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Measuring Space: Perimeter and Area - Area of a Rectangle

Grade 9CBSE

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

🔑Concepts

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The area of a rectangle is the measure of the total surface or space enclosed within its four boundaries. It is calculated by finding the product of its length and its breadth.

A rectangle showing length (l) and breadth (b) dimensions.
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The units of area are always expressed in square units, such as cm2cm^2, m2m^2, or km2km^2. To ensure a correct calculation, the length and breadth must be in the same unit of measurement before multiplication.

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The diagonal of a rectangle divides it into two congruent right-angled triangles. If the area is unknown but the diagonal and one side are given, Pythagoras theorem (l2+b2=d2l^2 + b^2 = d^2) can be used to find the missing dimension.

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When calculating the area of a path around a rectangular field, the total area is found by subtracting the area of the inner rectangle from the area of the outer rectangle (Inner area = l×bl \times b; Outer area = (l+2w)×(b+2w)(l+2w) \times (b+2w) where ww is the path width).

Diagram showing an outer rectangle containing an inner rectangle with a path between them.

📐Formulae

Area=l×bArea = l \times b

Perimeter=2(l+b)Perimeter = 2(l + b)

Diagonal(d)=l2+b2Diagonal (d) = \sqrt{l^2 + b^2}

Length(l)=AreabLength (l) = \frac{Area}{b}

Breadth(b)=ArealBreadth (b) = \frac{Area}{l}

💡Examples

Problem 1:

Calculate the area of a rectangular park whose length is 25 m25 \text{ m} and breadth is 18 m18 \text{ m}.

Solution:

Area=l×bArea = l \times b Area=25×18Area = 25 \times 18 Area=450 m2Area = 450 \text{ m}^2

Explanation:

Substitute the given length (l=25l = 25) and breadth (b=18b = 18) into the standard area formula for a rectangle.

Problem 2:

The diagonal of a rectangle is 13 cm13 \text{ cm} and its breadth is 5 cm5 \text{ cm}. Find its area.

Solution:

First, find the length (ll) using the Pythagoras theorem: l=d2−b2l = \sqrt{d^2 - b^2} l=132−52l = \sqrt{13^2 - 5^2} l=169−25l = \sqrt{169 - 25} l=144=12 cml = \sqrt{144} = 12 \text{ cm} Now, calculate the area: Area=l×bArea = l \times b Area=12×5=60 cm2Area = 12 \times 5 = 60 \text{ cm}^2

Explanation:

When the diagonal and one side are known, use the relationship d2=l2+b2d^2 = l^2 + b^2 to find the missing dimension before calculating the area.

Problem 3:

A rectangular room is 10 m10 \text{ m} long and 8 m8 \text{ m} wide. A carpet of size 7 m×5 m7 \text{ m} \times 5 \text{ m} is laid on the floor. Find the area of the floor left uncarpeted.

Solution:

Area of the floor: Af=10×8=80 m2A_f = 10 \times 8 = 80 \text{ m}^2 Area of the carpet: Ac=7×5=35 m2A_c = 7 \times 5 = 35 \text{ m}^2 Uncarpeted Area: Af−AcA_f - A_c 80−3545\begin{array}{r} 80 \\ - 35 \\ \hline 45 \end{array} The uncarpeted area is 45 m245 \text{ m}^2.

Explanation:

Calculate the total area of the floor and subtract the area occupied by the carpet to find the remaining space.

Problem 4:

Find the cost of tiling a rectangular courtyard 30 m30 \text{ m} long and 20 m20 \text{ m} wide at the rate of Rs 1515 per m2m^2.

Solution:

Step 1: Find the area. Area=30×20=600 m2Area = 30 \times 20 = 600 \text{ m}^2 Step 2: Calculate total cost. Total Cost=Area×RateTotal \text{ Cost} = Area \times Rate Total Cost = Rs 600×15600 \times 15 Total Cost = Rs 90009000

Explanation:

Multiply the calculated area of the rectangle by the cost per unit area to find the total expenditure.

Problem 5:

A rectangular plot has a length of 40 m40 \text{ m} and its area is 1200 m21200 \text{ m}^2. If a path of width 2 m2 \text{ m} is built inside the plot along its boundary, find the area of the path.

A rectangular plot of 40m by 30m with a 2m internal path.

Solution:

  1. First, find the breadth (bb) of the plot: b=Areal=120040=30 mb = \frac{Area}{l} = \frac{1200}{40} = 30 \text{ m}
  2. The path is inside, so the inner rectangle dimensions are: Inner length (lil_i) = 40−(2×2)=36 m40 - (2 \times 2) = 36 \text{ m} Inner breadth (bib_i) = 30−(2×2)=26 m30 - (2 \times 2) = 26 \text{ m}
  3. Area of inner rectangle: Areainner=36×26=936 m2Area_{inner} = 36 \times 26 = 936 \text{ m}^2
  4. Area of the path: Areapath=Areaouter−AreainnerArea_{path} = Area_{outer} - Area_{inner} Areapath=1200−936=264 m2Area_{path} = 1200 - 936 = 264 \text{ m}^2

Explanation:

To find the path's area, we subtract the area of the empty inner space from the total area. The inner dimensions are reduced by twice the path width because the path exists on both sides of the length and breadth.

Problem 6:

Calculate the area of a rectangle where the sum of the length and breadth is 25 cm25 \text{ cm} and the difference between them is 5 cm5 \text{ cm}.

Rectangle with calculated dimensions of 15 cm and 10 cm.

Solution:

  1. Let length be ll and breadth be bb. We are given: l+b=25l + b = 25 l−b=5l - b = 5
  2. Adding the two equations: (l+b)+(l−b)=25+5(l + b) + (l - b) = 25 + 5 2l=30  ⟹  l=15 cm2l = 30 \implies l = 15 \text{ cm}
  3. Substitute ll in the first equation: 15+b=25  ⟹  b=10 cm15 + b = 25 \implies b = 10 \text{ cm}
  4. Calculate the Area: Area=l×bArea = l \times b Area=15×10=150 cm2Area = 15 \times 10 = 150 \text{ cm}^2

Explanation:

We use simultaneous equations to solve for the individual dimensions first. Once the length and breadth are found, we multiply them to find the total area.