Grade 10 core mathematics Measurements and Geometry – Area of Polygons Notes

Specific learning outcomes

1. Identify and outline sub-sub-strands:
   • Area of a triangle using A = ½ab sin C
   • Area of a triangle using Heron's formula
   • Area of quadrilaterals (rectangle, square, parallelogram, trapezium, kite, irregular)
   • Area of regular and irregular polygons
   • Application of area of polygons in real life
2. Derive the formula for the area of a triangle given two sides and the included angle (A = ½ab sin C).
3. Work out the area of a triangle given two sides and included angle (worked examples and exercises).
4. Determine the area of a triangle using Heron's formula (derivation and practice).
5. Determine the area of various quadrilaterals in different situations (formulas and decomposition).
6. Work out the area of regular heptagon and octagon (use general formula and show numeric method).
7. Determine the area of irregular polygons (decompose into triangles, use the shoelace formula where appropriate).
8. Explore applications of polygon areas in real-life contexts (land, tiling, carpentry, design).

1. Area of a triangle using A = ½ab sin C — Derivation (Outcome b)

Consider triangle ABC with sides a = BC, b = AC and included angle C (angle between a and b). Drop a perpendicular from A to side BC (or its extension) to form height h.

Worked example 1 (Outcome c)

Find area of triangle with a = 8 m, b = 6 m and included angle C = 50°.

Use A = ½ ab sin C = 0.5 × 8 × 6 × sin 50° = 24 × 0.7660 ≈ 18.384 m². Answer ≈ 18.38 m².

(Use sin 50° ≈ 0.7660.)

2. Area using Heron's formula (Outcome d)

For triangle with sides a, b, c: let s = (a + b + c)/2 (semiperimeter). Heron's formula:

A = √[s(s − a)(s − b)(s − c)].

Derivation sketch

1. Use A = ½bc sin A. Express cos A from law of cosines: cos A = (b² + c² − a²)/(2bc).
2. Get sin²A = 1 − cos²A and substitute into A² = (¼)b²c² sin²A. After algebraic simplification, the expression reduces to s(s−a)(s−b)(s−c).

Worked example 2

Sides: 7 m, 8 m, 9 m. Compute area.

s = (7+8+9)/2 = 12. So A = √[12(12−7)(12−8)(12−9)] = √[12×5×4×3] = √[720] ≈ 26.833 m².

3. Area of quadrilaterals (Outcome e)

Common formulas and methods:

• Rectangle: A = length × width (A = lw).
• Square: A = side² (A = s²).
• Parallelogram: A = base × height (A = b × h).
• Trapezium (trapezoid): A = ½(h)(sum of parallel sides) = ½(a + b)h.
• Kite (with diagonals d1 and d2): A = ½ d1 d2.
• Irregular quadrilateral: split into two triangles (or use coordinates and the shoelace formula).

Example: trapezium

Parallel sides 10 m and 6 m, height 4 m. A = ½(10+6)×4 = 8×4 = 32 m².

Decomposition method

For irregular quadrilaterals, draw a diagonal to form two triangles, find areas (by ½bh, ½ ab sin C or Heron) and add.

4. Area of regular polygons (Outcome f)

General formula for a regular n-gon with side length s:

A = (n s²) / (4 tan(π/n)) or A = ½ × perimeter × apothem.

Regular octagon (n = 8)

Using known simplification: A = 2(1 + √2) s² ≈ (4.8284) s².

Example: Regular octagon with side s = 5 m: A = 2(1+√2)×25 ≈ 2(2.4142)×25 ≈ 120.71 m².

Regular heptagon (n = 7)

No simple radical form; use formula with tan or apothem:

A = (7 s²) / (4 tan(π/7)) ≈ 3.6339 s² (approx).

Example: s = 4 m → A ≈ 3.6339×16 ≈ 58.14 m².

(Use a calculator to evaluate tan(π/7) or the numeric factor.)

5. Area of irregular polygons (Outcome g)

Two standard methods:

1. Decompose into triangles (use ½ab sin C or Heron's formula for each triangle) and add areas.
2. Shoelace (coordinate) formula — useful when vertex coordinates are known:

Shoelace formula (for polygon with vertices (x1,y1),(x2,y2),...,(xn,yn) listed in order):

A = ½ |Σ (xi yi+1 − xi+1 yi)|, where xn+1 = x1, yn+1 = y1.

Short example (shoelace)

Vertices: (0,0), (4,0), (3,3), (0,2). Compute area:

Compute sum1 = 0×0 + 4×3 + 3×2 + 0×0 = 0 + 12 + 6 + 0 = 18.

Compute sum2 = 0×4 + 0×3 + 3×0 + 2×0 = 0 + 0 + 0 + 0 = 0.

A = ½ |18 − 0| = 9 units².

6. Applications & real-life situations (Outcome h)

Practical contexts where polygon area is needed (class activities and field tasks):

• Measuring land plots (many plots are irregular polygons): students measure sides/angles or GPS coordinates and use decomposition or shoelace to find area.
• Tiling a room (regular rectangles) or courtyard paving with paving stones (regular polygons) — compute area to estimate materials.
• Carpentry — cutting boards in polygonal shapes, determining paint or varnish needed.
• Design: school garden beds shaped as polygons (plan the area for planting).

Suggested field activity: In groups, measure a small school yard area that is not a simple rectangle. Record vertex coordinates (approx), use shoelace formula or divide into triangles, compute area, compare methods and discuss precision and sources of error.

7. Classroom suggested learning experiences & activities

1. Start with simple inspection: identify polygons around the school (tiles, windows, fields). List which are regular/irregular.
2. Derive A = ½ab sin C in class using a drawn triangle; have learners measure and verify with a protractor and ruler.
3. Practise Heron's formula with triangles of different sides; compare results with ½ab sin C where an angle is known.
4. Group practical: measure a plot (teachers may provide measuring tapes and protractors). Decompose into triangles and compute area. Discuss rounding and measurement error.
5. Introduce regular polygon area formula: have learners draw a regular octagon and compute area by dividing into 8 isosceles triangles (each triangle area = ½ r s, or use cot formula).
6. Shoelace practice: give coordinate vertices and ask learners to compute areas; then plot and verify with graph paper.
7. Project: design a small garden bed in the shape of an irregular polygon, calculate area and estimate soil/seed needed (costing exercise).

8. Practice exercises (with brief answers)

1. Triangle with sides 9 m and 7 m, included angle 60°. Find area. (A = ½×9×7×sin60° = 31.5×0.8660 ≈ 27.28 m².)
2. Triangle sides 5, 6, 7. Use Heron to find area. (s=9; A = √[9×4×3×2]=√216≈14.697 ≈14.70.)
3. Regular octagon side 3 m: area? (A = 2(1+√2) s² ≈ 4.8284×9 ≈ 43.456 m².)
4. Quadrilateral with vertices (0,0),(6,0),(5,4),(0,3) — find area by shoelace. (Compute → A = ½| (0×0+6×4+5×3+0×0) − (0×6+0×5+4×0+3×0)| = ½|0+24+15+0 − 0|=½×39 = 19.5.)
5. Trapezium with parallel sides 12 m and 8 m, height 5 m. Area? (A = ½(12+8)×5 = 10×5 = 50 m².)

Teachers: adapt numbers for class level and allow calculator use. Encourage showing full working steps.

9. Assessment ideas

• Written test: mixture of formula recall, derivation short answer, and computation (A = ½ab sin C, Heron, quadrilateral formulas, regular polygon area).
• Practical assessment: field measurement of an irregular plot and report showing method, calculations, and reflection on errors.
• Project: design and cost a tiling plan for a polygon-shaped area using area calculations and material rates.