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Surface Area Worksheets | Practice Questions and Answers | Cazoomy - Free Printable

Surface Area Worksheets | Practice Questions and Answers | Cazoomy

Educational worksheet: Surface Area Worksheets | Practice Questions and Answers | Cazoomy. Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Surface Area Worksheets | Practice Questions and Answers | Cazoomy
To solve the problem of finding the surface area of each figure (cylinders and cones), we will use the appropriate formulas for each shape. Let's go through each part step by step.

Formulas Needed:


1. Surface Area of a Cylinder:
\[
\text{Surface Area} = 2\pi r^2 + 2\pi rh
\]
- \( r \) is the radius.
- \( h \) is the height.

2. Surface Area of a Cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
- \( r \) is the radius.
- \( l \) is the slant height.

---

Solutions:



#### 1) Cylinder (8 ft radius, 12 ft height)
- Radius (\( r \)) = 8 ft
- Height (\( h \)) = 12 ft

Using the formula for the surface area of a cylinder:
\[
\text{Surface Area} = 2\pi r^2 + 2\pi rh
\]
\[
= 2\pi (8)^2 + 2\pi (8)(12)
\]
\[
= 2\pi (64) + 2\pi (96)
\]
\[
= 128\pi + 192\pi
\]
\[
= 320\pi
\]
\[
\approx 320 \times 3.1416 \approx 1005.31 \, \text{ft}^2
\]

#### 2) Cone (4 in radius, 6 in slant height)
- Radius (\( r \)) = 4 in
- Slant height (\( l \)) = 6 in

Using the formula for the surface area of a cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
\[
= \pi (4)^2 + \pi (4)(6)
\]
\[
= \pi (16) + \pi (24)
\]
\[
= 16\pi + 24\pi
\]
\[
= 40\pi
\]
\[
\approx 40 \times 3.1416 \approx 125.66 \, \text{in}^2
\]

#### 3) Cylinder (6 yd radius, 8 yd height)
- Radius (\( r \)) = 6 yd
- Height (\( h \)) = 8 yd

Using the formula for the surface area of a cylinder:
\[
\text{Surface Area} = 2\pi r^2 + 2\pi rh
\]
\[
= 2\pi (6)^2 + 2\pi (6)(8)
\]
\[
= 2\pi (36) + 2\pi (48)
\]
\[
= 72\pi + 96\pi
\]
\[
= 168\pi
\]
\[
\approx 168 \times 3.1416 \approx 527.79 \, \text{yd}^2
\]

#### 4) Cone (5 mm radius, 13 mm slant height)
- Radius (\( r \)) = 5 mm
- Slant height (\( l \)) = 13 mm

Using the formula for the surface area of a cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
\[
= \pi (5)^2 + \pi (5)(13)
\]
\[
= \pi (25) + \pi (65)
\]
\[
= 25\pi + 65\pi
\]
\[
= 90\pi
\]
\[
\approx 90 \times 3.1416 \approx 282.74 \, \text{mm}^2
\]

#### 5) Cone (6 mm radius, 14 mm slant height)
- Radius (\( r \)) = 6 mm
- Slant height (\( l \)) = 14 mm

Using the formula for the surface area of a cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
\[
= \pi (6)^2 + \pi (6)(14)
\]
\[
= \pi (36) + \pi (84)
\]
\[
= 36\pi + 84\pi
\]
\[
= 120\pi
\]
\[
\approx 120 \times 3.1416 \approx 376.99 \, \text{mm}^2
\]

#### 6) Cylinder (6 cm radius, 8 cm height)
- Radius (\( r \)) = 6 cm
- Height (\( h \)) = 8 cm

Using the formula for the surface area of a cylinder:
\[
\text{Surface Area} = 2\pi r^2 + 2\pi rh
\]
\[
= 2\pi (6)^2 + 2\pi (6)(8)
\]
\[
= 2\pi (36) + 2\pi (48)
\]
\[
= 72\pi + 96\pi
\]
\[
= 168\pi
\]
\[
\approx 168 \times 3.1416 \approx 527.79 \, \text{cm}^2
\]

#### 7) Cylinder (7 yd radius, 10 yd height)
- Radius (\( r \)) = 7 yd
- Height (\( h \)) = 10 yd

Using the formula for the surface area of a cylinder:
\[
\text{Surface Area} = 2\pi r^2 + 2\pi rh
\]
\[
= 2\pi (7)^2 + 2\pi (7)(10)
\]
\[
= 2\pi (49) + 2\pi (70)
\]
\[
= 98\pi + 140\pi
\]
\[
= 238\pi
\]
\[
\approx 238 \times 3.1416 \approx 748.69 \, \text{yd}^2
\]

#### 8) Cone (7 ft radius, 10 ft slant height)
- Radius (\( r \)) = 7 ft
- Slant height (\( l \)) = 10 ft

Using the formula for the surface area of a cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
\[
= \pi (7)^2 + \pi (7)(10)
\]
\[
= \pi (49) + \pi (70)
\]
\[
= 49\pi + 70\pi
\]
\[
= 119\pi
\]
\[
\approx 119 \times 3.1416 \approx 373.93 \, \text{ft}^2
\]

#### 9) Cone (4 cm radius, 6 cm slant height)
- Radius (\( r \)) = 4 cm
- Slant height (\( l \)) = 6 cm

Using the formula for the surface area of a cone:
\[
\text{Surface Area} = \pi r^2 + \pi r l
\]
\[
= \pi (4)^2 + \pi (4)(6)
\]
\[
= \pi (16) + \pi (24)
\]
\[
= 16\pi + 24\pi
\]
\[
= 40\pi
\]
\[
\approx 40 \times 3.1416 \approx 125.66 \, \text{cm}^2
\]

---

Final Answers:


1. \( \boxed{1005.31} \, \text{ft}^2 \)
2. \( \boxed{125.66} \, \text{in}^2 \)
3. \( \boxed{527.79} \, \text{yd}^2 \)
4. \( \boxed{282.74} \, \text{mm}^2 \)
5. \( \boxed{376.99} \, \text{mm}^2 \)
6. \( \boxed{527.79} \, \text{cm}^2 \)
7. \( \boxed{748.69} \, \text{yd}^2 \)
8. \( \boxed{373.93} \, \text{ft}^2 \)
9. \( \boxed{125.66} \, \text{cm}^2 \)

These are the rounded surface areas to the nearest hundredth.
Parent Tip: Review the logic above to help your child master the concept of volume and surface area worksheets grade 7.
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