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Area - Rectangle and Squares

Grade 8CBSE

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

🔑Concepts

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The area of a rectangle is the region enclosed by its four sides. It is calculated by multiplying the length (ll) by the breadth (bb). For a square, since length equals breadth (ss), the area is s×s=s2s \times s = s^2.

Diagram showing a rectangle with labeled length and breadth.
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The diagonal of a rectangle or square can be calculated using the Pythagorean theorem. For a rectangle, the diagonal d=l2+b2d = \sqrt{l^2 + b^2}. For a square, the diagonal d=s2d = s\sqrt{2}.

Rectangle with a diagonal line showing the relationship between length, breadth, and diagonal.
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Area of a Path: To find the area of a path around a rectangular field, calculate the area of the outer rectangle and subtract the area of the inner rectangle.

Concentric rectangles representing a field with a surrounding path.
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Converting units of area is crucial: 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.

📐Formulae

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

Perimeter of a Rectangle=2(l+b)Perimeter \text{ of a Rectangle} = 2(l + b), where ll is length and bb is breadth.

Area of a Square=s2=s×sArea \text{ of a Square} = s^2 = s \times s

Perimeter of a Square=4×sPerimeter \text{ of a Square} = 4 \times s, where ss is the side length.

Diagonal of a Rectangle=l2+b2Diagonal \text{ of a Rectangle} = \sqrt{l^2 + b^2}

Diagonal of a Square=s2Diagonal \text{ of a Square} = s\sqrt{2}

💡Examples

Problem 1:

Find the area of a square plot of land whose perimeter is 120 m120 \text{ m}.

Solution:

Step 1: Find the side of the square. Perimeter=4×s=120 mPerimeter = 4 \times s = 120 \text{ m} s=1204=30 ms = \frac{120}{4} = 30 \text{ m}

Step 2: Calculate the area. Area=s2=30×30=900 m2Area = s^2 = 30 \times 30 = 900 \text{ m}^2.

Explanation:

To find the area, we first determine the side length from the given perimeter using the formula P=4sP = 4s, then apply the area formula A=s2A = s^2.

Problem 2:

A rectangular room measures 15 m15 \text{ m} by 12 m12 \text{ m}. A carpet is laid on the floor leaving a margin of 1 m1 \text{ m} all around. Find the area of the carpet.

Solution:

Length of the room (LL) = 15 m15 \text{ m} Breadth of the room (BB) = 12 m12 \text{ m} Margin width = 1 m1 \text{ m}

Dimensions of the carpet: Length (ll) = 15−(1+1)=13 m15 - (1 + 1) = 13 \text{ m} Breadth (bb) = 12−(1+1)=10 m12 - (1 + 1) = 10 \text{ m}

Area of carpet=l×b=13×10=130 m2Area \text{ of carpet} = l \times b = 13 \times 10 = 130 \text{ m}^2.

Explanation:

Since a margin is left on all sides, we subtract twice the margin width from both the length and the breadth to find the dimensions of the carpeted area.

Problem 3:

A rectangular park is 40 m40 \text{ m} long and 30 m30 \text{ m} wide. A path 5 m5 \text{ m} wide is constructed outside the park. Find the area of the path.

Solution:

Area of the park (Inner Rectangle) = 40×30=1200 m240 \times 30 = 1200 \text{ m}^2.

Outer dimensions (including the path): Outer Length = 40+(5×2)=50 m40 + (5 \times 2) = 50 \text{ m} Outer Breadth = 30+(5×2)=40 m30 + (5 \times 2) = 40 \text{ m} Outer Area = 50×40=2000 m250 \times 40 = 2000 \text{ m}^2.

Area of Path = Outer Area - Inner Area: 2000−1200800\begin{array}{r} 2000 \\ - 1200 \\ \hline 800 \end{array} Area of path = 800 m2800 \text{ m}^2.

Explanation:

When the path is outside, we add twice the width of the path to the original dimensions to get the outer dimensions. The path area is the difference between the outer and inner rectangular areas.

Problem 4:

A square courtyard has a side of 25 m25 \text{ m}. A path 2.5 m2.5 \text{ m} wide is built inside it along the boundary. Find the area of the path.

A square within a square representing a 2.5m path inside a 25m courtyard.

Solution:

  1. Side of outer square (SS) = 25 m25 \text{ m}
  2. Area of outer square = S×S=25×25=625 m2S \times S = 25 \times 25 = 625 \text{ m}^2
  3. Width of path = 2.5 m2.5 \text{ m}. The inner square side (ss) = 25−(2×2.5)=25−5=20 m25 - (2 \times 2.5) = 25 - 5 = 20 \text{ m}
  4. Area of inner square = s×s=20×20=400 m2s \times s = 20 \times 20 = 400 \text{ m}^2
  5. Area of path = Outer Area - Inner Area = 625−400=225 m2625 - 400 = 225 \text{ m}^2

Explanation:

Since the path is inside the boundary, we subtract the width twice from the total length to find the inner dimension.

Problem 5:

Find the length of a rectangular plot whose area is 450 m2450 \text{ m}^2 and breadth is 15 m15 \text{ m}. Also, find the cost of fencing it at the rate of Rs 2020 per meter.

Rectangle labeled with area 450 sq.m and breadth 15m.

Solution:

  1. Area = l×bl \times b 450=l×15450 = l \times 15 l=45015=30 ml = \frac{450}{15} = 30 \text{ m}
  2. Perimeter = 2(l+b)2(l + b) Perimeter=2(30+15)=2(45)=90 mPerimeter = 2(30 + 15) = 2(45) = 90 \text{ m}
  3. Cost of fencing = Perimeter×RatePerimeter \times Rate Cost=90×20=1800Cost = 90 \times 20 = 1800 Total Cost = Rs 1800

Explanation:

We first use the area formula to find the missing length, then find the perimeter to calculate the total length of fencing required.