Math worksheet to calculate area and perimeter of different polygons.

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

---

1) Rectangle

- $ a = 88 \text{ cm},\ b = 59 \text{ cm} $
- Type: Rectangle (a quadrilateral with four right angles)

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

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

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

---

2) Rectangle

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

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

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

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

---

3) Isosceles Triangle

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

Area:
We need height. Use Pythagorean theorem to find height from base $ a = 40 $ ft.

Split base into two: $ 20 $ ft each.
$$
h = \sqrt{b^2 - (a/2)^2} = \sqrt{72^2 - 20^2} = \sqrt{5184 - 400} = \sqrt{4784} \approx 69.16 \text{ ft}
$$

Now,
$$
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 69.16 \approx 1383.2 \text{ ft}^2
$$

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

✔ Area: ≈ 1383.2 ft²
✔ Perimeter: 184 ft
✔ Type: Isosceles Triangle

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4) Isosceles Triangle

- $ a = 38 \text{ yds},\ b = 75 \text{ yds} $ (two equal sides)

Height:
Split base: $ 19 $ yds
$$
h = \sqrt{75^2 - 19^2} = \sqrt{5625 - 361} = \sqrt{5264} \approx 72.56 \text{ yds}
$$

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

---

5) Square

- $ s = 63 \text{ mm} $
- Type: Square (all sides equal, all angles 90°)

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

---

6) Square

- $ s = 65 \text{ yds} $

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

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

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

---

7) Parallelogram

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

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

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

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

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8) Trapezoid

- $ a1 = 100 \text{ cm},\ a2 = 43 \text{ cm} $ (bases)
- $ b1 = 68.28 \text{ cm},\ b2 = 58.8 \text{ cm} $ (legs)
- $ h = 56 \text{ cm} $
- Type: Trapezoid (one pair of parallel sides)

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

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

---

9) Parallelogram

- $ a = 61 \text{ inches},\ h = 55.28 \text{ inches} $
- Since it's a parallelogram, opposite sides are equal
- Base = 61, height = 55.28
- Type: Parallelogram

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

Perimeter:
We don’t have side length other than base. But since only one side is given and height, we can't compute perimeter unless we know slant side or angle.

Wait — the diagram shows a parallelogram with base $ a $ and height $ h $. But only one side (a) is given, and no other side.

But in a parallelogram, opposite sides are equal, but adjacent sides may differ.

We need the length of the non-base side to compute perimeter.

However, we’re not given that, but perhaps we can assume that the side shown is the same as the base? No — that would make it a rhombus, but not necessarily.

But here’s the issue: Only one side is labeled: $ a = 61 $ inches, and height $ h = 55.28 $. We cannot determine the other side without more info.

Wait — let's re-examine the diagram. It appears that both bottom and top are labeled $ a $, and the vertical height is $ h $. But the slanted side is not labeled.

So we cannot calculate perimeter unless we know the slant side.

But maybe it's intended to be a rectangle? But no, it’s drawn as a parallelogram.

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

But wait — if we use the height, and the side forming the angle, we might compute the other side?

No — unless we know the angle, we can’t.

But perhaps there's an assumption: maybe this is a rhombus? But only one side is given.

Wait — actually, looking at the label: it says $ a $ on the base and $ a $ on the top, and then $ h $ is the height. The left side is not labeled.

But in standard labeling, if it's a parallelogram, and only $ a $ is given, and height $ h $, then:

- Area = $ a \times h = 61 \times 55.28 = 3372.08 \text{ in}^2 $
- But for perimeter: $ P = 2(a + b) $, where $ b $ is the slant side.

But $ b $ is unknown.

Unless... is the side $ a $ meant to be the same as the slant side? But no — it’s labeled on the base.

Wait — perhaps the side $ a $ is the length of the slanted side, and the base is also $ a $? That would imply a rhombus.

But the diagram shows $ a $ on the base and $ a $ on the top, but the vertical height is $ h $, and the slanted side is not labeled.

But in many such problems, when they label the base as $ a $, and give height, and say "parallelogram", they expect you to compute area with $ A = a \times h $, but perimeter requires another side.

But here, no other side is given.

Wait — look again: it says $ a = 61 $ inches, and $ h = 55.28 $ inches. The diagram has the base labeled $ a $, and the height $ h $, and the slanted side is unlabeled.

So unless we assume something, we cannot compute perimeter.

But perhaps the slanted side is also $ a $? Then it would be a rhombus.

But that would mean all sides are 61 inches.

Then:
- Area = $ a \times h = 61 \times 55.28 = 3372.08 \text{ in}^2 $
- Perimeter = $ 4 \times 61 = 244 \text{ in} $
- Type: Rhombus

But is that valid?

Wait — in a rhombus, all sides are equal, and height is $ h = a \times \sin(\theta) $, so $ h = a \times \sin(\theta) $

Here, $ h = 55.28 $, $ a = 61 $, so $ \sin(\theta) = 55.28 / 61 \approx 0.9062 $, which is valid.

So yes, it's possible.

But is it labeled as a rhombus?

The diagram shows a parallelogram with only one side labeled $ a $, and the base labeled $ a $. So likely, both pairs of opposite sides are equal, and if base = $ a $, and slant side is also $ a $, then it's a rhombus.

But the problem doesn’t specify.

Alternatively, maybe it's just a general parallelogram, and we're missing data.

But in most textbook problems like this, if only one side is given and height, and it's a parallelogram, they often assume that the side is the base, and the other side is not given, so you cannot compute perimeter.

But here, the top is labeled $ a $, and the bottom is labeled $ a $, and the left side is not labeled.

But the slanted side is not labeled.

Wait — perhaps the label $ a $ applies to all sides? That would be a rhombus.

But the diagram shows $ a $ only on the base and top.

But the left and right sides are not labeled, so we don’t know.

But since only $ a $ and $ h $ are given, and no other side, we must assume that the non-base side is not given, so perimeter cannot be computed.

But that seems unlikely.

Alternative idea: maybe the side $ a $ is the length of the slanted side, and the base is different?

But the diagram labels the base as $ a $, and the top as $ a $, so base = $ a $, and top = $ a $, so the base is $ a $, and the slanted side is not labeled.

So unless the slanted side is also $ a $, we can't compute perimeter.

But perhaps it's a rhombus — all sides equal.

Given that $ h = 55.28 $, and $ a = 61 $, and $ h < a $, it's possible.

And since it's a common problem type, likely it's a rhombus.

So assume: All sides are 61 inches

Then:
- Area = $ \text{base} \times \text{height} = 61 \times 55.28 = 3372.08 \text{ in}^2 $
- Perimeter = $ 4 \times 61 = 244 \text{ in} $
- Type: Rhombus

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

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✔ Final Answers:

| # | Area | Perimeter | Type |
|---|------|-----------|------|
| 1 | 5192 cm² | 294 cm | Rectangle |
| 2 | 3312 in² | 236 in | Rectangle |
| 3 | ≈1383.2 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 |

> Note: For #9, assumed all sides are 61 inches (rhombus), since otherwise perimeter cannot be determined. This is a reasonable assumption based on typical problem design.

Let me know if you'd like the work in a printable format!