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Measuring Space: Perimeter and Area - Problems, Puzzles, and Paradoxes on Perimeter

Grade 9CBSE

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

🔑Concepts

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Perimeter is the total length of the boundary of a closed two-dimensional figure. For a polygon, it is the sum of the lengths of all its sides.

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The Isoperimetric Theorem states that for a fixed perimeter, the circle encloses the maximum possible area. Conversely, for a fixed area, the circle has the minimum perimeter.

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Relationship Paradox: Shapes with the same area can have vastly different perimeters. For instance, a very thin, long rectangle has a much larger perimeter than a square of the same area.

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The Boundary Paradox: It is possible for a shape to have a finite area but an infinitely long perimeter. A classic example studied in higher mathematics is the Koch Snowflake.

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Grid Puzzles: On a square grid, the perimeter of a shape depends on how the squares are connected. Removing a square from the middle of a shape increases the perimeter (inner boundary), while removing a square from a corner might leave the perimeter unchanged.

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The Staircase Paradox: A 'staircase' path consisting of many small horizontal and vertical steps tracking a diagonal line will always have a perimeter equal to the sum of the sides of the bounding rectangle (l+bl + b), regardless of how small the steps are, even though it visually approximates the diagonal.

📐Formulae

P=2(l+b)P = 2(l + b) (Perimeter of a Rectangle)

P=4sP = 4s (Perimeter of a Square)

C=2πrC = 2\pi r (Circumference of a Circle)

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

L=θ360∘×2πrL = \frac{\theta}{360^{\circ}} \times 2\pi r (Length of an Arc)

P=n×sP = n \times s (Perimeter of a Regular Polygon with nn sides of length ss)

💡Examples

Problem 1:

A square of side 10 cm10\text{ cm} has a small square of side 2 cm2\text{ cm} cut out from one of its corners. What is the perimeter of the new shape?

Solution:

Let the original square be ABCDABCD with side S=10 cmS = 10\text{ cm}. Original Perimeter = 4×10=40 cm4 \times 10 = 40\text{ cm}. When a corner square of side x=2 cmx = 2\text{ cm} is removed, the two outer edges of the corner are replaced by two inner edges of the same length (xx). New Perimeter = (10−2)+2+2+(10−2)+10+10(10 - 2) + 2 + 2 + (10 - 2) + 10 + 10 822810+1040\begin{array}{r} 8 \\ 2 \\ 2 \\ 8 \\ 10 \\ + 10 \\ \hline 40 \end{array} New Perimeter = 40 cm40\text{ cm}.

Explanation:

This is a perimeter puzzle showing that removing area from a corner does not necessarily change the perimeter because the boundary length is simply 'pushed' inward.

Problem 2:

Compare the areas of a square and a rectangle, both having a perimeter of 24 cm24\text{ cm}. The rectangle has a breadth of 4 cm4\text{ cm}.

Solution:

  1. For the square: 4s=24  ⟹  s=6 cm4s = 24 \implies s = 6\text{ cm}. Area = s2=62=36 cm2s^2 = 6^2 = 36\text{ cm}^2.
  2. For the rectangle: 2(l+4)=24  ⟹  l+4=12  ⟹  l=8 cm2(l + 4) = 24 \implies l + 4 = 12 \implies l = 8\text{ cm}. Area = l×b=8×4=32 cm2l \times b = 8 \times 4 = 32\text{ cm}^2.

Explanation:

This demonstrates that for a fixed perimeter, the square (being more 'regular') encloses a larger area than a non-square rectangle.

Problem 3:

Calculate the perimeter of a sector of a circle with radius 7 cm7\text{ cm} and a central angle of 90∘90^{\circ}. (Use π=227\pi = \frac{22}{7})

Solution:

Perimeter of sector = Arc Length+2×radiusArc\ Length + 2 \times radius L=90360×2×227×7=14×44=11 cmL = \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{1}{4} \times 44 = 11\text{ cm} Total Perimeter = 11+2(7)=11+14=25 cm11 + 2(7) = 11 + 14 = 25\text{ cm}.

Explanation:

The perimeter of a sector must include the curved arc length as well as the two straight radial boundaries.