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Mensuration - Surface Area and Volume of 3D Solids

Grade 9IGCSE

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

🔑Concepts

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Definition of Volume: The amount of 3D space an object occupies, measured in cubic units (cm3,m3cm^3, m^3).

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Total Surface Area (TSA): The sum of the areas of all faces of a solid.

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Curved Surface Area (CSA): The area of only the rounded surfaces (e.g., in cylinders, cones, and spheres).

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Uniform Cross-section: A prism has a constant cross-section along its length; Volume = Area of Cross-section × Length.

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Slant Height (ll): In cones and pyramids, the distance from the apex to the edge of the base, often found using Pythagoras theorem l=r2+h2l = \sqrt{r^2 + h^2}.

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Composite Solids: Solids made by combining two or more basic shapes (e.g., a hemisphere on top of a cylinder).

📐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=2πr(r+h)TSA = 2\pi r(r + h)

Prism: V=Area of cross-section×LV = \text{Area of cross-section} \times L

Sphere: V=43πr3V = \frac{4}{3} \pi r^3, SA=4πr2SA = 4\pi r^2

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

Pyramid: V=13×Base Area×hV = \frac{1}{3} \times \text{Base Area} \times h

Hemisphere: V=23πr3V = \frac{2}{3} \pi r^3, CSA=2πr2CSA = 2\pi r^2, TSA=3πr2TSA = 3\pi r^2

💡Examples

Problem 1:

A cylinder has a radius of 33 cm and a height of 77 cm. Calculate its Total Surface Area. (Take π=3.142\pi = 3.142)

Solution:

TSA=2πr2+2πrh=2(3.142)(32)+2(3.142)(3)(7)=56.556+131.964=188.52TSA = 2\pi r^2 + 2\pi rh = 2(3.142)(3^2) + 2(3.142)(3)(7) = 56.556 + 131.964 = 188.52 cm²

Explanation:

To find the Total Surface Area of a cylinder, you must add the area of the two circular bases (2πr22\pi r^2) to the area of the curved side (2πrh2\pi rh).

Problem 2:

A right-circular cone has a radius of 55 cm and a vertical height of 1212 cm. Find its Volume.

Solution:

V=13πr2h=13×π×52×12=100π≈314.16V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi \approx 314.16 cm³

Explanation:

The volume of a cone is exactly one-third the volume of a cylinder with the same radius and height. Plug the radius (55) and height (1212) into the formula.

Problem 3:

A metal sphere with a radius of 66 cm is melted down and recast into a solid cylinder with a radius of 44 cm. Find the height of the cylinder.

Solution:

Volume of Sphere = 43π(6)3=288π\frac{4}{3} \pi (6)^3 = 288\pi. Volume of Cylinder = π(4)2h=16πh\pi (4)^2 h = 16\pi h. Set volumes equal: 288π=16πh⇒h=288/16=18288\pi = 16\pi h \Rightarrow h = 288 / 16 = 18 cm.

Explanation:

In recasting problems, the volume remains constant. Calculate the volume of the original shape (sphere) and set it equal to the formula for the volume of the new shape (cylinder) to solve for the missing dimension.