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Mensuration: Area and Perimeter - Calculate perimeters and circumference in real-life measurement contexts

Grade 9CBSE

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

🔑Concepts

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The perimeter of a rectilinear figure is the total length of its boundary. For a rectangle with length ll and breadth bb, the perimeter is 2(l+b)2(l+b). In real-world applications, this represents the total length of fencing or edging required for a rectangular plot.

A rectangular field showing length and breadth.
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The circumference of a circle is the distance around it. For a circle with radius rr, the circumference is 2πr2\pi r. This is used to calculate the distance traveled by a wheel in one rotation or the length of a circular track.

A circle showing radius and the boundary representing circumference.
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For a semicircle, the total perimeter includes the curved boundary (arc) and the straight boundary (diameter). The formula is P=πr+2rP = \pi r + 2r. This is crucial for calculating materials for semi-circular windows or park sections.

A semicircle showing the arc and the diameter base.
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When a path is built around a circular region, two concentric circles are formed. The circumference of the outer circle and inner circle can be used to find the boundary lengths for curbing or railing.

Concentric circles representing a path around a circular garden.

📐Formulae

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

P=4×s (Perimeter of a Square)P = 4 \times s \text{ (Perimeter of a Square)}

C=2πr (Circumference of a Circle)C = 2\pi r \text{ (Circumference of a Circle)}

C=πd (Circumference in terms of Diameter)C = \pi d \text{ (Circumference in terms of Diameter)}

P=πr+2r (Perimeter of a Semicircle including diameter)P = \pi r + 2r \text{ (Perimeter of a Semicircle including diameter)}

💡Examples

Problem 1:

A farmer wants to fence a rectangular field of length 120 m120\text{ m} and breadth 80 m80\text{ m} with three rounds of wire. Calculate the total length of wire required.

Solution:

First, calculate the perimeter of the field using P=2(l+b)P = 2(l + b). P=2(120+80)=2(200)=400 mP = 2(120 + 80) = 2(200) = 400\text{ m} Since the farmer needs 3 rounds of wire, total length L=3×PL = 3 \times P: L=3×400=1200 mL = 3 \times 400 = 1200\text{ m} The total wire required is 1200 m1200\text{ m}.

Explanation:

The perimeter gives the length for one round. Multiplication by the number of rounds provides the total length.

Problem 2:

Find the distance covered by a wheel of radius 35 cm35\text{ cm} in 2020 complete rotations. (Use π=227\pi = \frac{22}{7})

Solution:

The distance covered in one rotation is equal to the circumference C=2πrC = 2\pi r. C=2×227×35C = 2 \times \frac{22}{7} \times 35 C=2×22×5=220 cmC = 2 \times 22 \times 5 = 220\text{ cm} Total distance for 20 rotations: D=220×20=4400 cmD = 220 \times 20 = 4400\text{ cm} Converting to meters: 4400/100=44 m4400 / 100 = 44\text{ m}.

Explanation:

A wheel covers a distance equal to its circumference in one full turn.

Problem 3:

A wire in the shape of a square of side 22 cm22\text{ cm} is bent into the form of a circle. Find the radius of the circle.

Solution:

The length of the wire remains constant. Thus, Perimeter of square = Circumference of circle. Psquare=4×22=88 cmP_{\text{square}} = 4 \times 22 = 88\text{ cm} Let the radius of the circle be rr. Then: 2πr=882\pi r = 88 2×227×r=882 \times \frac{22}{7} \times r = 88 447×r=88\frac{44}{7} \times r = 88 r=88×744=2×7=14 cmr = 88 \times \frac{7}{44} = 2 \times 7 = 14\text{ cm}

Explanation:

When a shape is reshaped (bent), its perimeter/length remains the same.

Problem 4:

A gardener has 500 m500\text{ m} of fencing material. He fences a circular garden and has some material left. If the garden circumference is 365 m365\text{ m}, how much material is left?

Solution:

We subtract the used material from the total available: 500−365135\begin{array}{r} 500 \\ - 365 \\ \hline 135 \end{array} The gardener has 135 m135\text{ m} of material left.

Explanation:

Simple subtraction in a vertical format is used to find the remaining length after application.

Problem 5:

A running track consists of two parallel straight sides each of length 100 m100\text{ m} and two semicircular ends with a diameter of 70 m70\text{ m}. Find the total distance covered by an athlete in one complete lap around the inner edge of the track. (Use π=227\pi = \frac{22}{7})

Diagram of a running track with straight sides and semicircular ends.

Solution:

  1. Identify components: The track has two straight lengths and two semicircular ends.
  2. Length of two straight sides = 100 m+100 m=200 m100\text{ m} + 100\text{ m} = 200\text{ m}.
  3. The two semicircular ends together form one full circle with diameter d=70 md = 70\text{ m}.
  4. Circumference of the two ends = πd=227×70=220 m\pi d = \frac{22}{7} \times 70 = 220\text{ m}.
  5. Total distance = 200 m+220 m=420 m200\text{ m} + 220\text{ m} = 420\text{ m}.

Explanation:

To find the total perimeter of a composite shape, we sum the lengths of all outer boundaries. Here, the two semicircles combine to form a full circle circumference, which is added to the straight edges.

Problem 6:

A rectangular park measures 50 m50\text{ m} by 30 m30\text{ m}. A path of uniform width 2 m2\text{ m} is built inside the park along its boundary. Find the total length of the outer fence and the inner boundary of the path.

A rectangle within a rectangle representing a park with an internal path.

Solution:

  1. Outer Perimeter: Pouter=2(l+b)=2(50+30)=2(80)=160 mP_{outer} = 2(l + b) = 2(50 + 30) = 2(80) = 160\text{ m}.
  2. Inner Dimensions: The width is reduced by 2 m2\text{ m} from both sides. Inner length = 50−(2+2)=46 m50 - (2 + 2) = 46\text{ m}. Inner breadth = 30−(2+2)=26 m30 - (2 + 2) = 26\text{ m}.
  3. Inner Perimeter: Pinner=2(46+26)=2(72)=144 mP_{inner} = 2(46 + 26) = 2(72) = 144\text{ m}.

Explanation:

When a path is inside a rectangle, the inner dimensions are found by subtracting twice the path width from the original length and breadth.