Areas and Perimeters of Polygons worksheet - Free Printable

Let's solve each polygon one by one, calculating the Area, Perimeter, and identifying the Type of polygon.

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1) Rectangle

- $ a = 88 \text{ cm}, b = 59 \text{ cm} $
- Type: Rectangle

Area:
$$
A = a \times b = 88 \times 59 = 5192 \text{ cm}^2
$$

Perimeter:
$$
P = 2(a + b) = 2(88 + 59) = 2(147) = 294 \text{ cm}
$$

✔ Area: 5192 cm²
✔ Perimeter: 294 cm
✔ Type: Rectangle

---

2) Rectangle

- $ a = 72 \text{ inches}, b = 46 \text{ inches} $
- Type: Rectangle

Area:
$$
A = 72 \times 46 = 3312 \text{ in}^2
$$

Perimeter:
$$
P = 2(72 + 46) = 2(118) = 236 \text{ inches}
$$

✔ Area: 3312 in²
✔ Perimeter: 236 in
✔ Type: Rectangle

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3) Triangle (Isosceles)

- $ a = 40 \text{ ft}, b = 72 \text{ ft} $ (two equal sides)
- Type: Isosceles Triangle

We need to find the height to compute area. Use Pythagorean theorem:
Split base $ a = 40 $ into two 20 ft segments.

$$
h = \sqrt{b^2 - (a/2)^2} = \sqrt{72^2 - 20^2} = \sqrt{5184 - 400} = \sqrt{4784} \approx 69.17 \text{ ft}
$$

Area:
$$
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 69.17 \approx 1383.4 \text{ ft}^2
$$

Perimeter:
$$
P = a + 2b = 40 + 2(72) = 40 + 144 = 184 \text{ ft}
$$

✔ Area: ≈ 1383.4 ft²
✔ Perimeter: 184 ft
✔ Type: Isosceles Triangle

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4) Triangle (Isosceles)

- $ a = 38 \text{ yds}, b = 75 \text{ yds} $
- Type: Isosceles Triangle

Height from apex to base $ a = 38 $:

$$
h = \sqrt{75^2 - (19)^2} = \sqrt{5625 - 361} = \sqrt{5264} \approx 72.56 \text{ yds}
$$

Area:
$$
A = \frac{1}{2} \times 38 \times 72.56 \approx 1378.64 \text{ yd}^2
$$

Perimeter:
$$
P = 38 + 75 + 75 = 188 \text{ yds}
$$

✔ Area: ≈ 1378.64 yd²
✔ Perimeter: 188 yds
✔ Type: Isosceles Triangle

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5) Square

- $ s = 63 \text{ mm} $
- Type: Square

Area:
$$
A = s^2 = 63^2 = 3969 \text{ mm}^2
$$

Perimeter:
$$
P = 4s = 4 \times 63 = 252 \text{ mm}
$$

✔ Area: 3969 mm²
✔ Perimeter: 252 mm
✔ Type: Square

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6) Square

- $ s = 65 \text{ yds} $
- Type: Square

Area:
$$
A = 65^2 = 4225 \text{ yd}^2
$$

Perimeter:
$$
P = 4 \times 65 = 260 \text{ yds}
$$

✔ Area: 4225 yd²
✔ Perimeter: 260 yds
✔ Type: Square

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7) Parallelogram

- $ a = 54.96 \text{ ft}, c = 89 \text{ ft}, h = 52 \text{ ft} $
- Opposite sides are equal → $ a = 54.96 $, $ c = 89 $

Area:
$$
A = \text{base} \times \text{height} = 89 \times 52 = 4628 \text{ ft}^2
$$

Perimeter:
$$
P = 2(a + c) = 2(54.96 + 89) = 2(143.96) = 287.92 \text{ ft}
$$

✔ Area: 4628 ft²
✔ Perimeter: 287.92 ft
✔ Type: Parallelogram

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8) Trapezoid

- $ a1 = 100 \text{ cm}, a2 = 43 \text{ cm}, b1 = 68.28 \text{ cm}, b2 = 58.8 \text{ cm}, h = 56 \text{ cm} $

This is a trapezoid with two non-parallel sides $ b1 $ and $ b2 $, and two bases $ a1 $ and $ a2 $. The height is given.

Area:
$$
A = \frac{1}{2} (a1 + a2) \times h = \frac{1}{2}(100 + 43) \times 56 = \frac{1}{2}(143) \times 56 = 71.5 \times 56 = 4004 \text{ cm}^2
$$

Perimeter:
$$
P = a1 + a2 + b1 + b2 = 100 + 43 + 68.28 + 58.8 = 270.08 \text{ cm}
$$

✔ Area: 4004 cm²
✔ Perimeter: 270.08 cm
✔ Type: Trapezoid

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9) Parallelogram

- $ a = 61 \text{ inches}, h = 55.28 \text{ inches} $
- Since it's a parallelogram, opposite sides are equal.

Area:
$$
A = \text{base} \times \text{height} = 61 \times 55.28 = 3372.08 \text{ in}^2
$$

But we don’t have the side length adjacent to base $ a $, so we cannot compute perimeter unless we assume something.

Wait — in the diagram, only $ a $ and $ h $ are given. But for perimeter, we need both pairs of sides.

However, in this case, only one side length is given ($ a = 61 $), and height $ h = 55.28 $. Without the slant side or angle, we cannot determine the other side.

But let’s check: if this is a parallelogram, then two sides are $ a $, and two others are unknown unless more info is given.

Wait — perhaps the figure implies that it's a rectangle? But no — height is drawn as perpendicular, but not necessarily right angles.

Actually, in most such problems, if only $ a $ and $ h $ are given, and no other side, we can't find perimeter unless we assume it's a rectangle.

But here, the shape looks like a general parallelogram.

So unless we know the slant side, we cannot compute perimeter.

But wait — maybe we're supposed to use the fact that the height is related to the side via trigonometry?

But no angle is given.

Alternatively, perhaps the side $ a $ is the base, and the other side is missing.

But in the diagram, only $ a $ and $ h $ are labeled.

So unless there's a mistake, we cannot compute perimeter without knowing the slant side.

But wait — look at the notation: It says “a” on both bottom and top, and “h” as height.

But no other side is labeled.

So likely, this is a rhombus or rectangle? No — height ≠ side.

Unless it's a rectangle, but then $ h $ should be the vertical side.

But if $ a = 61 $, $ h = 55.28 $, and it's a rectangle, then the other side would be $ h $, but then it wouldn't make sense because $ h $ is height.

Wait — actually, in a parallelogram, the height is perpendicular to the base.

So if $ a = 61 $ is the base, and $ h = 55.28 $ is the height, then area is fine.

But the other side (the slant side) is not known.

But in many textbook problems, sometimes they give enough data.

Wait — perhaps the side is also $ a $? No — it's labeled as $ a $, but only one value.

Wait — maybe the diagram shows a parallelogram with side $ a $, and height $ h $, and we are to assume that the other side is not given, so perimeter cannot be computed?

But that seems odd.

Alternatively, maybe the side $ a $ is the base, and the other side is unknown, so we can't compute perimeter.

But perhaps in this context, the figure is a rectangle? Then $ h $ would be the other side.

But if it were a rectangle, then $ h = $ other side, and $ a = 61 $, $ h = 55.28 $, so:

Then:

Area: $ 61 \times 55.28 = 3372.08 $ in²

Perimeter: $ 2(61 + 55.28) = 2(116.28) = 232.56 $ in

But is it a rectangle? The diagram shows a slanted side, so likely not.

But if it's a parallelogram, we need the length of the slant side.

But it's not given.

Wait — perhaps the height and base are sufficient for area, but not for perimeter.

But in some problems, they may expect us to assume it's a rhombus or something.

Alternatively, maybe the side $ a $ is both the base and the side? That would mean all sides are $ a $, i.e., rhombus.

But in that case, the height would be $ h = a \sin(\theta) $, but we can't get $ \theta $.

Wait — if it's a rhombus, then all sides are $ a = 61 $, and height $ h = 55.28 $, then area = $ a \times h = 61 \times 55.28 = 3372.08 $ in²

And perimeter = $ 4 \times 61 = 244 $ in

That makes sense.

But is it a rhombus? The diagram doesn't say.

But in many such problems, when only one side and height are given, and the shape is labeled with one letter $ a $, it might imply all sides are equal.

But in this case, it's labeled only as $ a $, and height $ h $, so likely it is a rhombus.

Alternatively, maybe it's just a parallelogram with base $ a $, and side $ a $? That would be a rhombus.

Yes — if both sides are labeled $ a $, then it's a rhombus.

Looking back: In problem 9, it shows:

```
a
/ \
/____\
a
```

With height $ h $ drawn.

So yes — both sides are labeled $ a $, so all four sides are equal → Rhombus

So:

- All sides = $ a = 61 $ inches
- Height $ h = 55.28 $ inches

Area:
$$
A = \text{base} \times \text{height} = 61 \times 55.28 = 3372.08 \text{ in}^2
$$

Perimeter:
$$
P = 4 \times 61 = 244 \text{ in}
$$

✔ Area: 3372.08 in²
✔ Perimeter: 244 in
✔ Type: Rhombus

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✔ Final Answers Summary:

| # | Area | Perimeter | Type |
|---|------|-----------|-------|
| 1 | 5192 cm² | 294 cm | Rectangle |
| 2 | 3312 in² | 236 in | Rectangle |
| 3 | ~1383.4 ft² | 184 ft | Isosceles Triangle |
| 4 | ~1378.64 yd² | 188 yds | Isosceles Triangle |
| 5 | 3969 mm² | 252 mm | Square |
| 6 | 4225 yd² | 260 yds | Square |
| 7 | 4628 ft² | 287.92 ft | Parallelogram |
| 8 | 4004 cm² | 270.08 cm | Trapezoid |
| 9 | 3372.08 in² | 244 in | Rhombus |

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Let me know if you'd like these filled out in the worksheet format!