I'll solve each triangle area problem step by step using the formula for the area of a triangle:
Area = (base × height) ÷ 2
Let me go through each triangle:
1.
Triangle 1: Base = 4.6 ft, Height = 7.1 ft
- Area = (4.6 × 7.1) ÷ 2 = 32.66 ÷ 2 =
16.33 ft²
2.
Triangle 2: Base = 13.3 m, Height = 14.2 m
- Area = (13.3 × 14.2) ÷ 2 = 188.86 ÷ 2 =
94.43 m²
3.
Triangle 3: Base = 14.2 mm, Height = 10.3 mm
- Area = (14.2 × 10.3) ÷ 2 = 145.66 ÷ 2 =
72.83 mm²
4.
Triangle 4: Base = 5.2 mm, Height = 7.5 mm
- Area = (5.2 × 7.5) ÷ 2 = 39 ÷ 2 =
19.5 mm²
5.
Triangle 5: This is an equilateral triangle with all sides 12.7 ft
- For equilateral triangles: Area = (s² × √3) ÷ 4 where s is side length
- Area = (12.7² × √3) ÷ 4 = (161.29 × 1.732) ÷ 4 ≈ 280.4 ÷ 4 =
70.1 ft²
- Alternatively, if we use base and height: Area = (12.7 × 12.7) ÷ 2 = 161.29 ÷ 2 =
80.645 ft²
- Wait, this doesn't match. Let me recalculate:
- Actually, since it's labeled as having height 12.7 ft, I'll use: Area = (12.7 × 12.7) ÷ 2 =
80.645 ft²
6.
Triangle 6: Base = 12.5 cm, Height = 7.8 cm
- Area = (12.5 × 7.8) ÷ 2 = 97.5 ÷ 2 =
48.75 cm²
7.
Triangle 7: This appears to be an equilateral triangle with all sides 14.8 mm
- Area = (14.8² × √3) ÷ 4 = (219.04 × 1.732) ÷ 4 ≈ 379.3 ÷ 4 =
94.83 mm²
8.
Triangle 8: Base = 6.5 in, Height = 5.7 in
- Area = (6.5 × 5.7) ÷ 2 = 37.05 ÷ 2 =
18.525 in²
9.
Triangle 9: Base = 10 mm, Height = 7.6 mm
- Area = (10 × 7.6) ÷ 2 = 76 ÷ 2 =
38 mm²
10.
Triangle 10: This appears to be an equilateral triangle with all sides 6.7 mm
- Area = (6.7² × √3) ÷ 4 = (44.89 × 1.732) ÷ 4 ≈ 77.76 ÷ 4 =
19.44 mm²
Final Answers:
1. 16.33 ft²
2. 94.43 m²
3. 72.83 mm²
4. 19.5 mm²
5. 80.645 ft²
6. 48.75 cm²
7. 94.83 mm²
8. 18.525 in²
9. 38 mm²
10. 19.44 mm²
The key concept here is that for any triangle, you need the base and the corresponding height (the perpendicular distance from the base to the opposite vertex). The formula is always (base × height) ÷ 2.
Parent Tip: Review the logic above to help your child master the concept of area of a triangle worksheet 6th grade.