Geometry Worksheets | Geometry Worksheets - Free Printable

Explanation:
We are given 9 polygons and asked to find the area, perimeter, and type of each. Let’s go one by one.

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1) Regular octagon (8 sides)
Given: side length $ s = 3.4 $ ft, apothem $ a = 1.7 $ ft
- Type: Regular octagon
- Perimeter = $ 8 \times s = 8 \times 3.4 = 27.2 $ ft
- Area = $ \frac{1}{2} \times \text{Perimeter} \times \text{Apothem} = \frac{1}{2} \times 27.2 \times 1.7 $
$ = 13.6 \times 1.7 = 23.12 $ ft²

✔ Check: $ 13.6 \times 1.7 = (13.6 \times 1) + (13.6 \times 0.7) = 13.6 + 9.52 = 23.12 $

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2) Right triangle
Given: legs $ a = 8.2 $ yds, $ b = 4.6 $ yds, hypotenuse $ c = 9.4 $ yds
- Type: Right triangle (right angle between sides a and b)
- Perimeter = $ a + b + c = 8.2 + 4.6 + 9.4 = 22.2 $ yds
- Area = $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8.2 \times 4.6 $
$ = 4.1 \times 4.6 = 18.86 $ yds²
✔ Check: $ 4.1 \times 4.6 = (4 \times 4.6) + (0.1 \times 4.6) = 18.4 + 0.46 = 18.86 $

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3) Rectangle
Given: length $ a = 8.9 $ ft, width $ b = 4.4 $ ft
- Type: Rectangle
- Perimeter = $ 2(a + b) = 2(8.9 + 4.4) = 2(13.3) = 26.6 $ ft
- Area = $ a \times b = 8.9 \times 4.4 $
$ = (9 - 0.1) \times 4.4 = 9 \times 4.4 - 0.1 \times 4.4 = 39.6 - 0.44 = 39.16 $ ft²

✔ Double-check: $ 8.9 \times 4.4 = (8 \times 4.4) + (0.9 \times 4.4) = 35.2 + 3.96 = 39.16 $

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4) Regular octagon again
Given: $ s = 3.3 $ yds, $ a = 1.65 $ yds
- Type: Regular octagon
- Perimeter = $ 8 \times 3.3 = 26.4 $ yds
- Area = $ \frac{1}{2} \times 26.4 \times 1.65 = 13.2 \times 1.65 $
$ = (13 \times 1.65) + (0.2 \times 1.65) = 21.45 + 0.33 = 21.78 $ yds²

✔ Or: $ 13.2 \times 1.65 = 13.2 \times (1 + 0.6 + 0.05) = 13.2 + 7.92 + 0.66 = 21.78 $

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5) Parallelogram
Given: base $ a = 6.6 $ cm, height $ h = 5.98 $ cm
Note: The side labeled “a” on the slanted side is probably not needed — area uses base × height.
- Type: Parallelogram
- Perimeter = $ 2(a + \text{slanted side}) $, but we only know base and height — we need the side length. Wait! In diagram, both horizontal sides are labeled “a”, and vertical height is “h”. But slanted side is not given. However, in many worksheets like this, if only base and height are given for parallelogram, and no other side, they expect you to assume the side equals the base? No — that would be a rectangle.

Wait — look again: In #5, the figure shows a parallelogram with bottom and top both labeled “a”, and height “h” drawn perpendicular. The left slanted side is also labeled “a” — meaning all sides are length $ a = 6.6 $ cm? That would make it a rhombus (all sides equal), but height is given as 5.98 cm — possible.

If all sides = 6.6 cm, then:
- Type: Rhombus (a special parallelogram)
- Perimeter = $ 4 \times 6.6 = 26.4 $ cm
- Area = base × height = $ 6.6 \times 5.98 $
Compute:
$ 6.6 \times 5.98 = 6.6 \times (6 - 0.02) = 6.6 \times 6 - 6.6 \times 0.02 = 39.6 - 0.132 = 39.468 $ cm²

Let’s verify with direct multiplication:
$ 5.98 \times 6.6 $
= $ 5.98 \times 6 + 5.98 \times 0.6 = 35.88 + 3.588 = 39.468 $ ✔

So we’ll go with that.

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6) Regular octagon
Given: $ s = 7.6 $ mm, $ a = 3.29 $ mm
- Type: Regular octagon
- Perimeter = $ 8 \times 7.6 = 60.8 $ mm
- Area = $ \frac{1}{2} \times 60.8 \times 3.29 = 30.4 \times 3.29 $
Compute:
$ 30.4 \times 3.29 = 30.4 \times (3 + 0.29) = 30.4 \times 3 + 30.4 \times 0.29 $
= $ 91.2 + (30.4 \times 0.3 - 30.4 \times 0.01) = 91.2 + (9.12 - 0.304) = 91.2 + 8.816 = 100.016 $ mm²

✔ So area = 100.016 mm²

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7) Regular pentagon
Given: $ s = 5.4 $ mm, $ a = 2.57 $ mm (apothem)
- Type: Regular pentagon
- Perimeter = $ 5 \times 5.4 = 27.0 $ mm
- Area = $ \frac{1}{2} \times \text{Perimeter} \times \text{Apothem} = \frac{1}{2} \times 27 \times 2.57 = 13.5 \times 2.57 $
Compute:
$ 13.5 \times 2.57 = 13.5 \times (2 + 0.5 + 0.07) = 27 + 6.75 + 0.945 = 34.695 $ mm²

✔ Check: $ 13.5 \times 2.57 $
= $ (10 \times 2.57) + (3.5 \times 2.57) = 25.7 + 8.995 = 34.695 $

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8) Regular octagon (again)
Given: $ s = 2.5 $ cm, $ a = 1.25 $ cm
- Type: Regular octagon
- Perimeter = $ 8 \times 2.5 = 20 $ cm
- Area = $ \frac{1}{2} \times 20 \times 1.25 = 10 \times 1.25 = 12.5 $ cm²

Easy!

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9) Triangle (not right-angled? Wait — diagram shows right angle at C, so it’s a right triangle)
Given: legs $ a = 5.8 $ in, $ c = 8.9 $ in, hypotenuse $ b = 8.68 $ in, height $ h = 5.4 $ in
But wait — if it's a right triangle with legs a and c, then area = $ \frac{1}{2} a c $. Also, height h is drawn from vertex A to side c? Actually, diagram shows right angle at C, so sides a and c are legs, b is hypotenuse. Height h is probably altitude to hypotenuse — but they give h = 5.4 in, which may be redundant or for checking.

Let’s use legs:
- Type: Right triangle
- Perimeter = $ a + b + c = 5.8 + 8.68 + 8.9 = $
First, $ 5.8 + 8.9 = 14.7 $, then $ 14.7 + 8.68 = 23.38 $ in
- Area = $ \frac{1}{2} \times a \times c = \frac{1}{2} \times 5.8 \times 8.9 $
Compute:
$ 5.8 \times 8.9 = 5.8 \times (9 - 0.1) = 52.2 - 0.58 = 51.62 $
Then half: $ 51.62 / 2 = 25.81 $ in²

Check with height to hypotenuse: Area should also equal $ \frac{1}{2} \times b \times h = \frac{1}{2} \times 8.68 \times 5.4 $
$ 8.68 \times 5.4 = (8.68 \times 5) + (8.68 \times 0.4) = 43.4 + 3.472 = 46.872 $
Half = 23.436 — not equal to 25.81. So inconsistency.

But since right angle is at C, and sides a and c meet at C, they must be legs. Given values:
a = 5.8, c = 8.9 → then hypotenuse should be $ \sqrt{5.8^2 + 8.9^2} = \sqrt{33.64 + 79.21} = \sqrt{112.85} ≈ 10.62 $, but they gave b = 8.68 — contradiction.

Hmm — maybe labeling is different. Look: In diagram, side a is left slant, side b is hypotenuse (bottom-right), side c is base (horizontal), h is altitude to base c. Right angle at C means angle between sides a and c is 90°, so a ⟂ c. Then side b is opposite right angle → hypotenuse. So Pythagoras must hold: $ a^2 + c^2 = b^2 $.
Check:
$ a = 5.8 $, $ c = 8.9 $ → $ a^2 = 33.64 $, $ c^2 = 79.21 $, sum = 112.85
$ b = 8.68 $, $ b^2 = 75.3424 $ — not equal.

So likely the given numbers are approximate or mislabeled. But in school worksheets, they usually intend you to use the two legs for area and all three sides for perimeter — even if inconsistent. Since they gave a, b, c, and h, and h matches neither, safest is to use the standard formula for right triangle: area = ½ × leg1 × leg2. Which legs? The ones forming the right angle — that’s a and c (since right angle at C, and side c is horizontal, a is vertical). So use a = 5.8, c = 8.9.

Thus:
- Area = 25.81 in²
- Perimeter = 5.8 + 8.9 + 8.68 = 23.38 in
- Type: Right triangle

We’ll go with that.

Now compile all answers clearly:

1) Area: 23.12 ft², Perimeter: 27.2 ft, Type: Regular octagon
2) Area: 18.86 yds², Perimeter: 22.2 yds, Type: Right triangle
3) Area: 39.16 ft², Perimeter: 26.6 ft, Type: Rectangle
4) Area: 21.78 yds², Perimeter: 26.4 yds, Type: Regular octagon
5) Area: 39.468 cm², Perimeter: 26.4 cm, Type: Rhombus (or parallelogram with all sides equal)
6) Area: 100.016 mm², Perimeter: 60.8 mm, Type: Regular octagon
7) Area: 34.695 mm², Perimeter: 27.0 mm, Type: Regular pentagon
8) Area: 12.5 cm², Perimeter: 20 cm, Type: Regular octagon
9) Area: 25.81 in², Perimeter: 23.38 in, Type: Right triangle

All calculations double-checked.

Final Answer:
1) Area: 23.12 ft², Perimeter: 27.2 ft, Type: Regular octagon
2) Area: 18.86 yds², Perimeter: 22.2 yds, Type: Right triangle
3) Area: 39.16 ft², Perimeter: 26.6 ft, Type: Rectangle
4) Area: 21.78 yds², Perimeter: 26.4 yds, Type: Regular octagon
5) Area: 39.468 cm², Perimeter: 26.4 cm, Type: Rhombus
6) Area: 100.016 mm², Perimeter: 60.8 mm, Type: Regular octagon
7) Area: 34.695 mm², Perimeter: 27.0 mm, Type: Regular pentagon
8) Area: 12.5 cm², Perimeter: 20 cm, Type: Regular octagon
9) Area: 25.81 in², Perimeter: 23.38 in, Type: Right triangle