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Mensuration - Surface area and volume of 3D solids

Grade 11IGCSE

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

🔑Concepts

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Difference between Total Surface Area (TSA) and Curved Surface Area (CSA).

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Prisms: Solids with a uniform cross-section where Volume = Area of cross-section × length.

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Pyramids and Cones: Solids tapering to a vertex where Volume is 1/3 of the corresponding prism.

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Compound Solids: Calculating volume and surface area by adding or subtracting basic 3D shapes.

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Similarity in 3D: If the linear scale factor is k, the area scale factor is k2k^2 and the volume scale factor is k3k^3.

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Units of conversion: 1m3=1,000,000cm31 m^3 = 1,000,000 cm^3 and 1liter=1,000cm31 liter = 1,000 cm^3.

📐Formulae

Cylinder Volume: V=πr2hV = \pi r^2 h

Cylinder Total Surface Area: TSA=2πrh+2πr2TSA = 2\pi rh + 2\pi r^2

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

Cone Curved Surface Area: CSA=πrlCSA = \pi rl (where ll is slant height)

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

Sphere Surface Area: A=4πr2A = 4\pi r^2

Pyramid Volume: V=13×base area×perpendicular heightV = \frac{1}{3} \times \text{base area} \times \text{perpendicular height}

Slant height of a cone: l=r2+h2l = \sqrt{r^2 + h^2}

💡Examples

Problem 1:

A solid metal cone has a radius of 5 cm and a perpendicular height of 12 cm. Calculate its total surface area. (Take π=3.142\pi = 3.142)

Solution:

l=52+122=25+144=13l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 cm. CSA=π×5×13=65πCSA = \pi \times 5 \times 13 = 65\pi. Base Area = π×52=25π\pi \times 5^2 = 25\pi. TSA=65π+25π=90π≈282.78TSA = 65\pi + 25\pi = 90\pi \approx 282.78 cm².

Explanation:

First, find the slant height (ll) using Pythagoras' theorem. Then, calculate the Curved Surface Area and the base area separately before adding them for the Total Surface Area.

Problem 2:

Two similar spheres have radii in the ratio 2:3. If the volume of the smaller sphere is 16π16\pi cm³, find the volume of the larger sphere.

Solution:

Linear scale factor k=32k = \frac{3}{2}. Volume scale factor =k3=(32)3=278= k^3 = (\frac{3}{2})^3 = \frac{27}{8}. Volume of larger sphere =16π×278=2π×27=54π= 16\pi \times \frac{27}{8} = 2\pi \times 27 = 54\pi cm³.

Explanation:

Use the property that the ratio of volumes of similar solids is the cube of the ratio of their corresponding lengths.

Problem 3:

A hemisphere has a radius of 6 cm. Calculate its volume in terms of π\pi.

Solution:

V=12×(43πr3)=23π(6)3=23π(216)=2π(72)=144πV = \frac{1}{2} \times (\frac{4}{3} \pi r^3) = \frac{2}{3} \pi (6)^3 = \frac{2}{3} \pi (216) = 2 \pi (72) = 144\pi cm³.

Explanation:

A hemisphere is half of a sphere. Use the sphere volume formula and divide by 2.