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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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The area of a circular ring (annulus) is the difference between the area of the larger outer circle and the smaller inner circle. If RR is the outer radius and rr is the inner radius, the area is π(R2−r2)\pi(R^2 - r^2).

Concentric circles showing outer radius R and inner radius r.
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A rectangular path built around a rectangle increases both the length and the breadth. If a path of width ww is built outside a rectangle of length ll and breadth bb, the new dimensions are L=l+2wL = l + 2w and B=b+2wB = b + 2w.

Rectangular path of width w outside an inner rectangle.
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Cross paths are two rectangular strips that intersect inside a rectangle. To find their area, sum the areas of the two strips and subtract the area of the central square common to both paths to avoid double counting: Area =(L×w)+(B×w)−w2= (L \times w) + (B \times w) - w^2.

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When a path is built inside a rectangle, the dimensions of the inner region are found by subtracting twice the width from the outer dimensions: l=L−2wl = L - 2w and b=B−2wb = B - 2w.

📐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.

Problem 3:

Two cross-roads, each of width 55 m, run at right angles through the centre of a rectangular park of length 7070 m and breadth 4545 m and parallel to its sides. Find the area of the roads.

Rectangular park with two perpendicular roads crossing at the center.

Solution:

  1. Area of the road parallel to the length =L×w=70 m×5 m=350 m2= L \times w = 70 \text{ m} \times 5 \text{ m} = 350 \text{ m}^2
  2. Area of the road parallel to the breadth =B×w=45 m×5 m=225 m2= B \times w = 45 \text{ m} \times 5 \text{ m} = 225 \text{ m}^2
  3. Area of the common central square =w×w=5 m×5 m=25 m2= w \times w = 5 \text{ m} \times 5 \text{ m} = 25 \text{ m}^2
  4. Total area of roads =350+225−25=550 m2= 350 + 225 - 25 = 550 \text{ m}^2

Explanation:

The paths overlap at the center of the park. By calculating the areas of the two rectangular strips, we count the middle 5 m×5 m5 \text{ m} \times 5 \text{ m} square twice. Thus, we must subtract it once.

Problem 4:

A circular pond has a diameter of 2828 m. A 1.41.4 m wide stone walk is built around it. Find the cost of gravelling the walk at Rs 5050 per square metre. (Take π=227\pi = \frac{22}{7})

Circular pond with diameter 28m and an outer path of width 1.4m.

Solution:

  1. Inner radius r=282=14 mr = \frac{28}{2} = 14 \text{ m}
  2. Outer radius R=14+1.4=15.4 mR = 14 + 1.4 = 15.4 \text{ m}
  3. Area of the walk =π(R2−r2)=227(15.42−142)= \pi(R^2 - r^2) = \frac{22}{7}(15.4^2 - 14^2)
  4. Area =227(237.16−196)=227×41.16=129.36 m2= \frac{22}{7}(237.16 - 196) = \frac{22}{7} \times 41.16 = 129.36 \text{ m}^2
  5. Cost =129.36×50=6468= 129.36 \times 50 = 6468 Total Cost = Rs 6468

Explanation:

To find the area of the path around a circular pond, we find the area of the larger circle (pond + path) and subtract the area of the pond. Finally, multiply the resulting area by the rate per square metre.