Recover a missing rectangle side and check the boundary

diagnostic. A rectangle covers 20 square cm and one side is 4 cm. What is the other side? Write a multiplication equation with a missing factor before solving it.

Answer: 5 cm, because 4 × 5 = 20.

Teacher response: A correct missing-factor equation distinguishes dividing the area by a known side from subtracting a length from an area.

model. A rectangular display must have area 48 square cm and width 6 cm. At most 30 cm of edging is available. What length is required, and does its complete boundary fit the edging limit?

1. Sketch a rectangle, label its known 6 cm width and write 6 × unknown length = 48. The area counts square-centimetre units; the missing side must be a length in centimetres.
2. Divide the area by the known side: 48 ÷ 6 = 8 cm. Label the opposite sides 8 cm and the remaining opposite side 6 cm, then verify 6 × 8 = 48 square cm.
3. Trace the whole boundary and record 8 + 6 + 8 + 6 = 28 cm, or 2 × (8 + 6) = 28 cm. Adding just one length and one width counts only half the boundary.
4. Compare the calculated 28 cm boundary with the separate 30 cm limit. It fits, with 30 − 28 = 2 cm left. The decision uses both the required area and the available edging.

Worked example: The length must be 8 cm because 48 ÷ 6 = 8. The perimeter is 2 × (8 + 6) = 28 cm, so the rectangle meets the 30 cm limit with 2 cm of edging left.

p1. Support: a rectangle has area 24 square cm and width 4 cm. Enter its missing length in cm. Draw four-square rows and check the completed area equation.

Answer: 6

Teacher response: Six rows of four unit squares cover 24 square cm. The length is 6 cm and 4 × 6 checks the required area.

p2. Core: a rectangle has area 54 square cm and width 6 cm. Recover its length, then enter its complete perimeter in cm. Label all four sides on your sketch.

Answer: 30

Teacher response: The length is 54 ÷ 6 = 9 cm. Both 9 cm sides and both 6 cm sides total 30 cm.

p3. Challenge: a rectangular plan requires area 36 square cm and width 4 cm, with at most 24 cm of edging. Enter how many extra centimetres of edging the required rectangle would need, and explain whether the plan meets all constraints.

Answer: 2

Teacher response: The required length is 9 cm and the boundary is 26 cm. It exceeds 24 cm by 2 cm, so the given area, width and edging limit cannot all be met by this rectangle.

e1. Independent check: a rectangle must cover 60 square cm and have width 5 cm. The edging limit is 34 cm. Enter the missing length in cm, then show the area check, complete perimeter and a decision about the limit.

Answer: 12

Look for:

• Writes a missing-factor equation and divides area by the known side.
• Checks the recovered side by multiplying to reconstruct the required area.
• Counts both pairs of opposite sides and labels perimeter in cm.
• Checks every stated constraint and distinguishes fitting exactly from having a positive margin.

Teacher response: The length is 60 ÷ 5 = 12 cm. Its area is 12 × 5 = 60 square cm and its perimeter is 2 × (12 + 5) = 34 cm. It meets the limit exactly, with no edging left.

transfer. A proposed repair to the worked plan uses length 7 cm and width 6 cm. Its author says the 26 cm perimeter is below the 30 cm limit, so it must meet the plan. Does it satisfy every requirement?

Answer: It meets the edging limit, since 2 × (7 + 6) = 26 cm, but its area is only 7 × 6 = 42 square cm. The required area is 48 square cm. Both constraints must be checked; the original 8 cm by 6 cm rectangle meets them with perimeter 28 cm.

teach-1. 5 minutes: Give the diagnostic and ask for its missing-factor equation. Route learners by the equation and explanation, then state which units describe the inside and which describe a side.

teach-2. 10 minutes: Build the worked display plan. Partners label opposite sides and take different jobs: one verifies the 48 square cm area, the other traces the 28 cm boundary. Switch jobs and record the feasibility decision with its 2 cm margin.

teach-3. 15 minutes: Support learners tile p1 and complete the 18-square-cm recheck. Ready learners solve p2 using separate side and boundary equations. Pairs test p3, then explain why a correct missing side can still produce a plan that exceeds its edging limit. Require a sketch and units with each report.

teach-4. 10 minutes: Allow four minutes for e1 independently, including its area and perimeter checks. Spend three minutes comparing the exact-limit decision, then three minutes on the transfer's faulty proposal. Re-teach unknown factors for a wrong side; re-trace the four edges for a correct side with a wrong perimeter.

misconception. The missing side is 48 − 6 = 42 cm, or the edging needed is just 8 + 6 = 14 cm.

Teacher response: Return to the two questions. To recover a side, ask how many 6-square-cm rows fill 48 square cm and check by multiplication. To find edging, mark and add every outside side, including the two opposite sides.

support. Use a grid whose known side has the stated number of unit squares. Build complete rows until the required area is filled, then use matching marks on opposite sides to count the boundary.

extend. For a 48-square-cm rectangle, test widths 4, 6 and 8 cm against the same 30 cm edging limit. The corresponding lengths are 12, 8 and 6 cm, and perimeters are 32, 28 and 28 cm. Explain why the 6-by-8 and 8-by-6 records describe rotations of one rectangle.

support-route. If the learner subtracts 4 from 20 or gives an unlabelled guess, arrange 20 drawn unit squares into rows of four. Count the five rows, connect 4 × 5 = 20 to 20 ÷ 4 = 5 and use p1. Recheck area 18 square cm with a 3 cm side: the missing side is 6 cm.

ready-route. If the learner states 5 cm and verifies 4 × 5 = 20, move to the worked plan and ask the learner to keep the area equation separate from the boundary equation.