Mathematics Standard • Year 11 • Module 2 • Lesson 2
Area of Basic Shapes
Apply the area toolkit (rectangle, triangle, parallelogram, trapezium, circle, composites) to realistic measurement scenarios.
Problem 1, L-shaped bathroom floor (tiling cost)
A bathroom is L-shaped. The main rectangle is 4 m by 3 m, with a smaller rectangle (1.5 m by 2 m) added at one corner.
Set up: What are we solving for?
(i) Plan: write "Area = rectangle 1 + rectangle 2" and calculate each separately. 2 marks
(ii) Calculate the total floor area in m². 1 mark
(iii) Tiles cost $52 per m². Find the total cost to tile the floor. 2 marks
Stuck? Revisit lesson § Composite Shapes, write the plan first ("Area = ___ + ___") before doing any arithmetic.Problem 2, Swimming pool with semicircular end (composite)
A swimming pool consists of a rectangle 12 m long and 6 m wide, with a semicircle attached to one short end (diameter = 6 m).
Set up: What are we solving for?
(i) Find the radius of the semicircle. 1 mark
(ii) Calculate the total surface area of the pool to 2 d.p. (rectangle + semicircle). 3 marks
(iii) A pool cover costs $24 per m². Find the cost of a cover for the whole pool surface. 2 marks
Stuck on (ii)? Asemi = ½πr². Keep π exact until the very last step.Problem 3, Square garden with circular pond (subtraction)
A square backyard garden is 8 m on each side. A circular pond of radius 1.5 m is built in the centre.
Set up: What are we solving for?
(i) Plan: write "Area of grass = square − circle" and calculate each to 2 d.p. 2 marks
(ii) Calculate the area of the grass (square − pond) to 2 d.p. 1 mark
(iii) Grass seed is sold in 1 kg bags that cover 25 m². How many bags must the gardener buy? Justify your rounding direction. 2 marks
Stuck on (iii)? Divide grass area by 25 to get the number of bags needed, then round in the direction that ensures full coverage.Problem 4, Trapezoidal block of land
A block of suburban land is shaped like a trapezium. The two parallel sides (front and back boundary) are 18 m and 26 m, and the perpendicular distance between them (the depth of the block) is 32 m.
Set up: What are we solving for?
(i) Label a, b and h, then write the formula. 1 mark
(ii) Calculate the area of the block in m². 2 marks
(iii) Land in this suburb sells for $1,850 per m². Find the value of the block. 2 marks
Stuck? Revisit lesson § Worked Example 3, Trapezium. h must be perpendicular to a and b (the parallel sides), not the slant boundary.Problem 5, Athletics oval (rectangle + two semicircles)
A school athletics oval has a rectangular infield 100 m long and 60 m wide, with a semicircle of grass attached to each short end (diameter = 60 m on each end).
Set up: What are we solving for?
(i) Plan: identify how many semicircles and how they combine into one full circle. 1 mark
(ii) Calculate the total grass area to 2 d.p. (rectangle + one full circle, since two equal semicircles combine into one). 3 marks
(iii) The school mows the grass weekly. The mower cuts 800 m² per hour. Find the time (in hours and minutes) to mow the whole oval, rounded UP to the nearest minute. 2 marks
Stuck on (iii)? Divide area by rate to get decimal hours, then convert the decimal part (× 60) to minutes.How did this worksheet feel?
What I'll revisit before next class:
Problem 1, L-shaped bathroom
Set up. We are splitting the L into two rectangles, summing the areas, then multiplying by the tile rate.
(i) Plan: Area = rectangle 1 + rectangle 2. Rectangle 1 = 4 × 3 = 12 m². Rectangle 2 = 1.5 × 2 = 3 m².
(ii) Total area = 12 + 3 = 15 m².
(iii) Cost = 15 × $52 = $780.
Problem 2, Pool with semicircular end
Set up. We are decomposing into rectangle + semicircle, summing, then multiplying by the cover rate.
(i) r = 6 ÷ 2 = 3 m.
(ii) Rectangle = 12 × 6 = 72 m². Semicircle = ½ × π × 3² = ½ × 9π = 4.5π = 14.137... m². Total = 72 + 14.14 = 86.14 m² (to 2 d.p.).
(iii) Cost = 86.14 × $24 = $2,067.36.
Problem 3, Garden with pond
Set up. We are subtracting the pond area from the square, then finding the number of seed bags needed.
(i) Plan: Areagrass = square − circle. Square = 8² = 64 m². Circle = π × 1.5² = 2.25π = 7.0685... m².
(ii) Grass = 64 − 7.07 = 56.93 m² (to 2 d.p.).
(iii) Bags needed = 56.93 ÷ 25 = 2.277... → round UP to 3 bags, because 2 bags would only cover 50 m², which is not enough to seed the full lawn.
Problem 4, Trapezoidal block
Set up. We are applying A = ½(a + b)h to a real block of land, then valuing it.
(i) a = 18, b = 26, h = 32. Formula: A = ½(a + b)h.
(ii) A = ½(18 + 26) × 32 = ½ × 44 × 32 = 704 m².
(iii) Value = 704 × $1,850 = $1,302,400.
Problem 5, Athletics oval
Set up. We are recognising that two equal semicircles make one full circle, summing rectangle + circle, then dividing by the mow rate.
(i) Two equal semicircles of radius 30 m combine into one full circle of radius 30 m.
(ii) Rectangle = 100 × 60 = 6000 m². Circle = π × 30² = 900π = 2827.43... m². Total = 6000 + 2827.43 = 8827.43 m² (to 2 d.p.).
(iii) Hours = 8827.43 ÷ 800 = 11.034... hr. Decimal part: 0.034 × 60 = 2.05 min → round UP to 3 min. Total ≈ 11 hours 3 minutes. (Marking note: rounding to "11 hours 2 minutes" is acceptable if the unrounded value is carried through. The key is rounding UP, not down, when the question demands full coverage.)