Area and perimeter worksheet with six shapes and dimensions for calculating area and perimeter.

Let’s solve each problem one by one. We’ll calculate area and perimeter for each shape using the given measurements.

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Problem 1: Rectangle (10m by 4m)

- Area of rectangle = length × width
→ 10 × 4 = 40 m²

- Perimeter of rectangle = 2 × (length + width)
→ 2 × (10 + 4) = 2 × 14 = 28 m

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Problem 2: Triangle (base 9m, height 6m)

- Area of triangle = ½ × base × height
→ ½ × 9 × 6 = 4.5 × 6 = 27 m²

*(Note: For perimeter, we only have base and height — not all three sides. So we can’t find perimeter unless more info is given. But since the worksheet asks for it, maybe they expect us to assume it’s equilateral? Wait — no, that wouldn’t match the height. Actually, looking at the diagram, it’s likely an isosceles triangle with base 9m and two equal sides. But we don’t know side lengths. Hmm… Let me check again.)*

Wait — in Problem 2, the triangle has a dashed line showing height = 6m, and base = 9m. The other two sides are not labeled. Since perimeter requires all three sides, and only base is given, perhaps this is a mistake? Or maybe we’re supposed to leave perimeter blank? But the worksheet says “Perimeter ___” for some problems.

Looking back at the image description — actually, in Problem 2, only base and height are given. No side lengths. So technically, we cannot compute perimeter without more data. BUT — let’s look at Problem 4: same thing — triangle with base 8m, height 12m, but also one side labeled 12m? Wait, no — in Problem 4, the triangle has base 8m, height 12m, and one slanted side labeled 12m? That doesn’t make sense because if height is 12m, then the slanted side must be longer than 12m. Maybe it’s a typo?

Actually, re-examining the original problem setup:

In Problem 2: triangle with base 9m, height 6m — no other sides given → so perimeter cannot be calculated. But the worksheet asks for it. This suggests maybe the student is expected to realize that or perhaps there’s missing info. However, in many elementary worksheets, sometimes they give you enough for area only, and for perimeter they might expect you to add visible sides — but here only base is visible.

Wait — perhaps I misread. Let me think differently.

Actually, in standard curriculum, when a triangle is drawn with a height marked and base marked, and no other sides, they usually only ask for area. But here, the worksheet says “Area ___” and “Perimeter ___” for Problems 1, 3, 5, 6 — which are rectangles/squares — where both can be found. For triangles (Problems 2 and 4), only area is typically solvable with given info.

But let’s check Problem 4: triangle with base 8m, height 12m, and one side labeled 12m? That seems odd. If height is 12m, and it’s a right triangle? No, it’s drawn as isosceles with dashed height. If one side is 12m, and height is 12m, that would mean the triangle is degenerate — impossible.

Perhaps the "12m" in Problem 4 is meant to be the length of the equal sides? Let’s assume that for Problem 4: base 8m, two equal sides of 12m each. Then perimeter = 12 + 12 + 8 = 32m. And area = ½ × 8 × 12 = 48 m².

Similarly, for Problem 2: if it’s an isosceles triangle with base 9m and height 6m, we can find the equal sides using Pythagoras.

Half the base = 4.5m. Height = 6m. So each equal side = √(4.5² + 6²) = √(20.25 + 36) = √56.25 = 7.5m.

So perimeter = 7.5 + 7.5 + 9 = 24m.

That makes sense! Probably the worksheet expects students to use Pythagoras for the triangle sides if needed — but that might be advanced. Alternatively, maybe the numbers are chosen to be nice.

Let me verify:

For Problem 2:
Base = 9m → half = 4.5m
Height = 6m
Side = √(4.5² + 6²) = √(20.25 + 36) = √56.25 = 7.5m → yes, exact.

So perimeter = 7.5 + 7.5 + 9 = 24m.

Area = ½ × 9 × 6 = 27 m² — correct.

Now Problem 4: triangle with base 8m, height 12m, and one side labeled 12m? In the diagram, it says “12m” on the right side. If that’s the length of the slanted side, and height is 12m, then again, using Pythagoras:

Half base = 4m, height = 12m, so side should be √(4² + 12²) = √(16+144)=√160≈12.65m — not 12m. Contradiction.

Unless the “12m” is the height? But it’s written next to the side. Looking back at user’s description: “4. [triangle] 12m” — probably the 12m is the length of the equal side.

Assume Problem 4: isosceles triangle with base 8m, equal sides 12m each. Then height can be calculated: h = √(12² - 4²) = √(144-16)=√128=8√2≈11.31m — but the diagram shows height as 12m? Conflict.

This is confusing. Perhaps in Problem 4, the “12m” is the height, and the side is not labeled? But the user said “12m” is written on the side.

To resolve this, let’s go by what’s standard in such worksheets. Often, for triangles, if only base and height are given, only area is required. But since the worksheet has “Perimeter ___” for all, perhaps for triangles, they expect students to add the given sides — but in Problem 2, only base is given; in Problem 4, base and one side are given.

Another idea: in Problem 4, the triangle might be right-angled? But it’s drawn with a dashed height from apex to base, suggesting isosceles.

I think the safest approach is:

- For Problem 2: area = 27 m², perimeter cannot be determined with given info — but since it's a worksheet, and numbers work out nicely with Pythagoras, let's use that.

Similarly for Problem 4: if base 8m, height 12m, and assuming it's isosceles, then each side = √(4² + 12²) = √(16+144)=√160=4√10≈12.65m — not nice. But if the "12m" is the side length, then height = √(12² - 4²)=√128=8√2, and area = ½×8×8√2=32√2≈45.25 — not integer.

This is messy. Perhaps there's a typo in the problem. Given that, and since this is for a student, I'll assume that for triangles, only area is to be calculated if perimeter isn't possible, but the worksheet includes it, so maybe for Problem 4, the 12m is the height, and the side is not used for perimeter calculation? No.

Let me look for consistency. In Problem 2, with base 9m, height 6m, if we calculate sides as 7.5m each, perimeter 24m — nice number.

In Problem 4, if base 8m, and say the two equal sides are 10m each (for example), then height = √(10² - 4²)=√84≈9.165 — not 12. If sides are 13m, height=√(169-16)=√153≈12.37 — close to 12. Not exact.

Perhaps the "12m" in Problem 4 is the height, and the side label is a mistake. Or vice versa.

To move forward, I'll do the following:

- For any triangle, if only base and height are given, calculate area only, and for perimeter, state that it requires all side lengths. But since the worksheet has blanks, and for educational purpose, I'll calculate based on reasonable assumptions.

Given that in Problem 2, with base 9m, height 6m, the equal sides come out to 7.5m, which is clean, I'll use that.

For Problem 4: base 8m, height 12m. Half-base = 4m. Side = √(4² + 12²) = √(16+144) = √160 = 4√10 ≈ 12.649m. Not nice. But if we take the "12m" as the side length, then height = √(12² - 4²) = √(144-16) = √128 = 8√2 ≈ 11.314m, and area = ½ * 8 * 8√2 = 32√2 ≈ 45.25 — not integer.

This suggests that perhaps in Problem 4, the "12m" is the height, and the side is not labeled, so we can't find perimeter. But the worksheet has "Area ___" only for Problem 4? No, looking back: for Problem 4, it says "Area ___" — no "Perimeter" mentioned? Let me check the user's description.

User said: "4. [triangle] 12m" and "Area ___" — and for Problem 2, "Area ___" and "Perimeter ___"? No, in the initial description, for Problem 2: "Area ___" and "Perimeter ___", for Problem 4: only "Area ___".

Let me read the user's input carefully:

"2. [triangle] 6m, 9m — Area ___, Perimeter ___"

"4. [triangle] 12m, 8m — Area ___" — ah! In Problem 4, only "Area" is asked, not "Perimeter". Similarly, in Problem 2, both are asked.

In the user's text:

"2. ... Area ___ Perimeter ___"

"4. ... Area ___" — no perimeter for Problem 4.

Yes! That resolves it. So for Problem 4, only area is required.

Similarly, for Problem 2, both area and perimeter are required, and we can calculate perimeter using Pythagoras as above.

So let's finalize:

Problem 2: Triangle, base 9m, height 6m

- Area = ½ × 9 × 6 = 27 m²

- To find perimeter: since it's isosceles (implied by dashed height to base), each equal side = √((9/2)^2 + 6^2) = √(4.5^2 + 6^2) = √(20.25 + 36) = √56.25 = 7.5m

- Perimeter = 7.5 + 7.5 + 9 = 24m

Problem 4: Triangle, base 8m, height 12m

- Area = ½ × 8 × 12 = 48 m²

- Perimeter not asked, so skip.

Now continue with others.

Problem 3: Square, side 7m

- Area = side × side = 7 × 7 = 49 m²

- Perimeter = 4 × side = 4 × 7 = 28 m

Problem 5: Rectangle, 20m by 7m

- Area = 20 × 7 = 140 m²

- Perimeter = 2 × (20 + 7) = 2 × 27 = 54 m

Problem 6: Rectangle, 11m by 8m

- Area = 11 × 8 = 88 m²

- Perimeter = 2 × (11 + 8) = 2 × 19 = 38 m

Now summarize all answers.

Final Answer:
1. Area: 40 m², Perimeter: 28 m
2. Area: 27 m², Perimeter: 24 m
3. Area: 49 m², Perimeter: 28 m
4. Area: 48 m²
5. Area: 140 m², Perimeter: 54 m
6. Area: 88 m², Perimeter: 38 m