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Lesson 6 of 7

Perimeter and Area · Lesson 6 of 7

Area Puzzles and Floor Plans

“Use shared edges and missing dimensions to solve floor plans, area mazes and practical layout problems.”

Learning Objectives

• Infer missing dimensions from aligned room boundaries. • Find areas of rectangular and L-shaped rooms. • Check room areas against the whole plot. • Compare equal-area plots with different perimeters. • Solve area mazes and layout problems using multiplication, division and subtraction.

Read a plan through its shared edges

A floor plan is a flat drawing showing how rooms fit inside a plot. Room boundaries that line up give useful relationships between lengths. You do not need to measure the picture with a ruler: use its stated dimensions and shared edges. First identify the whole shape, then work through the smaller regions.

Here lengths are in feet, written ft, and areas are in square feet, written ft². Treat the drawing as a mathematical partition with no extra wall thickness. A room might be rectangular or have an inward corner, so inspect its whole outline before choosing a formula.

Hall5 × 5Master15 × 15 ftToiletUtilityKitchen15 × 12 ftSmall bedroom15 × 12 ftGardenParkingCharan: 35 ft across30 ftAll rooms share edges; the outside plot is one rectangle.
Charan’s house plot— Use shared alignments to find missing dimensions. The dashed line separates the hall into 20 × 12 and 5 × 5 ft rectangles.

Charan’s plot: find the missing measurements

Charan’s plot has three horizontal widths across its top: 15 ft for the master bedroom, 5 ft for the toilet and 15 ft for the kitchen. Its total width is therefore 35 ft. The stated height is 30 ft. These outside measurements help us recover the lower and upper strips.

Example — The small bedroom, utility and lower strip

Problem
Find the missing depths of the small bedroom, utility, garden and parking.

  1. 1.The small bedroom has area 180 ft² and width 15 ft. Its depth is 180 ÷ 15 = 12 ft.
  2. 2.The master bedroom above it is 15 ft deep. Together they use 15 + 12 = 27 ft of the 30 ft height. The bottom strip is 30 − 27 = 3 ft deep.
  3. 3.The kitchen is 12 ft deep and ends at the master bedroom’s bottom level, 15 ft below the top. Its utility strip above is 15 − 12 = 3 ft deep.
  4. 4.The garden spans 15 + 5 = 20 ft, so its area is 20 × 3 = 60 ft². Parking spans 15 ft, so its area is 15 × 3 = 45 ft². The utility is also 15 × 3 = 45 ft².
Example — The L-shaped hall

Problem
Find the area of Charan’s hall, including the narrow part below the toilet.

  1. 1.The main lower rectangle is 20 ft across and 12 ft deep: area 20 × 12 = 240 ft².
  2. 2.The toilet ends 10 ft below the top, while the master bedroom and kitchen end 15 ft below the top. Between these levels, the hall has an extra strip 5 ft wide and 5 ft deep.
  3. 3.That extra part has area 5 × 5 = 25 ft².
  4. 4.Total hall area = 240 + 25 = 265 ft². Treating the whole hall as only the lower rectangle would miss this region.
RegionDimensions or decompositionArea
Master bedroom15 × 15 ft225 ft²
Toilet5 × 10 ft50 ft²
Kitchen15 × 12 ft180 ft²
Small bedroom15 × 12 ft180 ft²
Utility15 × 3 ft45 ft²
Hall20 × 12 ft plus 5 × 5 ft265 ft²
Garden20 × 3 ft60 ft²
Parking15 × 3 ft45 ft²

Adding the regions gives 225 + 50 + 180 + 180 + 45 + 265 + 60 + 45 = 1,050 ft². The whole rectangular plot also gives 35 × 30 = 1,050 ft². This agreement checks that no part was missed or counted twice. The total is plot area including garden and parking; the other six regions together occupy 945 ft².

Common mistake

An L-shaped room cannot be measured using only its largest-looking rectangle. Account for every part. Also say whether a total includes outside regions such as a garden or parking; “plot area” and “indoor room area” can refer to different regions.

Sharan’s plot: a different arrangement

Sharan’s plan has another combination of shared boundaries. The left side contains a 15 ft-deep master bedroom above a 10 ft-deep small bedroom, giving total height 25 ft. The stated total width is 42 ft. Use these complete dimensions to subtract known room widths and depths.

Master12 × 15 ftSmall bedroom12 × 10 ftToiletKitchen18 × 10 ftUtilityHall23 × 15 ftEntranceSharan: 42 ft across25 ftAll rooms share edges; the outside plot is one rectangle.
Sharan’s house plot— Use shared alignments to find missing dimensions. The top-right strip is 10 ft deep and the lower-right strip is 15 ft deep.
Example — Recovering Sharan’s missing room dimensions

Problem
Find the toilet, utility, hall and entrance dimensions.

  1. 1.The lower hall is 23 ft wide, starting after the 12 ft-wide bedrooms. The remaining width for the entrance is 42 − 12 − 23 = 7 ft. The utility above it is also 7 ft wide.
  2. 2.The utility area is 70 ft², so its depth is 70 ÷ 7 = 10 ft, matching the kitchen’s depth.
  3. 3.The top band contains the toilet, kitchen and utility after the 12 ft bedroom width. Toilet width = 42 − 12 − 18 − 7 = 5 ft, and its depth is 10 ft.
  4. 4.The hall and entrance occupy the remaining height 25 − 10 = 15 ft. Thus the hall is 23 × 15 ft and the entrance is 7 × 15 ft.
RegionDimensionsArea
Master bedroom12 × 15 ft180 ft²
Small bedroom12 × 10 ft120 ft²
Toilet5 × 10 ft50 ft²
Kitchen18 × 10 ft180 ft²
Utility7 × 10 ft70 ft²
Hall23 × 15 ft345 ft²
Entrance7 × 15 ft105 ft²
Example — Compare the whole plots

Problem
Compare Charan’s and Sharan’s plot areas and outside perimeters.

  1. 1.Charan: area = 35 × 30 = 1,050 ft²; outside perimeter = 2 × (35 + 30) = 130 ft.
  2. 2.Sharan: area = 42 × 25 = 1,050 ft²; outside perimeter = 2 × (42 + 25) = 134 ft.
  3. 3.The areas match, but Sharan’s outside perimeter is 134 − 130 = 4 ft longer.
  4. 4.The two different room arrangements reinforce that equal enclosed area need not give equal boundary length.

Area mazes: let one region unlock the next

In an area maze, neighbouring rectangles share some dimensions. A known area and a known side can reveal the other side through division. Then aligned boundaries let us add or subtract lengths to reach another region. Follow this chain carefully and keep area units separate from length units.

132615Aa: areas in cm²1010B3 cm3 cm2 cm2 cmb: areas in cm²C42603 cm5 cm6 cm15 cmc: areas in cm²38 ccm²18 ccm²4 cm5 cmD = ? cmd: shared lower level
Four area mazes— All given areas are in square centimetres and lengths are in centimetres. Use shared widths and heights to find each unknown.
Example — Maze a: a shared height gives a ratio

Problem
Find missing area A in square centimetres.

  1. 1.The top rectangles share a height. Their areas are 13 cm² and 26 cm², so the right width is twice the left width.
  2. 2.The bottom rectangles share a different height but retain those same column widths.
  3. 3.The bottom-right area is therefore twice 15 cm²: A = 30 cm².
  4. 4.No individual side lengths are needed; the shared-height comparison carries the ratio down.
Example — Maze b: recover width, then height

Problem
Find missing area B in square centimetres.

  1. 1.The bottom rectangle has area 10 cm² and height 2 cm, so its full width is 10 ÷ 2 = 5 cm.
  2. 2.The bottom rectangle extends 3 cm left of the vertical rectangle. That vertical rectangle therefore has width 5 − 3 = 2 cm.
  3. 3.Its area is 10 cm², so its height is 10 ÷ 2 = 5 cm. The pink region on the right ends 2 cm above the bottom of this vertical rectangle, so its height is 5 − 2 = 3 cm.
  4. 4.The pink region is 3 cm by 3 cm: B = 9 cm².
Example — Maze c: subtract two heights

Problem
Find missing area C in square centimetres.

  1. 1.The middle rectangle has area 42 cm² and height 6 cm. Its width is 42 ÷ 6 = 7 cm.
  2. 2.The pink top width is 7 − 3 = 4 cm. The bottom width is 7 + 5 = 12 cm.
  3. 3.The bottom area is 60 cm², so its height is 60 ÷ 12 = 5 cm.
  4. 4.Total height is 15 cm; the pink height is 15 − 6 − 5 = 4 cm. Thus C = 4 × 4 = 16 cm².
Example — Maze d: a length that is not a whole number on the way

Problem
Find missing left width D. The right rectangle has area 18 cm² and width 5 cm; the left rectangle has area 38 cm² and extends 4 cm higher.

  1. 1.The right height is 18 ÷ 5 = 3.6 cm. Here 3.6 means 3 whole centimetres and six tenths of another centimetre. You can check: 5 × 3.6 = 18.
  2. 2.The left height is 4 + 3.6 = 7.6 cm.
  3. 3.Its width is 38 ÷ 7.6 = 5 cm, checked by 5 × 7.6 = 38.
  4. 4.Thus D = 5 cm. As a second check, the 4 cm upper strip then has area 5 × 4 = 20 cm², and its lower part has the same 18 cm² area as the right rectangle: 20 + 18 = 38.

Combining areas and designing borders

The same tools solve practical layout questions. Work out which quantity is being requested before choosing an operation. Adding two areas gives an area, while drawing a border requires lengths and perimeter. Margins on opposite sides must both be removed from a page dimension.

Example — One rectangle with a combined area

Problem
Find dimensions of a rectangle whose area equals those of a 5 m by 10 m rectangle and a 2 m by 7 m rectangle together.

  1. 1.The two areas are 5 × 10 = 50 m² and 2 × 7 = 14 m².
  2. 2.Combined area = 50 + 14 = 64 m².
  3. 3.An 8 m by 8 m rectangle has that area. A 4 m by 16 m rectangle is another answer.
  4. 4.We are matching the total area; the question does not require the two uncut pieces to fit directly into the new rectangle.
Example — Space left for a lawn

Problem
Four 2 m by 1 m flower beds lie in the corners of a 15 m by 12 m garden. Find the lawn area.

  1. 1.The garden area is 15 × 12 = 180 m².
  2. 2.One bed has area 2 × 1 = 2 m². Four separate beds occupy 8 m².
  3. 3.The available lawn area is 180 − 8 = 172 m².
Example — A page border with unequal margins

Problem
A page is 24 cm high and 18 cm wide. A border is 1 cm from top and bottom, and 1.5 cm from left and right. Find the border perimeter.

  1. 1.Remove both vertical margins: border height = 24 − 1 − 1 = 22 cm.
  2. 2.Remove both horizontal margins: border width = 18 − 1.5 − 1.5 = 15 cm. The two 1.5 cm margins together use 3 cm.
  3. 3.Border perimeter = 2 × (22 + 15) = 74 cm.
  4. 4.For another page, measure its height and width first; do not assume this example’s dimensions.

In general, a page L cm high and W cm wide gives a border height L − 2 and width W − 3 under these margins. Its perimeter is 2 × [(L − 2) + (W − 3)] cm. The page must be large enough for the margins and the remaining border dimensions to be positive.

Quiz

Quick check

What is Charan’s whole plot width?

Quick check

What is Charan’s complete hall area?

Quick check

Which comparison of the two plots is correct?

Quick check

A rectangle has area 42 cm² and height 6 cm. What is its width?

Quick check

Maze a has top areas 13 and 26 cm², and lower-left area 15 cm². What is the lower-right area?

Quick check

A 24 cm by 18 cm page has the margins described above. What are the border dimensions?

Practice Problems

Practice Problems
  1. Find the sum of all regions in each floor plan and compare it with the whole rectangle area.
  2. In Charan’s plan, explain why the hall has an extra 5 ft by 5 ft section.
  3. A rectangular garden has length 50 m and area 1,000 m². Find its width and perimeter.
  4. A 5 m by 4 m floor contains a square carpet of side 3 m. Find the carpet area and uncovered area, and explain why neither is a perimeter.
  5. Redraw all four mazes and label each length as soon as you find it.
  6. Measure a real book page and draw the specified border. Calculate its perimeter from your measurements.
  7. Create a two-room rectangular plan with one room area and two useful lengths given. Exchange it with a partner and check the total areas.

Key Takeaways

Key Takeaways

• Use stated dimensions and aligned boundaries to infer missing room sizes. • Split an L-shaped room into rectangles and include every part. • Sum all region areas and check against the complete plot. • Equal-area plots can have different outside perimeters. • Known area divided by a known side gives the missing rectangle side. • Subtract margins on both sides before calculating a border perimeter.

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Finding Triangle Areas by Rearranging

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Chapter Summary and Practice