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Geometry and Trigonometry - Surface area and volume of prisms, cylinders, pyramids, cones, and spheres

Grade 10IB

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. The volume is calculated by multiplying the area of the base (AbaseA_{base}) by the height (hh). For a cylinder, which is a circular prism, the base area is πr2\pi r^2.

A diagram of a prism showing the base area and the perpendicular height.
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Pyramids and cones share a common property: their volume is exactly one-third of the volume of a prism or cylinder with the same base and height (V=13AbasehV = \frac{1}{3} A_{base} h).

A square-based pyramid showing vertical height from the apex to the center of the base.
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The surface area of a sphere is 4πr24\pi r^2, which is exactly four times the area of its great circle. The volume is 43πr3\frac{4}{3}\pi r^3.

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Slant height (ll) vs. Vertical height (hh): In cones and pyramids, the slant height is the distance from the apex down the face to the edge. They are related to the radius or base segment by the Pythagorean theorem: l2=h2+r2l^2 = h^2 + r^2.

📐Formulae

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

Surface Area of a Rectangular Prism: SA=2(lw+lh+wh)SA = 2(lw + lh + wh)

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

Total Surface Area of a Cylinder: SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh

Volume of a Cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

Total Surface Area of a Cone: SA=πr2+πrlSA = \pi r^2 + \pi rl (where ll is slant height)

Volume of a Pyramid: V=13×Abase×hV = \frac{1}{3} \times A_{base} \times h

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Surface Area of a Sphere: SA=4πr2SA = 4\pi r^2

💡Examples

Problem 1:

A right cone has a base radius of 66 cm and a slant height of 1010 cm. Calculate the volume of the cone. (Leave your answer in terms of π\pi)

Solution:

  1. Identify the given values: r=6r = 6 cm, l=10l = 10 cm.
  2. We need the vertical height hh for the volume formula. Use the Pythagorean theorem: r2+h2=l2r^2 + h^2 = l^2.
  3. 62+h2=102⇒36+h2=1006^2 + h^2 = 10^2 \Rightarrow 36 + h^2 = 100.
  4. h2=100−36=64⇒h=64=8h^2 = 100 - 36 = 64 \Rightarrow h = \sqrt{64} = 8 cm.
  5. Apply the volume formula: V=13πr2hV = \frac{1}{3} \pi r^2 h.
  6. V=13×π×62×8V = \frac{1}{3} \times \pi \times 6^2 \times 8.
  7. V=13×π×36×8=12×8×π=96πV = \frac{1}{3} \times \pi \times 36 \times 8 = 12 \times 8 \times \pi = 96\pi cm3^3.

Explanation:

To find the volume of a cone, the vertical height is required. Since only the slant height and radius were provided, the first step was to form a right triangle and solve for hh before substituting all values into the cone volume formula.

Problem 2:

Calculate the total surface area of a cylinder with a diameter of 1414 cm and a height of 1010 cm. Use π≈227\pi \approx \frac{22}{7}.

Solution:

  1. Find the radius: r=diameter2=142=7r = \frac{diameter}{2} = \frac{14}{2} = 7 cm.
  2. Identify the height: h=10h = 10 cm.
  3. Use the Surface Area formula: SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh.
  4. Calculate the area of the two circular bases: 2×227×72=2×227×49=2×22×7=3082 \times \frac{22}{7} \times 7^2 = 2 \times \frac{22}{7} \times 49 = 2 \times 22 \times 7 = 308 cm2^2.
  5. Calculate the lateral surface area: 2×227×7×10=2×22×10=4402 \times \frac{22}{7} \times 7 \times 10 = 2 \times 22 \times 10 = 440 cm2^2.
  6. Add the parts together: SA=308+440=748SA = 308 + 440 = 748 cm2^2.

Explanation:

The total surface area of a cylinder includes the top and bottom circles plus the rectangular side. By using the radius (half the diameter), we calculate these parts separately and sum them for the final result.

Problem 3:

A metal sphere has a radius of 33 cm. It is melted down and recast into a solid cylinder with a radius of 22 cm. Calculate the height of the cylinder.

Diagram showing a sphere of radius 3 and a cylinder of radius 2 and height h.

Solution:

Volume of Sphere=43πr3=43π(3)3=43π(27)=36π cm3\text{Volume of Sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi \text{ cm}^3 Volume of Cylinder=πr2h=π(2)2h=4πh\text{Volume of Cylinder} = \pi r^2 h = \pi (2)^2 h = 4\pi h Equating volumes: 4πh=36π\text{Equating volumes: } 4\pi h = 36\pi h=36π4π=9 cmh = \frac{36\pi}{4\pi} = 9 \text{ cm}

Explanation:

Since the sphere is recast into a cylinder, the volume remains constant. We calculate the sphere's volume first, then set it equal to the cylinder's volume formula to solve for the unknown height.

Problem 4:

Find the total surface area of a solid hemisphere with a radius of 1010 cm. (Use π≈3.14\pi \approx 3.14)

Diagram of a hemisphere showing the curved top and the flat base radius.

Solution:

Curved Surface Area=12(4πr2)=2πr2\text{Curved Surface Area} = \frac{1}{2}(4\pi r^2) = 2\pi r^2 Base Area (Circle)=πr2\text{Base Area (Circle)} = \pi r^2 Total SA=2πr2+πr2=3πr2\text{Total SA} = 2\pi r^2 + \pi r^2 = 3\pi r^2 Total SA=3×3.14×(10)2\text{Total SA} = 3 \times 3.14 \times (10)^2 Total SA=3×3.14×100=942 cm2\text{Total SA} = 3 \times 3.14 \times 100 = 942 \text{ cm}^2

Explanation:

A solid hemisphere has two surfaces: the curved top and the flat circular base. The total area is the sum of half the sphere's surface area plus the area of the circular base.