Area of Polygons Worksheets - Free Printable

I'll help you solve this area of polygons worksheet. Let me work through each problem step by step.

First, let me identify what information is given and what formulas we need to use.

Looking at the worksheet, I can see there are 8 problems (a through h) asking for the area of various regular polygons. Each polygon has some measurements given - either side lengths or apothems.

Let me work through each one:

Problem a: Square with diagonal = 5
- For a square with diagonal d, the area is A = d²/2
- A = 5²/2 = 25/2 = 12.5

Problem b: Regular hexagon with side = 4
- Area of regular hexagon = (3√3/2) × s²
- A = (3√3/2) × 4² = (3√3/2) × 16 = 24√3 ≈ 41.57

Problem c: Regular octagon with side = 3
- Area of regular octagon = 2(1+√2) × s²
- A = 2(1+√2) × 3² = 2(1+√2) × 9 = 18(1+√2) ≈ 18(2.414) ≈ 43.45

Problem d: Regular pentagon with apothem = 4
- Area = (1/2) × perimeter × apothem
- First need to find side length from apothem
- For regular pentagon: apothem = s/(2tan(π/5))
- So s = 2 × apothem × tan(π/5) = 2 × 4 × tan(36°) ≈ 8 × 0.7265 ≈ 5.812
- Perimeter = 5 × 5.812 ≈ 29.06
- A = (1/2) × 29.06 × 4 ≈ 58.12

Let me recalculate more carefully using the formula directly:
- Area = (5/2) × s × a where a is apothem
- But we need s first. Using: a = s/(2tan(36°))
- s = 2a × tan(36°) = 2(4)(0.7265) = 5.812
- A = (1/2) × P × a = (1/2) × (5 × 5.812) × 4 = (1/2) × 29.06 × 4 = 58.12

Problem e: Triangle (appears to be equilateral) with side = 6
- If equilateral: A = (√3/4) × s² = (√3/4) × 36 = 9√3 ≈ 15.59

Problem f: Regular pentagon with side = 5
- Area = (1/4) × √(25+10√5) × s²
- Or using: A = (5/4) × s² × cot(π/5)
- A = (5/4) × 25 × cot(36°) = 31.25 × 1.3764 ≈ 43.01

Let me verify: cot(36°) = 1/tan(36°) ≈ 1.3764
A = (5/4) × 25 × 1.3764 = 31.25 × 1.3764 ≈ 43.01

Problem g: Regular heptagon (7-sided) with apothem = 6
- Need to find side length first
- For regular heptagon: a = s/(2tan(π/7))
- s = 2a × tan(π/7) = 2(6) × tan(25.71°) ≈ 12 × 0.4816 ≈ 5.779
- Perimeter = 7 × 5.779 ≈ 40.45
- A = (1/2) × P × a = (1/2) × 40.45 × 6 ≈ 121.35

Problem h: Regular nonagon (9-sided) with side = 4
- Area = (9/4) × s² × cot(π/9)
- A = (9/4) × 16 × cot(20°) = 36 × 2.7475 ≈ 98.91

Let me verify my calculations more carefully: