Surface-areas-volumes-year-8-10 - Areas and volumes of regular and composite solids

Sep 18, 2026

Uploaded by Jiven

A structured math lesson for Years 8 to 10 on surface area and volume of regular and composite 3D solids. It introduces key formulae for spheres, cylinders, and cones, including total surface area and component-based calculations. The lesson explains how minor and major domes relate to a sphere when the minor dome subtends 120° at the centre. Worked examples combine frustums, domes, and cylinders to compute curved areas, TSA, and volumes, followed by an exercise to find TSA and volume for composite solids.

LanguageEnglish

CategoryExam

FormatPDF

Pages3

Reading Time8 min

AccessFree to download

Surface area formulae

Sphere, cylinder, cone

Composite solids using domes, frustums, and cylinders

Volume formulae

Sphere, cylinder, cone

Composite-solid volumes

Exercise

Find TSA and volume for composite solids

Math Lesson (Suitable for Years 8 to 10) Areas and volumes of regular and composite solids © itute 2018 Surface area formulae Sphere: SA = 4πr2 (SA: surface area) Cylinder: Two ends + curved surface TSA = 2πr2 + 2πrl = 2πr(r + l) (TSA: total surface area) Cone: Base + curved surface TSA = πr2 + πrs r l r s r Do you know the following facts about a sphere of radius r? The above is true only when the minor dome subtendsan angle of 120o at the centre of the sphere. The following examples involve the use of these two particular domes, a cylinder and a cone to form composite solids. Preliminary: Consider a sphere of radius 23 units. It is divided into a major dome and a minor dome such that the minor dome subtends an angle of 120o at the centre of the sphere. The triangle in the above drawing is similar to the special 30-60-90 triangle shown below. It is enlarged by a scale factor of 3 . Multiply each side of the 30-60-90 triangle by 3 3 × 3 = 3 30o 1 × 3 = 3 2 × 3 = 23 60o Example 1 Find the area of the curved surface of the frustum of the cone. (Measurements in cm) Area = π rs (large cone)– π rs (small cone)= π (6)(12)– π (3)(6) = 90π cm2 Example 2 Find the surface areas of the minor dome and the major dome. (Measurements in cm) The radius of the sphere is 23 cm (refer to preliminary) . Area of minor dome = π r2 = π (23 )2 = 12π cm2 Area of major dome = 3π r2 = 3π(23 )2 = 36π cm2 Areas and volumes of regular and composite solids 1 © itute 2018 Example 3 Find the TSA of the composite solid formed by placing the minor dome in Example 2 ontop of the frustum of the cone in Example 1. TSA = minor dome + frustum of cone + base = 12π + 90π + π (6) 2 = 138π cm2 Example 4 Find the TSA of the composite solid formed by placing a major dome discussed in Example 2 on each end of a 10 cm long cylinder. TSA = 2 major domes + cylindrical curved surface = 2 × 36π + 2π rl = 2 × 36π + 2π(3)(10) = 132π cm2 Volume formulae Sphere: V = ~~ ~~43 π r3 Cylinder: V = π r2l Cone: V = ~~ ~~13π r2h l h r r Do you know the following facts about a sphere of radius r? The above is true only when the minor dome subtendsan angle of 120o at the centre of the sphere. Example 5 Find the volume of the frustum of the cone. (Measurements are in cm) The heights of the cones were obtained by the method outlined in the preliminary, using scale factors 3 and 6. Volume = large cone – small cone = 1 π (6) 2 (6~~ ~~3 ) − 1 π (3) 2 (3 3 ) = 63 3 π cm3 3 3 Example 6 Find the volumes of the minor dome and the major dome. (Measurements are in cm) Minor dome: V = 5 × 􀀆􀀄 4 π r3 􀀃􀀁 = 5 × π(23 )3 = 53 π 32 􀀅3 􀀂 24 Major dome: V = 27 × 􀀆􀀄 4 π r3 􀀃􀀁 = 9 × π(23 )3 = 273 π 32 􀀅3 􀀂 8 cm3 cm3 Areas and volumes of regular and composite solids 2 © itute 2018 Example 7 Find the volume of material to be cut from an appropriate cone to form the composite solid discussed in Example 3. Volume of material to be removed = volume of small cone – volume of minor dome = 1 π (3)2 (3 3 )− 53 π = 4~~ ~~3 π cm3 3 Example 8 Find the volume of material to be cut from an appropriate cylinder to form the composite solid discussed in example 4. 33 10 33 Radius of the large cylinder = 23 cm Length of the large cylinder = 63 +10 cm Volume of material to be removed = Large cylinder – small cylinder – 2 major domes = π (23 )2 (63 + 10)− π(3)2 (10) − 2(273 π ) = 6π(33 + 5) cm3 Exercise Find the TSA and volume of each of the following three-dimensional composite solids. (1) (2) (3) Areas and volumes of regular and composite solids 3 © itute 2018

FAQs

What are the surface area formulae for a sphere, cylinder, and cone?

The lesson gives SA for a sphere as SA = 4πr². For a cylinder it uses TSA = 2πr² + 2πrl, and for a cone it uses TSA = πr² + πrs (base plus curved surface).

How are the minor dome and major dome defined in the sphere examples?

They are defined for a sphere of radius r so that the minor dome subtends an angle of 120° at the centre. This allows the lesson to compute minor dome area as πr² and major dome area as 3πr².

How do you calculate the TSA of composite solids in the worked examples?

The lesson forms TSA by adding the relevant components, such as a minor dome plus a frustum of a cone and a base, or two major domes plus a cylinder’s curved surface. Each example substitutes the given measurements into the surface area formulae.

Sep 18, 2026

Uploaded by Jiven