The idea in brief. A border takes its width off BOTH ends of every side, so each dimension changes by twice the border; and a tile that has to be cut is not a tile, so tiles are counted along each side by itself and never by dividing the areas.
Lesson video, with the working
Area of a border or path
4 min 27 s
The film teaches 2 questions in turn. Each question shows first; its working is written as the film explains it.
A rectangular garden is 32 m long and 18 m wide. A path 2 m wide is laid all round the outside of the garden. Find the area covered by the path.
Length of the larger rectangle = 32 m + 2 m + 2 m = 36 m
Breadth of the larger rectangle = 18 m + 2 m + 2 m = 22 m
Area of the larger rectangle = 36 m × 22 m = 792 m²
Area of the garden = 32 m × 18 m = 576 m²
Area of the path = 792 m² − 576 m² = 216 m²
Ans: 216 m²
Example 1
Mei Ling pastes a photograph on a rectangular card 30 cm long and 24 cm wide, leaving a border 3 cm wide all round the photograph. Find the area of the border.
Length of the photograph = 30 cm − 3 cm − 3 cm = 24 cm
Breadth of the photograph = 24 cm − 3 cm − 3 cm = 18 cm
Area of the photograph = 24 cm × 18 cm = 432 cm²
Area of the card = 30 cm × 24 cm = 720 cm²
Area of the border = 720 cm² − 432 cm² = 288 cm²
Ans: 288 cm²
How to solve it
An outside path adds its breadth at both ends. The outer length is 32 plus 2 plus 2, which is 36 metres.
Outer breadth: 18 plus 2 plus 2 is 22 metres. Outer area: 36 times 22 is 792 square metres.
Garden area: 32 times 18 is 576 square metres. Path area: 792 minus 576 is 216 square metres.
For the photograph, subtract both borders: 30 minus 3 minus 3 is 24 centimetres; 24 minus 3 minus 3 is 18 centimetres.
Photograph area: 24 times 18 is 432 square centimetres. Card area: 30 times 24 is 720 square centimetres.
Subtract the photograph area from the card area: 720 minus 432 is 288 square centimetres of border.
Change both ends of each dimension. Then subtract the inner area from the outer area.
Practice questions with worked solutions
Question 1 is free to try. The other 6 open in the free trial, each with its full working.
Question 1 of 7 · Intermediate · free to tryRaju lays four identical rectangular mats on a rectangular floor 14 m long…Raju lays four identical…
Raju lays four identical rectangular mats on a rectangular floor 14 m long and 10 m wide, in two rows of two. The mats cover the middle of the floor exactly and leave a border 1 m wide all round them. Find the area of one mat.
Length of the covered rectangle = 14 m − 1 m − 1 m = 12 m
Breadth of the covered rectangle = 10 m − 1 m − 1 m = 8 m
Length of one mat = 12 m ÷ 2 = 6 m
Breadth of one mat = 8 m ÷ 2 = 4 m
Area of one mat = 6 m × 4 m = 24 m²
The area of one mat is 24 m².
Answer: 24 m²
For parents
How can I support my child with borders and tiling questions?
Borders and tiled surfaces give rectangle and square calculations a clear purpose.
What your child learns here
Borders and tiled surfaces give rectangle and square calculations a clear purpose. Your child may need the area around an inner region or the number of complete tiles that fit in a stated arrangement. These are applications of the Primary 4 outcome on composite figures, so understanding which region is required and which lengths belong to it comes before choosing multiplication or subtraction.
Distinguish the dimensions of an inner rectangle from those of the outer rectangle surrounding it, including border thickness on both ends.
Find a rectangular border's area as the outer area minus the inner area, keeping the units consistent throughout.
Count complete tiles along each side when tiles are arranged in aligned rows and columns, then multiply the two counts.
Use the question's conditions about cuts, gaps, overlaps and orientation when deciding what a tile count represents.
Does the answer make sense?
For a border surrounding an inner rectangle, the proposed border area plus the inner area must equal the outer area. Your child can use this addition to check the subtraction and also confirm that the border has not been counted twice at its corners.
How to practise these questions
The Basic question finds the area of a path round a garden. Intermediate work covers mats inside a border, whole tiles in aligned rows, uncovered floor area, crossing strips and the border round a photograph. Challenge questions consider how rectangular labels can be turned and how a path's pieces combine within a park. Read the arrangement conditions before calculating: a fixed row direction and permission to turn pieces lead to different counting choices. Your child should record the number fitting along a side separately from the resulting area. In a layout question, check that the proposed pieces really fit without overlap, rather than relying on the total available area alone.
Questions parents ask
Why does a border add its thickness twice to a dimension? A uniform border around all four sides extends beyond both ends of the inner length and both ends of the inner breadth. Each outer dimension therefore includes two border thicknesses. If the path lies along only selected sides, the given arrangement determines which additions are needed. Ask your child to identify the two endpoints of the dimension being found.
Is dividing total area by tile area always enough? That division checks the area requirement, but it may not prove that whole tiles fit the given dimensions. For aligned rows of equal tiles, your child should check how many fit along each side as well. A leftover strip can prevent another complete row even when the remaining surface area seems large enough for more tiles.
How can I help my child distinguish path area from path length? Ask what is being measured or covered. The surface to pave needs an area in square units, while a distance travelled along a specified route needs a length unit. A question about fencing can refer to an inner or outer boundary, so your child should identify that boundary explicitly rather than assume every garden question asks for the same measurement.
