Year 10 Standard · Unit 5 · Lesson 3 · teacher page

Teaching: spheres and composite solids

Students can find the area of a circle and the surface of a cylinder. This lesson gives the surface of a sphere through comparison with circles and then asks the question that decides most composite-solid errors: which faces are actually on the outside?

Geometry and TrigonometryGDC optional

At a glance. One lesson of about 60 minutes. It assumes the sphere volume from lesson 2 and slant height from lesson 1. Students find the surface area of spheres and hemispheres and the surface area and volume of solids made from a cylinder with a hemisphere or cone.

Calculator: GDC optional. The lesson is done by hand with π left in the answer. The optional section shows GDC users how to enter the whole expression once, and warns that a GDC cannot decide which faces are outside.

What students will learn

Know

  • Sphere: surface area 4πr². Solid hemisphere: total 3πr².
  • A composite solid's surface is its outside faces only.
  • A composite solid's volume is the sum of the parts.

Understand

  • The sphere's surface is four times its great circle.
  • Joining solids hides faces: they are not part of the surface.

Do

  • Find the surface area of spheres and hemispheres, in exact form and rounded.
  • Work back from a surface area to a radius.
  • Find the surface area and volume of a cylinder with a hemisphere or cone on top.

MYP assessment criteria

Wording paraphrased. What this lesson gives you evidence for, and what to look for:

The one thing to do with the figure

Settle the hook with the ratio. Ask students to read the sphere and circle values at two different radii and say what stays the same.

In the second figure, switch between hemisphere and cone and ask: which line changed, and which did not? Then ask where the missing circle went.

A possible order

5 minWarm-up from the student page, in silence, then answers.
5 minHook: the orange peel. Collect votes and reasons.
10 minFirst figure; key idea; settle the hook.
10 minSecond figure; worked examples; pairs finish the faded example.
5 minQuestion 1, the check. If many choose A, return to the flat face.
20 minYour turn, Levels 1 to 3, at each student's own pace.
5 minSomething to think about, then the exit ticket, question 9.

The answers

Warm-up48π cm²; 36π cm³; 25π cm².
Hook4 circles: the surface is 4πr², four times πr².
Worked examples900π ≈ 2827.4 cm²; 300π ≈ 942.5 cm²; 215π ≈ 675.4 cm².
Faded example44π cm².
1. A solid hemisphere's surfaceB. A leaves out the flat face; C is a whole sphere; D puts 50 in place of r².
2. Surface area of a sphere256. 4 × 64.
3. Surface area of a solid hemisphere147. 3 × 49.
4. Surface area of a ball, rounded1521 cm². 484π.
5. Working back to a radius9. r² = 81.
6. Surface area of a cylinder with a dome168. 36 + 60 + 72.
7. Volume of the same solid324. 180 + 144.
8. Surface area of a cylinder with a cone116. 16 + 80 + 20.
9. A ball in a tubeA. B uses the volume ratio; C counts the lids; D halves the area.

What each wrong answer tells you

They giveWhat it usually meansWhat to do
A, C or D in question 1Left out the flat face, used a whole sphere, or took 50 as r².Write 2πr² = 50π, then find πr² before adding.
1024 or 64 in question 2Used the diameter as the radius, or gave r².Write r = ____ on the first line.
98 or 196 in question 3Left out the flat face, or used a whole sphere.List the faces of a solid hemisphere: one curved, one flat.
380 or 5575 in question 4Gave one circle, or used the volume formula.Ask: surface is a square measure, so which formula has r²?
81 or 18 in question 5Stopped at r², or forgot to divide by 4.Solve 4r² = 324 in two clear steps.
204 or 132 in question 6Counted the join, or left out the base.Sketch the solid and shade only the faces you could touch.
180 or 468 in question 7Gave only the cylinder, or used a whole sphere.Find each volume on its own line, then add.
132 or 108 in question 8Counted the cone's base, or used its vertical height.Mark the cone's slant side and find it by Pythagoras.
B, C or D in question 9Used the volume ratio, counted the lids, or halved.Work out both areas as numbers and compare.

Questions worth asking

Differentiation

A way in for anyone stuck

Stretch

Students learning in English as an additional language

“Surface” is the outside skin of a solid. “Dome” is the curved top of a hemisphere. “Join” is where two solids meet. Say “four pi r squared” aloud and point to the circle it is built from.

WordPlain meaningExample
spherea perfectly round solid, every point the same distance from the centrea ball
hemispherehalf of a spherehalf a ball
great circlethe biggest circle you can draw on a sphere, through its centrethe equator
composite solida solid made by joining simpler solidsa toy with a cone fixed to a cylinder
outside surfacethe faces you can touch from outside the solidbase, side and top
domethe curved top of a hemispherethe top of a mosque

Sentence starters for explaining aloud or in writing:

Two things not to say

“Add up the surface areas of the parts.” The joined faces are hidden: ask which faces you can touch.

“It is a sphere, so use 4πr².” A hemisphere's curved surface is half of that, and it may have a flat face.

Questions to set

Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.

1Fluency

The idea on its own, with friendly numbers. Set these first.

  1. SphereFind the surface area of a sphere of radius 5 cm in terms of π.
    100π cm².
  2. Earlier work (lesson 2)Find the volume of a hemisphere of radius 3 cm in terms of π.
    18π cm³.

2In context

The same idea inside a situation, where the first job is deciding what to calculate.

  1. ContextA glass dome on a building in Istanbul is a hemisphere of radius 10 m. Find the area of glass needed, to the nearest whole number.
    200π ≈ 628 m².
  2. Working backA solid hemisphere has total surface area 108π cm². Find its radius.
    3r² = 108, so r = 6 cm.

3Challenge

Reasoning, generalizing and working backwards.

  1. ReasoningA cone has radius 5 cm and slant height 13 cm, and a cylinder has radius 5 cm and height 12 cm. Which has the larger curved surface?
    The cone has π × 5 × 13 = 65π and the cylinder has 2π × 5 × 12 = 120π, so the cylinder.
  2. OpenA sphere and a solid hemisphere have the same total surface area. How do their radii compare?
    4πR² = 3πr², so r = 2R ÷ √3, about 1.15 times R.

Practicalities

Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.

Both figures work with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.

No device? An orange, a marker and a sheet of paper will do the first figure. The With a GDC section is optional.