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Mensuration - Area of Rings and Paths

Grade 7ICSE

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

🔑Concepts

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A circular ring, also known as an annulus, is the region bounded by two concentric circles. Visually, this resembles a 'donut' shape where one smaller circle is placed perfectly in the center of a larger circle.

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Concentric circles are defined as two or more circles that share the same center point but have different radii. In a diagram, the distance between the boundaries of these two circles represents the width of the path or ring.

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The area of a path or ring is always calculated by finding the difference between the outer area and the inner area. Imagine taking a large circular or rectangular sheet and cutting out a smaller similar shape from the middle; the remaining border is the path.

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For rectangular paths built outside a given area, the outer dimensions are calculated by adding twice the width (ww) of the path to the original length (LL) and breadth (BB). Visually, the path extends outwards from all four sides, so Outer Length = L+2wL + 2w and Outer Breadth = B+2wB + 2w.

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For rectangular paths built inside a given area, the inner dimensions are calculated by subtracting twice the width (ww) of the path from the original length (LL) and breadth (BB). In this case, the path is 'stolen' from the existing field area, making the inner garden or field smaller: Inner Length = L−2wL - 2w and Inner Breadth = B−2wB - 2w.

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Cross paths are two rectangular paths that intersect each other at right angles within a larger rectangle, often forming a '+' shape. To find the total area of cross paths, you calculate the area of both paths and then subtract the area of the common overlapping square or rectangle in the middle to avoid counting it twice.

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The width of a circular ring is the difference between the outer radius (RR) and the inner radius (rr). In a visual representation, this is the straight-line distance from the edge of the inner circle to the edge of the outer circle along the radius.

📐Formulae

Area of a Circle = πr2\pi r^2

Area of a Ring = πR2−πr2=π(R2−r2)\pi R^2 - \pi r^2 = \pi(R^2 - r^2) (where RR is the outer radius and rr is the inner radius)

Width of a Ring = R−rR - r

Area of a Rectangular Path = (L×B)−(l×b)(L \times B) - (l \times b) (where L,BL, B are outer dimensions and l,bl, b are inner dimensions)

Area of Cross Paths = (L×w)+(B×w)−w2(L \times w) + (B \times w) - w^2 (where ww is the uniform width of the paths crossing a rectangle of length LL and breadth BB)

💡Examples

Problem 1:

A circular park has a radius of 2121 m. A uniform path of width 77 m is constructed outside the park. Find the area of the path. (Take π=227\pi = \frac{22}{7})

Solution:

  1. Inner radius of the park (rr) = 2121 m.
  2. Width of the path = 77 m.
  3. Outer radius (RR) = r+width=21+7=28r + \text{width} = 21 + 7 = 28 m.
  4. Area of the path = Area of outer circle - Area of inner circle
  5. Area = πR2−πr2=π(R2−r2)\pi R^2 - \pi r^2 = \pi(R^2 - r^2)
  6. Area = 227×(282−212)\frac{22}{7} \times (28^2 - 21^2)
  7. Using a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b), Area = 227×(28+21)×(28−21)\frac{22}{7} \times (28 + 21) \times (28 - 21)
  8. Area = 227×49×7=22×49=1078\frac{22}{7} \times 49 \times 7 = 22 \times 49 = 1078 m2m^2.

Explanation:

To find the area of the path built outside, we first determine the outer radius by adding the path width to the inner radius. Then, we use the formula for the area of a ring by subtracting the smaller circle's area from the larger one.

Problem 2:

A rectangular lawn measures 5050 m by 3030 m. A path 2.52.5 m wide is constructed all around it on the inside. Find the area of the path.

Solution:

  1. Outer length (LL) = 5050 m, Outer breadth (BB) = 3030 m.
  2. Width of the path (ww) = 2.52.5 m.
  3. Inner length (ll) = L−2w=50−2(2.5)=50−5=45L - 2w = 50 - 2(2.5) = 50 - 5 = 45 m.
  4. Inner breadth (bb) = B−2w=30−2(2.5)=30−5=25B - 2w = 30 - 2(2.5) = 30 - 5 = 25 m.
  5. Area of outer lawn = 50×30=150050 \times 30 = 1500 m2m^2.
  6. Area of inner rectangular portion = 45×25=112545 \times 25 = 1125 m2m^2.
  7. Area of the path = Outer Area - Inner Area
  8. Area of path = 1500−1125=3751500 - 1125 = 375 m2m^2.

Explanation:

Since the path is inside the lawn, we subtract twice the width from both the length and breadth to find the dimensions of the inner rectangle. The area of the path is the difference between the total area of the lawn and the area of the remaining inner portion.