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

Grade 9IB

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

🔑Concepts

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Uniform Prisms and Cross-sections: A prism is a 3D shape with a constant cross-section along its entire length. This means if you slice the solid parallel to its base at any point, the resulting shape is identical to the base. For example, a triangular prism looks like a chocolate Toblerone bar; every vertical slice produces an identical triangle.

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Understanding Surface Area through Nets: The total surface area (SA) of a 3D solid is the sum of the areas of all its exterior faces. Visualizing this involves 'unfolding' the shape into a 2D drawing called a net. For instance, the net of a cube consists of six squares arranged in a 'cross' or 'T' shape, while the net of a cylinder consists of two circles and one rectangle.

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Volume as Capacity: Volume measures the amount of 3D space an object occupies. For any solid with a uniform cross-section (like prisms and cylinders), the volume is calculated by finding the area of the base (AbaseA_{base}) and multiplying it by the height (hh). Imagine stacking thin 2D 'slices' of the base area on top of each other until they reach the total height.

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Cylinders: A cylinder is technically a circular prism. It consists of two congruent circular bases. The curved side, when unrolled (its lateral surface), forms a flat rectangle. The length of this rectangle is equal to the circumference of the circle (2πr2\pi r), and its width is the height (hh) of the cylinder.

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Pyramids and Cones: These are 'pointed' solids where all points on the perimeter of a base are connected to a single point called the apex. A right cone or pyramid has its apex directly above the center of the base. Visually, these occupy exactly 13\frac{1}{3} of the volume of a prism or cylinder with the same base area and height.

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Slant Height vs. Vertical Height: In cones and pyramids, there is a distinction between the vertical height (hh), which is the perpendicular distance from the apex to the center of the base, and the slant height (ll), which is the distance from the apex to the edge of the base. These segments, along with the radius (rr) or base apothem, form a right-angled triangle, meaning we can use the Pythagorean theorem: l2=r2+h2l^2 = r^2 + h^2.

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Spheres: A sphere is a perfectly round geometrical object in 3D space that is the surface of a completely round ball. Every point on the surface is an equal distance (rr) from the center point. Unlike prisms, spheres do not have flat faces or edges, so their formulas rely solely on the radius.

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Composite Solids: Many real-world objects are combinations of these basic shapes. To find the volume of a composite solid (like a silo made of a cylinder and a cone), you add the volumes of the individual parts. To find the surface area, you must be careful to only count the 'exposed' outer surfaces and subtract the areas where the shapes touch.

📐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.