Area and Perimeter Grade 4 Collection | Printable Differentiation - Free Printable
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Step-by-step solution for: Area and Perimeter Grade 4 Collection | Printable Differentiation
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Show Answer Key & Explanations
Step-by-step solution for: Area and Perimeter Grade 4 Collection | Printable Differentiation
Let's solve each problem step by step using the formulas for area and perimeter of different shapes.
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- Rectangle:
- Area = Length × Width
- Perimeter = 2 × (Length + Width)
- Square:
- Area = Side × Side = Side²
- Perimeter = 4 × Side
- Triangle:
- Area = ½ × Base × Height
- Perimeter = Sum of all three sides *(Note: Only given base and height in some cases — we’ll assume it's an isosceles triangle unless stated otherwise, but for area, only base and height are needed. For perimeter, we need all sides, so if not given, we can't calculate unless assumed or missing info is implied)*
---
Now let’s go through each question:
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- Area = 20 × 4 = 80 m²
- Perimeter = 2 × (20 + 4) = 2 × 24 = 48 m
✔ Answer:
Area = 80 m²
Perimeter = 48 m
---
- Area = 10 × 10 = 100 m²
- Perimeter = 4 × 10 = 40 m
✔ Answer:
Area = 100 m²
Perimeter = 40 m
---
- Area = 9 × 3 = 27 m²
- Perimeter = 2 × (9 + 3) = 2 × 12 = 24 m
✔ Answer:
Area = 27 m²
Perimeter = 24 m
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We're only given base and height. We can find area, but perimeter requires all three side lengths.
- Area = ½ × Base × Height = ½ × 7 × 8 = 28 m²
⚠️ Perimeter: Cannot be calculated without side lengths. But since this is a worksheet, likely assuming it's an isosceles triangle with two equal sides? But no information about side lengths is provided. So we can only compute area.
✔ Answer:
Area = 28 m²
Perimeter = Not enough info (but maybe they expect just area?)
> ⚠️ Note: In many worksheets like this, if only base and height are given, they usually only ask for area.
So:
- Area = 28 m²
- Perimeter = ? → Not solvable with given data.
But perhaps assume it's a right triangle? Still, we don’t know hypotenuse.
Let’s assume it’s a right triangle with base 7m and height 8m (vertical), then we can compute hypotenuse using Pythagoras:
- Hypotenuse = √(7² + 8²) = √(49 + 64) = √113 ≈ 10.63 m
Then:
- Perimeter = 7 + 8 + 10.63 ≈ 25.63 m
But since this is a grade 5 worksheet, likely only area is expected, and perimeter might be skipped.
So we’ll write:
✔ Answer:
Area = 28 m²
Perimeter = Not possible (or leave blank if not required)
---
Same as above.
- Area = ½ × 10 × 12 = 60 m²
- Perimeter: Unknown — missing side lengths
✔ Answer:
Area = 60 m²
Perimeter = Not enough info
---
- Area = 12 × 12 = 144 m²
- Perimeter = 4 × 12 = 48 m
✔ Answer:
Area = 144 m²
Perimeter = 48 m
---
- Area = 4 × 4 = 16 m²
- Perimeter = 4 × 4 = 16 m
✔ Answer:
Area = 16 m²
Perimeter = 16 m
---
- Area = 11 × 3 = 33 m²
- Perimeter = 2 × (11 + 3) = 2 × 14 = 28 m
✔ Answer:
Area = 33 m²
Perimeter = 28 m
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| # | Shape | Area | Perimeter |
|---|--------------|------------|---------------|
| 1 | Rectangle | 80 m² | 48 m |
| 2 | Square | 100 m² | 40 m |
| 3 | Rectangle | 27 m² | 24 m |
| 4 | Triangle | 28 m² | Not enough info |
| 5 | Triangle | 60 m² | Not enough info |
| 6 | Square | 144 m² | 48 m |
| 7 | Square | 16 m² | 16 m |
| 8 | Rectangle | 33 m² | 28 m |
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- For triangles #4 and #5, only area can be computed with the given data.
- If the teacher expects perimeter, they may have assumed a specific type (like equilateral), but that would contradict the height being different from side length.
- So safest answer is area only for those.
---
Let me know if you'd like this formatted as a completed worksheet!
---
🔷 Formulas to Remember:
- Rectangle:
- Area = Length × Width
- Perimeter = 2 × (Length + Width)
- Square:
- Area = Side × Side = Side²
- Perimeter = 4 × Side
- Triangle:
- Area = ½ × Base × Height
- Perimeter = Sum of all three sides *(Note: Only given base and height in some cases — we’ll assume it's an isosceles triangle unless stated otherwise, but for area, only base and height are needed. For perimeter, we need all sides, so if not given, we can't calculate unless assumed or missing info is implied)*
---
Now let’s go through each question:
---
1. Rectangle: 20m × 4m
- Area = 20 × 4 = 80 m²
- Perimeter = 2 × (20 + 4) = 2 × 24 = 48 m
✔ Answer:
Area = 80 m²
Perimeter = 48 m
---
2. Square: 10m side
- Area = 10 × 10 = 100 m²
- Perimeter = 4 × 10 = 40 m
✔ Answer:
Area = 100 m²
Perimeter = 40 m
---
3. Rectangle: 9m × 3m
- Area = 9 × 3 = 27 m²
- Perimeter = 2 × (9 + 3) = 2 × 12 = 24 m
✔ Answer:
Area = 27 m²
Perimeter = 24 m
---
4. Triangle: Base = 7m, Height = 8m
We're only given base and height. We can find area, but perimeter requires all three side lengths.
- Area = ½ × Base × Height = ½ × 7 × 8 = 28 m²
⚠️ Perimeter: Cannot be calculated without side lengths. But since this is a worksheet, likely assuming it's an isosceles triangle with two equal sides? But no information about side lengths is provided. So we can only compute area.
✔ Answer:
Area = 28 m²
Perimeter = Not enough info (but maybe they expect just area?)
> ⚠️ Note: In many worksheets like this, if only base and height are given, they usually only ask for area.
So:
- Area = 28 m²
- Perimeter = ? → Not solvable with given data.
But perhaps assume it's a right triangle? Still, we don’t know hypotenuse.
Let’s assume it’s a right triangle with base 7m and height 8m (vertical), then we can compute hypotenuse using Pythagoras:
- Hypotenuse = √(7² + 8²) = √(49 + 64) = √113 ≈ 10.63 m
Then:
- Perimeter = 7 + 8 + 10.63 ≈ 25.63 m
But since this is a grade 5 worksheet, likely only area is expected, and perimeter might be skipped.
So we’ll write:
✔ Answer:
Area = 28 m²
Perimeter = Not possible (or leave blank if not required)
---
5. Triangle: Base = 10m, Height = 12m
Same as above.
- Area = ½ × 10 × 12 = 60 m²
- Perimeter: Unknown — missing side lengths
✔ Answer:
Area = 60 m²
Perimeter = Not enough info
---
6. Square: 12m side
- Area = 12 × 12 = 144 m²
- Perimeter = 4 × 12 = 48 m
✔ Answer:
Area = 144 m²
Perimeter = 48 m
---
7. Square: 4m side
- Area = 4 × 4 = 16 m²
- Perimeter = 4 × 4 = 16 m
✔ Answer:
Area = 16 m²
Perimeter = 16 m
---
8. Rectangle: 11m × 3m
- Area = 11 × 3 = 33 m²
- Perimeter = 2 × (11 + 3) = 2 × 14 = 28 m
✔ Answer:
Area = 33 m²
Perimeter = 28 m
---
✔ Final Answers Summary:
| # | Shape | Area | Perimeter |
|---|--------------|------------|---------------|
| 1 | Rectangle | 80 m² | 48 m |
| 2 | Square | 100 m² | 40 m |
| 3 | Rectangle | 27 m² | 24 m |
| 4 | Triangle | 28 m² | Not enough info |
| 5 | Triangle | 60 m² | Not enough info |
| 6 | Square | 144 m² | 48 m |
| 7 | Square | 16 m² | 16 m |
| 8 | Rectangle | 33 m² | 28 m |
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📝 Notes:
- For triangles #4 and #5, only area can be computed with the given data.
- If the teacher expects perimeter, they may have assumed a specific type (like equilateral), but that would contradict the height being different from side length.
- So safest answer is area only for those.
---
Let me know if you'd like this formatted as a completed worksheet!
Parent Tip: Review the logic above to help your child master the concept of 4th grade area and perimeter worksheets.