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Mensuration - Surface Area and Volume of 3D Solids

Grade 12IGCSE

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

🔑Concepts

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Understanding the difference between Total Surface Area (TSA) and Lateral/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 point (apex) where Volume is exactly 1/3 of the corresponding prism/cylinder.

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Spheres and Hemispheres: Calculating properties based solely on the radius.

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Composite Solids: Calculating total volume or surface area by adding or subtracting basic geometric shapes.

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Scale Factors: Relationship between linear ratio (k), area ratio (k²), and volume ratio (k³) for mathematically similar solids.

📐Formulae

Cuboid: V=l×w×hV = l \times w \times h; TSA=2(lw+wh+lh)TSA = 2(lw + wh + lh)

Cylinder: V=πr2hV = \pi r^2 h; CSA=2πrhCSA = 2\pi rh; TSA=2πr2+2πrhTSA = 2\pi r^2 + 2\pi rh

Cone: V=13πr2hV = \frac{1}{3}\pi r^2 h; CSA=πrlCSA = \pi rl (where ll is slant height r2+h2\sqrt{r^2 + h^2})

Sphere: V=43πr3V = \frac{4}{3}\pi r^3; SurfaceArea=4πr2Surface Area = 4\pi r^2

Pyramid: V=13×Base Area×Perpendicular HeightV = \frac{1}{3} \times \text{Base Area} \times \text{Perpendicular Height}

Similar Solids: V1V2=(l1l2)3\frac{V_1}{V_2} = (\frac{l_1}{l_2})^3 and A1A2=(l1l2)2\frac{A_1}{A_2} = (\frac{l_1}{l_2})^2

💡Examples

Problem 1:

A solid toy is made of a hemisphere of radius 3 cm topped by a cone of the same radius and a height of 4 cm. Calculate the total volume of the toy.

Solution:

Vtotal=Vhemisphere+Vcone=(23πr3)+(13πr2h)=(23π×33)+(13π×32×4)=18π+12π=30π≈94.25 cm3V_{total} = V_{hemisphere} + V_{cone} = (\frac{2}{3} \pi r^3) + (\frac{1}{3} \pi r^2 h) = (\frac{2}{3} \pi \times 3^3) + (\frac{1}{3} \pi \times 3^2 \times 4) = 18\pi + 12\pi = 30\pi \approx 94.25 \text{ cm}^3

Explanation:

To find the volume of a composite solid, calculate the volume of each component part separately and sum them. Note that the volume of a hemisphere is half that of a sphere.

Problem 2:

Two mathematically similar cylinders have heights of 5 cm and 10 cm. If the smaller cylinder has a surface area of 40 cm², find the surface area of the larger cylinder.

Solution:

k=h2h1=105=2k = \frac{h_2}{h_1} = \frac{10}{5} = 2. Therefore, Area Ratio =k2=22=4= k^2 = 2^2 = 4. A2=A1×4=40×4=160 cm2A_2 = A_1 \times 4 = 40 \times 4 = 160 \text{ cm}^2

Explanation:

When objects are similar, the ratio of their areas is the square of the ratio of their corresponding linear dimensions (the scale factor kk).

Problem 3:

A cone has a radius of 5 cm and a perpendicular height of 12 cm. Find its curved surface area.

Solution:

l=r2+h2=52+122=25+144=13 cml = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 \text{ cm}. CSA=πrl=π×5×13=65π≈204.2 cm2CSA = \pi r l = \pi \times 5 \times 13 = 65\pi \approx 204.2 \text{ cm}^2

Explanation:

To find the Curved Surface Area of a cone, you must first find the slant height (ll) using Pythagoras' theorem with the radius (rr) and the vertical height (hh).