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 \times b = 88 \times 59 = 5192 \text{ cm}^2 $
- Perimeter = $ 2(a + b) = 2(88 + 59) = 2(147) = 294 \text{ cm} $

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

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2) Rectangle

- $ a = 72 \text{ inches},\ b = 46 \text{ inches} $
- Type: Rectangle
- Area = $ 72 \times 46 = 3312 \text{ in}^2 $
- Perimeter = $ 2(72 + 46) = 2(118) = 236 \text{ inches} $

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

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3) Triangle (Isosceles)

- $ a = 40 \text{ ft},\ b = 72 \text{ ft} $
- Two sides equal → Isosceles triangle
- Type: Isosceles Triangle
- Perimeter = $ a + b + b = 40 + 72 + 72 = 184 \text{ ft} $
- Area: Use Heron’s formula or base-height. But we don’t have height. We can use Heron’s formula:
- $ s = \frac{a + b + b}{2} = \frac{40 + 72 + 72}{2} = \frac{184}{2} = 92 $
- Area = $ \sqrt{s(s-a)(s-b)(s-b)} = \sqrt{92(92-40)(92-72)(92-72)} $
- $ = \sqrt{92 \times 52 \times 20 \times 20} $
- $ = \sqrt{92 \times 52 \times 400} $
- First: $ 92 \times 52 = 4784 $
- Then: $ 4784 \times 400 = 1,913,600 $
- $ \sqrt{1,913,600} \approx 1383.3 \text{ ft}^2 $

✔ Answer:
- Area: ≈ 1383.3 ft²
- Perimeter: 184 ft
- Type: Isosceles Triangle

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4) Triangle (Isosceles)

- $ a = 38 \text{ yds},\ b = 75 \text{ yds} $
- Two sides equal → Isosceles triangle
- Type: Isosceles Triangle
- Perimeter = $ 38 + 75 + 75 = 188 \text{ yds} $
- Area: Use Heron’s formula
- $ s = \frac{38 + 75 + 75}{2} = \frac{188}{2} = 94 $
- Area = $ \sqrt{94(94-38)(94-75)(94-75)} = \sqrt{94 \times 56 \times 19 \times 19} $
- $ = \sqrt{94 \times 56 \times 361} $
- $ 94 \times 56 = 5264 $
- $ 5264 \times 361 \approx 1,900,000 $ (approx)
- Let's compute more accurately:
- $ 5264 \times 361 = 5264 \times (300 + 60 + 1) = 5264×300 = 1,579,200; 5264×60=315,840; 5264×1=5,264 $
- Total = $ 1,579,200 + 315,840 + 5,264 = 1,900,304 $
- $ \sqrt{1,900,304} \approx 1378.5 \text{ yd}^2 $

✔ Answer:
- Area: ≈ 1378.5 yd²
- Perimeter: 188 yds
- Type: Isosceles Triangle

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5) Square

- $ s = 63 \text{ mm} $
- Type: Square
- Area = $ s^2 = 63^2 = 3969 \text{ mm}^2 $
- Perimeter = $ 4s = 4 \times 63 = 252 \text{ mm} $

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

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6) Square

- $ s = 65 \text{ yds} $
- Type: Square
- Area = $ 65^2 = 4225 \text{ yd}^2 $
- Perimeter = $ 4 \times 65 = 260 \text{ yds} $

✔ Answer:
- 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 equal → two sides are 54.96 ft, two are 89 ft
- Type: Parallelogram
- Area = base × height = $ 89 \times 52 = 4628 \text{ ft}^2 $
- Perimeter = $ 2(a + c) = 2(54.96 + 89) = 2(143.96) = 287.92 \text{ ft} $

✔ Answer:
- 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} $
- Type: Trapezoid (non-parallel sides given)
- Area = $ \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 = $ a1 + a2 + b1 + b2 = 100 + 43 + 68.28 + 58.8 = 270.08 \text{ cm} $

✔ Answer:
- 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 equal.
- Type: Parallelogram
- Area = base × height = $ 61 \times 55.28 $
- Compute: $ 60 \times 55.28 = 3316.8 $, $ 1 \times 55.28 = 55.28 $ → total = $ 3372.08 \text{ in}^2 $
- Perimeter: Need side length. But only one side and height given. However, unless we know angle or other side, we cannot determine perimeter.

Wait — we're missing information for perimeter. The diagram shows two adjacent sides as 'a', but only one side is labeled 'a' and height is given. But since it's a parallelogram, both pairs of opposite sides are equal.

But we only know one side (a = 61 in), and height (h = 55.28 in). To find perimeter, we need both adjacent sides.

However, if this is a rectangle, then height would be the other side. But here, it's a general parallelogram.

But wait — the figure shows height from side a, so base = a = 61 in, height = 55.28 in → area = $ 61 \times 55.28 = 3372.08 \text{ in}^2 $

But perimeter requires knowing the other side. Since it's not given, perhaps the figure implies that the other side is unknown, so we cannot calculate perimeter.

Wait — maybe the label "a" appears on both adjacent sides? That would make it a rhombus?

But no — the diagram shows only one side labeled "a", and height drawn to it.

So unless the other side is also 61 in, we can't assume.

But in standard labeling, if both sides are labeled 'a', then it's a rhombus. But here, only one side is labeled 'a'.

Looking back at image: It says "a" on the bottom and left side — probably indicating both sides are equal? But that would be unusual.

Wait — let's recheck: In problem 9, the diagram shows:

```
a
/|\
/ | \
a | h
```

Wait — actually, the base is labeled 'a', and the slanted side is also labeled 'a'. So both adjacent sides are 'a' → this is a rhombus!

So all sides = 61 inches

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

✔ Answer:
- Area: 3372.08 in²
- Perimeter: 244 inches
- 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 | ≈1383.3 ft² | 184 ft | Isosceles Triangle |
| 4 | ≈1378.5 yd² | 188 yd | Isosceles Triangle |
| 5 | 3969 mm² | 252 mm | Square |
| 6 | 4225 yd² | 260 yd | 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 formatted for printing or want exact values instead of approximations!