Grade 5 Perimeter and Area of Polygons

What this problem type covers

Twenty-six problem types across five levels cover rectangle perimeters, the area formulas for five polygons, missing dimensions, scale factors, equal perimeters, square-metre and square-kilometre conversion, related triangles, composite rectangles, wire bent into a rectangle, indented boundaries, hidden heights, tiled rectangles, comparing several perimeters, joined squares, shaded and overlapping regions, clusters and staircases of squares, rectangles with fixed area, cut-out boundaries, roads through a field, nested squares, overlapping rhombi, midpoint partitions, symmetric kites, and equal-area pieces in an equilateral triangle.

Example problems

One example from each difficulty band, with worked solutions.

These problems are generated automatically. The answers are derived as each problem is built and are checked by tests, but mistakes are still possible — please tell us if you find one.

1. 1 · ★1
   Both dimensions of a square are enlarged by a factor of 2. How many times the original perimeter and area are the new perimeter and area?
   • ①perimeter ×4, area ×4
   • ②perimeter ×4, area ×2
   • ③perimeter ×3, area ×4
   • ④perimeter ×2, area ×4
   • ⑤perimeter ×1, area ×4
   Show solution

   ④perimeter ×2, area ×4

   1. Scaling every side by 2 scales the perimeter by 2.
   2. Area uses both dimensions, so the scale factor is multiplied twice.
   3. 2×2=4, so the area is 4 times as large.
2. 2 · ★2
   A right-angled compound shape splits into rectangles measuring 13 cm by 9 cm and 10 cm by 4 cm. Find its total area.
   • ①157 cm²
   • ②117 cm²
   • ③299 cm²
   • ④40 cm²
   • ⑤314 cm²
   Show solution

   ①157 cm²

   1. The first rectangle has area 117 cm².
   2. The second rectangle has area 40 cm².
   3. Adding them gives 157 cm².
3. 3 · ★3
   A 56 cm wire is bent into a rectangle with no overlap. Its length is 6 cm greater than its width. Find the rectangle's area.
   • ①188 cm²
   • ②11 cm²
   • ③28 cm²
   • ④22 cm²
   • ⑤187 cm²
   Show solution

   ⑤187 cm²

   1. The length and width add to 28 cm.
   2. The width is (28−6)÷2=11 cm, so the length is 17 cm.
   3. The area is 17×11=187 cm².
4. 4 · ★4
   A trapezoid has area 279 cm². Unshaded triangular pieces have areas 32 cm² and 2 cm². Find the shaded area.
   • ①245 cm²
   • ②279 cm²
   • ③490 cm²
   • ④34 cm²
   • ⑤311 cm²
   Show solution

   ①245 cm²

   1. The whole area is 279 cm².
   2. The unshaded pieces total 34 cm².
   3. Subtracting gives 245 cm².
5. 5 · ★5
   A 28 m by 20 m field has two vertical paths 3 m wide and two horizontal paths 2 m wide. Find the area left.
   • ①352 m²
   • ②364 m²
   • ③448 m²
   • ④560 m²
   • ⑤440 m²
   Show solution

   ①352 m²

   1. The remaining horizontal length is 22 m.
   2. The remaining vertical length is 16 m.
   3. 22×16=352, so the area left is 352 m².

Where people usually go wrong

• Adding the length and width only once when a rectangle has two of each.
• Using a perimeter formula for an area question, or attaching square units to a length.
• Forgetting to divide a triangle, trapezoid, or rhombus product by two.
• Scaling area by the side-length factor instead of by the square of that factor.
• Converting kilometres to metres after multiplying, which loses a factor of one million in area.
• Counting a shared edge as part of the outside perimeter of a joined figure.
• Subtracting an overlap twice when it was counted twice and should remain once.
• Assuming every cut-out changes the perimeter; a corner cut can replace two equal lengths without changing the total.