cm2Area = \frac{1}{2} \times 100 = 50 \text{ cm}^2

Explanation:

When the diagonal of a square is known, we can find the area directly without finding the side length first by using the formula derived from the Pythagorean theorem.