Area of Rectangular Path (Frame) with Worked Examples

Area of Rectangular Path (Frame) is found by subtracting the area of the inner rectangle from the area of the outer rectangle. The formula changes according to whether the path lies outside or inside the rectangle. This page covers both cases, equal-width paths, solved calculations and useful shortcuts for aptitude problems.

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What Is the Area of a Rectangular Path (Frame)?

The area of a rectangular path or frame is the area occupied between two rectangles. It equals the area of the larger rectangle minus the area of the smaller rectangle.

If a path of uniform width w surrounds a rectangle of length L and breadth B, the outer dimensions are L + 2w and B + 2w. Therefore, the outside path area is (L + 2w)(B + 2w) − LB. If the path is inside the rectangle, the inner dimensions become L − 2w and B − 2w, so the path area is LB − (L − 2w)(B − 2w).

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Area of Paths, Rings and Borders

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Area of Rectangular Path (Frame) Formula & Tricks

Important Formulas

Outside rectangular path
Area = (L + 2w)(B + 2w) − LB = 2w(L + B) + 4w²

Use this when a path of width w is constructed outside a rectangle of length L and breadth B.

Inside rectangular path
Area = LB − (L − 2w)(B − 2w) = 2w(L + B) − 4w²

Use this when a path of width w is made inside the boundary of a rectangle.

Path using perimeter
Outside path area = w × perimeter of inner rectangle + 4w²; Inside path area = w × perimeter of inner rectangle − 4w²

Since the inner rectangle perimeter is 2(L + B), these forms are equivalent to the standard formulas.

Quick Tricks

Subtract the two rectangle areas

Find the outer and inner dimensions first, calculate both rectangle areas, and subtract the smaller area from the larger area. This avoids sign errors.

Example: For an outside path around 20 m × 12 m with width 2 m, area = 24 × 16 − 20 × 12 = 384 − 240 = 144 m².
Use the perimeter shortcut

For a uniform path of width w, start with w times the perimeter of the original rectangle. Add 4w² for an outside path and subtract 4w² for an inside path.

Example: For L = 20 m, B = 12 m and w = 2 m, w × 2(L + B) = 2 × 64 = 128. Outside area = 128 + 16 = 144 m².

Area of Rectangular Path (Frame) Concepts

Path Constructed Outside a Rectangle

For an outside path of width w, add 2w to both the length and breadth before subtracting the original area.

The outer rectangle has length L + 2w and breadth B + 2w. Hence, area of the path = (L + 2w)(B + 2w) − LB. The width is added twice in each dimension because the path extends on both opposite sides.

Example: If L = 15 cm, B = 10 cm and w = 2 cm, path area = (15 + 4)(10 + 4) − 15 × 10 = 19 × 14 − 150 = 116 cm².

Path Constructed Inside a Rectangle

For an inside path of width w, subtract 2w from both the length and breadth before subtracting the inner area from the original area.

The remaining inner rectangle has length L − 2w and breadth B − 2w. Therefore, area of the path = LB − (L − 2w)(B − 2w). The condition L > 2w and B > 2w must hold so that the remaining rectangle has positive dimensions.

Example: For a 20 m × 14 m rectangle with an inside path 2 m wide, path area = 20 × 14 − 16 × 10 = 280 − 160 = 120 m².

Difference Between Outside and Inside Path Formulas

The outside formula contains +4w², while the inside formula contains −4w².

Expanding the outside expression gives (L + 2w)(B + 2w) − LB = 2w(L + B) + 4w². Expanding the inside expression gives LB − (L − 2w)(B − 2w) = 2w(L + B) − 4w². The sign depends on whether the corner squares are added outside or removed inside.

Example: With L = 10 m, B = 6 m and w = 1 m, outside area = 2(1)(16) + 4 = 36 m², while inside area = 2(1)(16) − 4 = 28 m².

Finding Path Width from Areas

If the original dimensions and path area are known, form a quadratic equation in the width w.

For an outside path, use A = 2w(L + B) + 4w². For an inside path, use A = 2w(L + B) − 4w². Solve the resulting equation and retain the positive width that is physically possible.

Example: For a 10 m × 8 m rectangle and an outside path area of 96 m², 4w² + 36w − 96 = 0. Dividing by 4 gives w² + 9w − 24 = 0, so w = 2 m is the positive solution.

Area of Rectangular Path (Frame) Visual Guide & Data

Use the table and chart below to understand the data better.

Outside and Inside Rectangular Paths

Comparison of dimensions and area formulas for a uniform path of width w.

Path position Outer or remaining dimensions Area of path
Outside rectangle L + 2w, B + 2w (L + 2w)(B + 2w) − LB
Inside rectangle L − 2w, B − 2w LB − (L − 2w)(B − 2w)

Note: L and B are the original length and breadth; w is the uniform path width.

Rectangular Frame with an Outside Path

Visual guide: Rectangular Frame with an Outside Path
Original length: L Original breadth: B Path width on each side: w Outer length: L + 2w Outer breadth: B + 2w
The shaded frame lies between the original rectangle and the enlarged outer rectangle.

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Quick Revision Notes

Area of Rectangular Path (Frame) Revision Points

Use the difference between the outer and inner rectangular areas, taking care with the path position and units.

  • Outside path dimensions are L + 2w and B + 2w.
  • Inside path dimensions are L − 2w and B − 2w.
  • Outside path area = 2w(L + B) + 4w².
  • Inside path area = 2w(L + B) − 4w².
  • A path area is expressed in square units, such as cm² or m².
  • For an inside path, both L > 2w and B > 2w must be satisfied.
  • Convert all dimensions to the same unit before applying the formula.

Area of Rectangular Path (Frame) FAQs

What is the formula for an outside rectangular path?

For length L, breadth B and path width w, area = (L + 2w)(B + 2w) − LB = 2w(L + B) + 4w².

What is the formula for an inside rectangular path?

Area = LB − (L − 2w)(B − 2w) = 2w(L + B) − 4w².

A rectangle is 25 m long and 15 m broad. Find the area of an outside path 2 m wide.

Outer dimensions are 29 m and 19 m. Path area = 29 × 19 − 25 × 15 = 551 − 375 = 176 m².

Why is 2w added to each dimension for an outside path?

The path extends by w on both opposite sides of the rectangle. Thus, the total increase in length and breadth is w + w = 2w.

Why is 4w² added in the outside formula but subtracted in the inside formula?

Expanding the outside area gives 2w(L + B) + 4w² because four corner squares are added. For an inside path, those four corner portions are removed, giving 2w(L + B) − 4w².

Can the length or breadth be smaller than twice the inside path width?

No. For an inside path, L − 2w and B − 2w must be positive. Therefore, both L > 2w and B > 2w are required.

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