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Mensuration - Surface Area and Volume of Prisms, Pyramids, Cones, and Spheres

Grade 10IGCSE

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

🔑Concepts

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Definition of a Prism: A solid with a uniform cross-section throughout its length.

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Perpendicular Height (h) vs. Slant Height (l): In pyramids and cones, volume uses perpendicular height, while curved surface area uses slant height.

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Curved Surface Area (CSA) vs. Total Surface Area (TSA): CSA only includes the side surfaces, while TSA includes the base(s).

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Pythagoras' Theorem in Cones: The relationship between radius (r), perpendicular height (h), and slant height (l) is l2=r2+h2l^2 = r^2 + h^2.

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Composite Solids: Calculating the total volume or surface area by adding or subtracting standard shapes.

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Units Conversion: Understanding that 1extcm3=1extml1 ext{ cm}^3 = 1 ext{ ml} and 1000extcm3=1extlitre1000 ext{ cm}^3 = 1 ext{ litre}.

📐Formulae

Volume of any Prism = Area of cross-section×length\text{Area of cross-section} \times \text{length}

Volume of a Cylinder = πr2h\pi r^2 h

Total Surface Area of a Cylinder = 2πr2+2πrh2\pi r^2 + 2\pi rh

Volume of a Pyramid = 13×Base Area×h\frac{1}{3} \times \text{Base Area} \times h

Volume of a Cone = 13πr2h\frac{1}{3} \pi r^2 h

Curved Surface Area of a Cone = πrl\pi r l (where ll is slant height)

Volume of a Sphere = 43πr3\frac{4}{3} \pi r^3

Surface Area of a Sphere = 4πr24\pi r^2

Surface Area of a Hemisphere = 3πr23\pi r^2 (Curved area 2πr22\pi r^2 + Base area πr2\pi r^2)

💡Examples

Problem 1:

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

Solution:

l=52+122=25+144=169=13 cml = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}. TSA = πr2+πrl=(3.142×52)+(3.142×5×13)=78.55+204.23=282.78 cm2\pi r^2 + \pi r l = (3.142 \times 5^2) + (3.142 \times 5 \times 13) = 78.55 + 204.23 = 282.78 \text{ cm}^2.

Explanation:

First, use Pythagoras' theorem to find the slant height (ll). Then, calculate the base area (circle) and the curved surface area separately before adding them for the Total Surface Area.

Problem 2:

A metal sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Calculate 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. Setting them equal: 16πh=288π⇒16h=288⇒h=18 cm16\pi h = 288\pi \Rightarrow 16h = 288 \Rightarrow h = 18 \text{ cm}.

Explanation:

When a shape is melted and recast, the volume remains constant. Equate the volume of the sphere to the volume of the cylinder and solve for hh.

Problem 3:

Calculate the volume of a square-based pyramid with a base side of 10 cm and a perpendicular height of 15 cm.

Solution:

Base Area = 10×10=100 cm210 \times 10 = 100 \text{ cm}^2. Volume = 13×Base Area×h=13×100×15=500 cm3\frac{1}{3} \times \text{Base Area} \times h = \frac{1}{3} \times 100 \times 15 = 500 \text{ cm}^3.

Explanation:

Identify the base area first. For a square-based pyramid, Base Area = side2\text{side}^2. Then apply the pyramid volume formula.