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Geometry and Measurement - Volume and Surface Area of 3D Shapes (Prisms and Cylinders)

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A prism is a 3D shape with a constant cross-section throughout its length. The volume VV is found by multiplying the area of the base AbaseA_{base} by the height (or length) hh.

A triangular prism showing the base area and the height (length).
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A cylinder is a special type of prism with a circular base. The Total Surface Area (TSATSA) consists of the two circular ends and the rectangular 'label' (curved surface) that wraps around the height.

Cylinder diagram showing radius r and height h.
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The Surface Area of any prism is the sum of the areas of all its faces. This is calculated as SA=(2×Area of Base)+(Perimeter of Base×height)SA = (2 \times \text{Area of Base}) + (\text{Perimeter of Base} \times \text{height}).

Flowchart for calculating the total surface area of a prism.
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To find the volume of composite 3D shapes, divide the object into simpler prisms, calculate each individual volume, and add them together.

📐Formulae

Volume of a General Prism: V=Abase×hV = A_{base} \times h

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Total Surface Area of a General Prism: SA=(2×Abase)+(Pbase×h)SA = (2 \times A_{base}) + (P_{base} \times h) (where PP is perimeter)

Curved Surface Area (Lateral Area) of a Cylinder: CSA=2πrhCSA = 2 \pi r h

Total Surface Area of a Cylinder: TSA=2πr2+2πrhTSA = 2 \pi r^2 + 2 \pi r h

Area of a Triangle (for triangular prisms): A=12bhA = \frac{1}{2} b h

Circumference of a Circle: C=2πrC = 2 \pi r

💡Examples

Problem 1:

Calculate the volume and total surface area of a triangular prism with a base triangle of base 6 cm6\text{ cm} and height 4 cm4\text{ cm}. The length (height) of the prism is 12 cm12\text{ cm}, and the other two sides of the triangle are 5 cm5\text{ cm} each.

Solution:

  1. Base Area (AbaseA_{base}): A=12×6×4=12 cm2A = \frac{1}{2} \times 6 \times 4 = 12\text{ cm}^2
  2. Volume: V=Abase×h=12×12=144 cm3V = A_{base} \times h = 12 \times 12 = 144\text{ cm}^3
  3. Base Perimeter (PbaseP_{base}): P=6+5+5=16 cmP = 6 + 5 + 5 = 16\text{ cm}
  4. Surface Area: SA=(2×Abase)+(Pbase×h)=(2×12)+(16×12)=24+192=216 cm2SA = (2 \times A_{base}) + (P_{base} \times h) = (2 \times 12) + (16 \times 12) = 24 + 192 = 216\text{ cm}^2

Explanation:

To find the volume, we first find the area of the triangular cross-section and multiply by the prism's length. For the surface area, we add the areas of the two triangular ends to the area of the three rectangular sides (calculated using Perimeter ×\times Height).

Problem 2:

A cylinder has a radius of 3 cm3\text{ cm} and a height of 10 cm10\text{ cm}. Find its volume and total surface area. (Use π≈3.14\pi \approx 3.14)

Solution:

  1. Volume: V=πr2h=3.14×32×10=3.14×9×10=282.6 cm3V = \pi r^2 h = 3.14 \times 3^2 \times 10 = 3.14 \times 9 \times 10 = 282.6\text{ cm}^3
  2. Surface Area: TSA=2πr2+2πrhTSA = 2\pi r^2 + 2\pi r h TSA=(2×3.14×32)+(2×3.14×3×10)TSA = (2 \times 3.14 \times 3^2) + (2 \times 3.14 \times 3 \times 10) TSA=(2×3.14×9)+(188.4)TSA = (2 \times 3.14 \times 9) + (188.4) TSA=56.52+188.4=244.92 cm2TSA = 56.52 + 188.4 = 244.92\text{ cm}^2

Explanation:

For volume, we calculate the area of the circular base (πr2\pi r^2) and multiply by the height. For the surface area, we calculate the area of the two circular lids (2πr22\pi r^2) and add the area of the curved side, which is a rectangle when flattened (2πr×h2\pi r \times h).

Problem 3:

A concrete L-shaped prism has a cross-sectional area as shown. The prism is 5 m5\text{ m} long. Calculate the volume of the prism. The base is an L-shape with a total width of 4 m4\text{ m}, total height of 6 m6\text{ m}, and uniform thickness of 2 m2\text{ m}.

An L-shaped prism with dimensions labeled.

Solution:

  1. Calculate the area of the L-shaped cross-section by splitting it into two rectangles:
  • Rectangle 1 (Bottom): 4 m×2 m=8 m24\text{ m} \times 2\text{ m} = 8\text{ m}^2
  • Rectangle 2 (Top): (6 m−2 m)×2 m=4 m×2 m=8 m2(6\text{ m} - 2\text{ m}) \times 2\text{ m} = 4\text{ m} \times 2\text{ m} = 8\text{ m}^2
  • Total Abase=8+8=16 m2A_{base} = 8 + 8 = 16\text{ m}^2
  1. Calculate the volume: V=Abase×lengthV = A_{base} \times \text{length} V=16×5=80 m3V = 16 \times 5 = 80\text{ m}^3

Explanation:

To find the volume of any prism, find the area of the uniform cross-section first, then multiply by the length (or height) of the prism.

Problem 4:

A cylindrical water tank is open at the top. The internal radius is 2 m2\text{ m} and the height is 7 m7\text{ m}. Calculate the total internal surface area that needs to be waterproofed (the base and the side wall). (Use π=227\pi = \frac{22}{7})

An open cylindrical tank with radius 2m and height 7m.

Solution:

  1. Area of the base (one circle): Abase=πr2=227×22=887≈12.57 m2A_{base} = \pi r^2 = \frac{22}{7} \times 2^2 = \frac{88}{7} \approx 12.57\text{ m}^2
  2. Curved Surface Area (CSACSA): CSA=2πrh=2×227×2×7=88 m2CSA = 2 \pi r h = 2 \times \frac{22}{7} \times 2 \times 7 = 88\text{ m}^2
  3. Total Surface Area (Open Top): SA=Abase+CSA=12.57+88=100.57 m2SA = A_{base} + CSA = 12.57 + 88 = 100.57\text{ m}^2

Explanation:

Since the tank is 'open at the top', we only calculate the area of the circular base and the curved lateral surface, excluding the top circle.