Area of Paths and Borders

Find the area of a path, frame or border between two rectangles, the length of its outer edge, the number of tiles it takes, and the width of a path from its area or perimeter.

Before you start

The area of a shaded region can be found by subtracting: the area of the whole shape minus the area of the part removed (as in the lesson on perimeter and area of composite figures). A card 16 cm by 10 cm with a square of side 3 cm cut out has area $16 \times 10 - 3 \times 3 = 160 - 9 = 151\text{ cm}^{2}$ .
The perimeter of a rectangle $l$ by $b$ is $2(l + b)$ : a rectangle 24 m by 16 m has perimeter $2 \times (24 + 16) = 80$ m.
Change lengths to one unit before you multiply: 50 cm = 0.5 m and 0.6 m = 60 cm. Never change an area; change the lengths first.
An equation such as $6x + 20 = 50$ is solved by subtracting 20 from both sides, $6x = 30$ , then dividing both sides by 6: $x = 5$ .
Expanding a bracket and collecting like terms (algebraic manipulation): $3(5 + 2x) = 15 + 6x$ , and $15 + 6x + 4x = 15 + 10x$ .

Paths and Borders Round a Rectangle

A path of the same width all the way round a lawn, a frame round a picture and a border of tiles round a pool are each the region between two rectangles, one inside the other. The area of such a region is found by subtracting: the outer rectangle minus the inner one.

Take care with the outer rectangle. A path 2 m wide round a lawn adds 2 m at both ends of every side, so a 20 m side becomes $20 + 2 + 2 = 24$ m. A path 2 m wide inside the edge of a field takes 2 m from both ends of every side.

To check, cut the path into four strips: two long strips the full length of the outer rectangle, and two short strips that fit between them.

Paths and borders

Area of a path or border $=$ area of the outer rectangle $-$ area of the inner rectangle.

A path $w$ m wide outside a rectangle $l$ m by $b$ m: the outer rectangle is $(l + 2w)$ m by $(b + 2w)$ m.

A path $w$ m wide inside the edge of a rectangle $l$ m by $b$ m: the inner rectangle is $(l - 2w)$ m by $(b - 2w)$ m.

A path $w$ m wide along one long side and one short side of a rectangle $l$ m by $b$ m: the rectangle and the path together are $(l + w)$ m by $(b + w)$ m. The width is added once each way, and the corner square is counted once.

Where two paths cross, the part they share is counted only once.

Example 1: A path round a lawn

A rectangular lawn in a neighbourhood park is 20 m by 12 m. A path 2 m wide runs all round the outside of the lawn.

(a) Find the area of the path.

(b) The path is paved at $28 per $\text{m}^{2}$ , and a low fence is put along its outer edge at $12.50 per metre. Find the cost of each.

Solution

(a) The outer rectangle is $(20 + 2 + 2)$ m by $(12 + 2 + 2)$ m, that is 24 m by 16 m.

Area of path $= 24 \times 16 - 20 \times 12 = 384 - 240 = 144\text{ m}^{2}$

Check by adding strips: two long strips 24 m by 2 m, and two short strips 12 m by 2 m between them: $2 \times (24 \times 2) + 2 \times (12 \times 2) = 96 + 48 = 144\text{ m}^{2}$

(b) Paving: $144 \times 28 = 4032$ , so the paving costs $4032.

The outer edge is the 24 m by 16 m rectangle: $P = 24 + 16 + 24 + 16 = 80$ m. Fence: $80 \times 12.50 = 1000$ , so the fence costs $1000.

(a) The area of the path is $144\text{ m}^{2}$ . (b) The paving costs $4032 and the fence costs $1000.

A path along only two sides that meet, one long side and one short side, is different. It adds its width at one end of each side, not both, so the lawn and the path together make a rectangle $(l + w)$ m by $(b + w)$ m. The two strips of path share the square in the corner: count it once.

Example 2: A path along two sides

A rectangular lawn beside a void deck is 12 m by 8 m. A path 1.5 m wide is laid outside it along one long side and one short side, as shown. Find the area of the path.

Solution

The path runs along two sides only, so it adds 1.5 m once each way. The lawn and the path together make a rectangle $(12 + 1.5)$ m by $(8 + 1.5)$ m, that is 13.5 m by 9.5 m (not $12 + 3$ by $8 + 3$ ).

Area of path $= 13.5 \times 9.5 - 12 \times 8 = 128.25 - 96 = 32.25\text{ m}^{2}$

Check by strips: along the long side $12 \times 1.5 = 18$ , along the short side $8 \times 1.5 = 12$ , and the corner square $1.5 \times 1.5 = 2.25$ , counted once: $18 + 12 + 2.25 = 32.25\text{ m}^{2}$

The area of the path is $32.25\text{ m}^{2}$ .

When the path runs round the inside of the edge, the outer rectangle is the one you are given, and the inner rectangle is smaller by twice the width each way.

Example 3: A path inside the edge of a school field

A school field is a rectangle 100 m by 60 m. A path 5 m wide runs round the inside of its edge, and the rest of the field is grass. Find the area of (a) the grass and (b) the path.

Solution

(a) The path takes 5 m from both ends of every side: the grass is $(100 - 5 - 5)$ m by $(60 - 5 - 5)$ m, that is 90 m by 50 m.

Area of grass $= 90 \times 50 = 4500\text{ m}^{2}$

(b) Area of path $= 100 \times 60 - 4500 = 6000 - 4500 = 1500\text{ m}^{2}$

Check by adding strips: two strips 100 m by 5 m along the long sides, and two strips 50 m by 5 m between them: $2 \times (100 \times 5) + 2 \times (50 \times 5) = 1000 + 500 = 1500\text{ m}^{2}$

(a) The grass covers $4500\text{ m}^{2}$ . (b) The path covers $1500\text{ m}^{2}$ .

Paths That Cross

When two straight paths cross a lawn, one parallel to each side, each path is a rectangle as long as the lawn is long or wide. The square where they cross belongs to both paths, so adding the two paths counts it twice: take it away once.

The lawn left can be checked by sliding its pieces together, as if the paths were pushed out to the edges: they make one rectangle, shorter each way by the width of a path.

Example 4: Two paths crossing a lawn

A lawn 30 m by 20 m has two straight paths, each 2 m wide, crossing it at right angles, one parallel to each side. Find (a) the area of the paths and (b) the area of the lawn left.

Solution

(a) The path along the lawn is 30 m by 2 m, and the path across it is 20 m by 2 m.

Adding them, $30 \times 2 + 20 \times 2 = 60 + 40 = 100\text{ m}^{2}$ , counts the square where they cross, 2 m by 2 m, twice. Take it away once:

Area of paths $= 100 - 2 \times 2 = 96\text{ m}^{2}$

(b) Area of lawn left $= 30 \times 20 - 96 = 600 - 96 = 504\text{ m}^{2}$

Check: slide the four pieces of lawn together, as if the paths were pushed out to the edges. They make one rectangle $(30 - 2)$ m by $(20 - 2)$ m: $28 \times 18 = 504\text{ m}^{2}$

(a) The paths cover $96\text{ m}^{2}$ . (b) The lawn left is $504\text{ m}^{2}$ .

Borders of Tiles

When the tiles fit exactly (each side and the width of the border a whole number of tiles), the number of square tiles a border needs is found by dividing the area of the border by the area of one tile, with every length in the same unit. For a border one tile wide, check by counting: the tiles along each side, and one more in each corner.

Example 5: Tiles round a pool

A children's pool at a community club is a rectangle 6 m by 4 m. A border one tile wide is laid round the outside of the pool with square tiles of side 50 cm. How many tiles are needed?

Solution

Change the tile's side to metres first: 50 cm = 0.5 m.

The border adds 0.5 m at both ends of every side: the outer rectangle is $(6 + 0.5 + 0.5)$ m by $(4 + 0.5 + 0.5)$ m, that is 7 m by 5 m.

Area of border $= 7 \times 5 - 6 \times 4 = 35 - 24 = 11\text{ m}^{2}$

Area of one tile $= 0.5 \times 0.5 = 0.25\text{ m}^{2}$

Number of tiles $= 11 \div 0.25 = 44$

Check by counting: $6 \div 0.5 = 12$ tiles along each long side, $4 \div 0.5 = 8$ tiles along each short side, and 1 tile in each of the 4 corners: $2 \times 12 + 2 \times 8 + 4 = 44$ .

44 tiles are needed.

Finding the Width of a Path

When the width of a path is not known, call it $x$ m, write the area (or the perimeter of the outer edge) in terms of $x$ , and solve the equation. Check the answer by putting it back.

Example 6: The width of a path from its area

A straight path crosses a rectangular lawn 18 m by 12 m from one long side to the other, parallel to the short sides. The path is $x$ m wide and its area is $30\text{ m}^{2}$ . Find (a) the width of the path and (b) the area of the lawn left.

Solution

(a) The path is a rectangle 12 m long and $x$ m wide, so its area is $12x\text{ m}^{2}$ :

$12x = 30$

Divide both sides by 12: $x = 2.5$

Check: $12 \times 2.5 = 30$ .

(b) Area of lawn left $= 18 \times 12 - 30 = 216 - 30 = 186\text{ m}^{2}$

Check: the two pieces of lawn pushed together make a rectangle $(18 - 2.5)$ m by 12 m: $15.5 \times 12 = 186\text{ m}^{2}$

(a) The path is $2.5\text{ m}$ wide. (b) The lawn left is $186\text{ m}^{2}$ .

Example 7: The width of a path from its outer edge

A rectangular flower bed in a community garden is 10 m by 6 m. A path $x$ m wide runs all round the outside of it, and the outer edge of the path is 44 m long. Find (a) the value of $x$ and (b) the area of the path.

Solution

(a) The path adds $x$ m at both ends of every side, so the outer edge is a rectangle $(10 + 2x)$ m by $(6 + 2x)$ m. Its perimeter is 44 m:

$2(10 + 2x) + 2(6 + 2x) = 44$

Expand the brackets: $20 + 4x + 12 + 4x = 44$

Collect like terms: $8x + 32 = 44$

Subtract 32 from both sides: $8x = 12$

Divide both sides by 8: $x = 1.5$

Check: the outer rectangle is 13 m by 9 m, and $2 \times (13 + 9) = 44$ m.

(b) Area of path $= 13 \times 9 - 10 \times 6 = 117 - 60 = 57\text{ m}^{2}$

Check by strips: $2 \times (13 \times 1.5) + 2 \times (6 \times 1.5) = 39 + 18 = 57\text{ m}^{2}$

(a) $x = 1.5$ : the path is 1.5 m wide. (b) The area of the path is $57\text{ m}^{2}$ .

Common mistakes

Adding the width of a path once. A path 2 m wide round a 20 m side makes the outer side $20 + 2 + 2 = 24$ m, not 22 m. Inside the edge, take the width off twice: $100 - 5 - 5 = 90$ m, not 95 m.
Counting the crossing twice. Where two paths cross, the square they share is in both paths; take it away once (Example 4).
Losing a corner. A path along two sides that meet has a square in the corner (Example 2); a border of tiles has a tile in each corner. Count each corner once: not twice, and not leaving it out.
Using the wrong rectangle for a fence. A fence along the outer edge of a path goes round the outer rectangle (Example 1).
Multiplying lengths in different units. Change 50 cm to 0.5 m (or 6 m to 600 cm) before you multiply. Never change an area: change the lengths, then multiply.

Practice

Every path, frame and border in these questions has the same width all the way along. Work each question before you open its solution.

Practice 1: A picture frame

Hui Min's painting for the school art exhibition is 0.6 m by 0.4 m. It is put in a frame 5 cm wide all round. Find (a) the area of the frame and (b) the length of the outer edge of the frame.

ShowHide solution

Change to centimetres first: 0.6 m = 60 cm and 0.4 m = 40 cm.

The outer rectangle is $(60 + 5 + 5)$ cm by $(40 + 5 + 5)$ cm, that is 70 cm by 50 cm.

(a) Area of frame $= 70 \times 50 - 60 \times 40 = 3500 - 2400 = 1100\text{ cm}^{2}$

Check by adding strips: $2 \times (70 \times 5) + 2 \times (40 \times 5) = 700 + 400 = 1100\text{ cm}^{2}$

(b) Outer edge $= 70 + 50 + 70 + 50 = 240\text{ cm}$

(a) $1100\text{ cm}^{2}$ (b) $240\text{ cm}$ , that is 2.4 m

Practice 2: A path along two sides

A rectangular lawn behind a community centre is 15 m by 10 m. A path 2 m wide is laid outside it along one long side and one short side, as shown. (a) Find the area of the path. (b) The path is paved at $30 per $\text{m}^{2}$ . Find the cost.

ShowHide solution

(a) The lawn and the path together make a rectangle $(15 + 2)$ m by $(10 + 2)$ m, that is 17 m by 12 m. The path adds 2 m at one end of each side only, because it runs along two sides.

Area of path $= 17 \times 12 - 15 \times 10 = 204 - 150 = 54\text{ m}^{2}$

Check by strips: along the long side $15 \times 2 = 30$ , along the short side $10 \times 2 = 20$ , and the corner square $2 \times 2 = 4$ , counted once: $30 + 20 + 4 = 54\text{ m}^{2}$

(b) $54 \times 30 = 1620$ , so the paving costs $1620.

(a) $54\text{ m}^{2}$ (b) $1620

Practice 3: A path inside a garden plot

Mrs Goh's plot in a community garden is a rectangle 18 m by 12 m. A path 1 m wide runs round the inside of its edge. (a) Find the area of the path. (b) The path is paved at $32 per $\text{m}^{2}$ . Find the cost. (c) A fence is put round the outer edge of the plot at $14.50 per metre. Find the cost.

ShowHide solution

(a) The inner rectangle is $(18 - 1 - 1)$ m by $(12 - 1 - 1)$ m, that is 16 m by 10 m.

Area of path $= 18 \times 12 - 16 \times 10 = 216 - 160 = 56\text{ m}^{2}$

Check by adding strips: $2 \times (18 \times 1) + 2 \times (10 \times 1) = 36 + 20 = 56\text{ m}^{2}$

(b) $56 \times 32 = 1792$ , so the paving costs $1792.

(a) $56\text{ m}^{2}$ (b) $1792 (c) $870

Practice 4: A border inside a void deck

The floor of a void deck is a rectangle 24 m by 15 m. A border 1.5 m wide round the inside of its edge is laid with dark tiles, and the rest with light tiles. (a) Find the area of the border. (b) The dark tiles cost $35 per $\text{m}^{2}$ . Find their cost.

ShowHide solution

(a) The border takes 1.5 m from both ends of every side: the inner rectangle is $(24 - 1.5 - 1.5)$ m by $(15 - 1.5 - 1.5)$ m, that is 21 m by 12 m.

Area of border $= 24 \times 15 - 21 \times 12 = 360 - 252 = 108\text{ m}^{2}$

Check by strips: $2 \times (24 \times 1.5) + 2 \times (12 \times 1.5) = 72 + 36 = 108\text{ m}^{2}$

(b) $108 \times 35 = 3780$ , so the dark tiles cost $3780.

(a) $108\text{ m}^{2}$ (b) $3780

Practice 5: Paths across a lawn

A lawn in a park is 40 m by 25 m. Two straight paths, each 3 m wide, cross it at right angles, one parallel to each side. (a) Find the area of the paths. (b) The rest of the lawn is covered with turf at $5.50 per $\text{m}^{2}$ . Find the cost.

ShowHide solution

(a) The paths are 40 m by 3 m and 25 m by 3 m, and they share a square 3 m by 3 m where they cross:

Area of paths $= 40 \times 3 + 25 \times 3 - 3 \times 3 = 120 + 75 - 9 = 186\text{ m}^{2}$

(b) Area of turf $= 40 \times 25 - 186 = 1000 - 186 = 814\text{ m}^{2}$

Check: pushed together, the pieces of lawn make a rectangle $(40 - 3)$ m by $(25 - 3)$ m: $37 \times 22 = 814\text{ m}^{2}$

$814 \times 5.50 = 4477$ , so the turf costs $4477.

(a) $186\text{ m}^{2}$ (b) $4477

Practice 6: The cost of a frame

A mirror 0.8 m by 0.5 m has a wooden frame 10 cm wide all round it. (a) Find the area of the frame in $\text{m}^{2}$ . (b) The wood costs $60 per $\text{m}^{2}$ , and a rubber strip is fixed round the outer edge of the frame at $2.50 per metre. Find the total cost.

ShowHide solution

Change to metres first: 10 cm = 0.1 m.

The outer rectangle is $(0.8 + 0.1 + 0.1)$ m by $(0.5 + 0.1 + 0.1)$ m, that is 1 m by 0.7 m.

(a) Area of frame $= 1 \times 0.7 - 0.8 \times 0.5 = 0.7 - 0.4 = 0.3\text{ m}^{2}$

(b) Wood: $0.3 \times 60 = 18$ , so the wood costs $18.

Outer edge $= 2 \times (1 + 0.7) = 3.4$ m. Rubber strip: $3.4 \times 2.50 = 8.50$ , so the strip costs $8.50.

Total: $18 + 8.50 = 26.50$ , so the total cost is $26.50.

(a) $0.3\text{ m}^{2}$ (b) $26.50

Practice 7: Tiles round a pond

A rectangular pond at a community club is 3 m by 1.5 m. A border one tile wide is laid round the outside of the pond with square tiles of side 25 cm. (a) How many tiles are needed? (b) The tiles cost $1.20 each. Find the cost.

ShowHide solution

(a) Change to metres: 25 cm = 0.25 m. The outer rectangle is $(3 + 0.25 + 0.25)$ m by $(1.5 + 0.25 + 0.25)$ m, that is 3.5 m by 2 m.

Area of border $= 3.5 \times 2 - 3 \times 1.5 = 7 - 4.5 = 2.5\text{ m}^{2}$

Area of one tile $= 0.25 \times 0.25 = 0.0625\text{ m}^{2}$

Number of tiles $= 2.5 \div 0.0625 = 40$

Check by counting: $3 \div 0.25 = 12$ tiles along each long side, $1.5 \div 0.25 = 6$ along each short side, and 4 corner tiles: $2 \times 12 + 2 \times 6 + 4 = 40$ .

(b) $40 \times 1.20 = 48$ , so the tiles cost $48.

(a) 40 tiles (b) $48

Practice 8: A border two tiles wide

A wading pool 4 m by 3 m has a border 1 m wide round its outside, laid with square tiles of side 50 cm. (a) How many tiles are needed? (b) The tiles cost $2.40 each. Find the cost.

ShowHide solution

(a) Change to metres: 50 cm = 0.5 m. The outer rectangle is $(4 + 1 + 1)$ m by $(3 + 1 + 1)$ m, that is 6 m by 5 m.

Area of border $= 6 \times 5 - 4 \times 3 = 30 - 12 = 18\text{ m}^{2}$

Area of one tile $= 0.5 \times 0.5 = 0.25\text{ m}^{2}$

Number of tiles $= 18 \div 0.25 = 72$

Check by counting: the outer rectangle would take $(6 \div 0.5) \times (5 \div 0.5) = 12 \times 10 = 120$ tiles and the pool would take $(4 \div 0.5) \times (3 \div 0.5) = 8 \times 6 = 48$ , so the border takes $120 - 48 = 72$ .

(b) $72 \times 2.40 = 172.80$ , so the tiles cost $172.80.

(a) 72 tiles (b) $172.80

Practice 9: A width from the outer edge

A rectangular pond is 12 m by 8 m. A path $x$ m wide runs all round the outside of it, and the outer edge of the path is 56 m long. (a) Find $x$ . (b) Find the area of the path.

ShowHide solution

(a) The outer edge is a rectangle $(12 + 2x)$ m by $(8 + 2x)$ m. Its perimeter is 56 m:

$2(12 + 2x) + 2(8 + 2x) = 56$

Expand the brackets: $24 + 4x + 16 + 4x = 56$

Collect like terms: $8x + 40 = 56$

Subtract 40 from both sides: $8x = 16$

Divide both sides by 8: $x = 2$

Check: the outer rectangle is 16 m by 12 m, and $2 \times (16 + 12) = 56$ m.

(b) Area of path $= 16 \times 12 - 12 \times 8 = 192 - 96 = 96\text{ m}^{2}$

(a) $x = 2$ : the path is 2 m wide. (b) $96\text{ m}^{2}$

Practice 10: A lawn from the area of its path

A rectangular lawn is 20 m long and $b$ m wide. A path 2 m wide runs all round the outside of it, and the area of the path is $152\text{ m}^{2}$ . Find $b$ .

ShowHide solution

The outer rectangle is $(20 + 4)$ m by $(b + 4)$ m, that is 24 m by $(b + 4)$ m.

Area of path $=$ outer rectangle $-$ lawn:

$24(b + 4) - 20b = 152$

Expand the bracket: $24b + 96 - 20b = 152$

Collect like terms: $4b + 96 = 152$

Subtract 96 from both sides: $4b = 56$

Divide both sides by 4: $b = 14$

Check: the outer rectangle is $24 \times 18 = 432\text{ m}^{2}$ , the lawn is $20 \times 14 = 280\text{ m}^{2}$ , and $432 - 280 = 152\text{ m}^{2}$ .

$b = 14$ : the lawn is 14 m wide.

Key points

Area of a path or border $=$ area of the outer rectangle $-$ area of the inner rectangle.
A path $w$ m wide outside a rectangle $l$ m by $b$ m: the outer rectangle is $(l + 2w)$ m by $(b + 2w)$ m. Inside the edge: the inner rectangle is $(l - 2w)$ m by $(b - 2w)$ m.
A path $w$ m wide along one long side and one short side only: the rectangle and the path together are $(l + w)$ m by $(b + w)$ m (the width added once each way; the corner square counted once).
Check by strips: two long strips the full length, and two short strips between them.
Where two paths $w$ m wide cross, one parallel to each side, the shared square is counted once; pushed together, the pieces of lawn left make a rectangle $(l - w)$ m by $(b - w)$ m.
When the tiles fit exactly (each side and the width a whole number of tiles): number of tiles $=$ area of the border $\div$ area of one tile, with lengths in one unit; check by counting along the sides and the corners.
An unknown width or side is found by writing an equation for the area or the perimeter and solving it.
Cost $=$ area $\times$ price per $\text{m}^{2}$ , or length of edge $\times$ price per metre.