Cones – Part 2Slideshow 48, MathematicsMr Richard SasakiRoom 307

Objectives• Review the properties of a cone and

formulae we have learned about their properties

• Be able to make calculations about attributes of cones regarding both their bases and lateral surfaces

Answers

𝑏=9𝑎=5𝑎=3

𝑏=1𝜋 𝑜𝑟 𝜋−1

𝑏=1𝑏=

450𝜋

𝑎=300𝜋

𝑏=18𝜋

The Net for a ConeLet’s have a quick review.

Sector(s)

Radius / Radii

Central angle(s)

Arc Length

𝑟 𝑟𝑎𝑜

𝑎𝑜 (Area)

(Area)𝑙 𝑙

Lateral Surface

Base

The Net for a ConeThe net for a cone is formed by two things.A and a .circle sectorIf we have information regarding both…what is one common piece of information with the same name? The radiusBecause of this we will think of a cone as just a cone, not as a circle and a sector.

The ConeWhen the cone is produced, what do we call the radius of the sector?

Slant height -

𝑎𝑜

𝑟h

The slant heightTwo pieces of information (not labelled) are identical. What are they?The circumference of the base and arc length of the sector

Finding Things…Have a look at this cone.

5𝑐𝑚

2𝑐𝑚We’re given two pieces of information.𝑟 −The radius of the base.𝑙−The slant height.

How could we calculate the circumference at the base?2𝜋 ∙2¿ 4𝜋𝑐𝑚

This is also

.

the lateral surface’s arc length

Finding Things…5𝑐𝑚

2𝑐𝑚4𝜋𝑐𝑚

How about the angle at the apex formed by the lateral surface?

𝑎𝑜

Do you remember the formula we need?We need the formula to find the circumference of the base. 𝐶=

2𝜋 𝑙𝑎360

4𝜋=2𝜋 ∙5 𝑎360⟹4 𝜋=

𝜋 𝑎36⟹144𝜋=𝜋 𝑎⟹𝑎=144𝑜

Answers (Section 1)𝐶=2𝜋 ∙4=8𝜋 𝑐𝑚𝐶=

2𝜋 𝑙𝑎360 ⟹ 8𝜋=

2𝜋 ∙5𝑎360

⟹2880𝜋=10𝜋 𝑎⟹𝑎=288𝑜𝐶=2𝜋 ∙7=14 𝜋𝑐𝑚𝐶=

2𝜋 𝑙𝑎360 ⟹14𝜋=

2𝜋 ∙13𝑎360

⟹5040 𝜋=26𝜋 𝑎⟹𝑎=( 252013 )𝑜

𝐴=𝜋 ∙42=16𝜋 𝑐𝑚2 𝐶=2𝜋 ∙4=8𝜋 𝑐𝑚𝐶=

2𝜋 𝑙𝑎360 ⟹ 8𝜋=

2𝜋 ∙8𝑎360⟹2880 𝜋=16𝜋 𝑎⟹𝑎=180𝑜

𝑆=𝜋 𝑙2𝑎360

=𝜋 ∙82 ∙180360

=32𝜋𝑐𝑚2

+

Shortcuts…We know and and these are fine to use to calculate the surface area when we are given and . But it’s a long process, right?

Let’s check our process with Q3.

𝐴=𝜋 𝑟2Surface Area

𝑆=𝜋 𝑙2𝑎360

𝐶=2𝜋 𝑙 𝑎360

𝐶=2𝜋 𝑟⟹𝑎=¿

360𝐶2𝜋 𝑙

Shortcuts…𝐴=𝜋 𝑟2Surface Area

𝑆=𝜋 𝑙2𝑎360

𝐶=2𝜋 𝑟𝑎=360𝐶2𝜋 𝑙

𝑎=360 ∙2𝜋𝑟2𝜋 𝑙 ¿

360𝑟𝑙

𝑆=𝜋 𝑙2360∙( 360𝑟𝑙 )¿ 𝜋 𝑙 ∙360𝑟360

¿𝜋 𝑙𝑟

Surface Area (cone) =

ExampleKnowing shortcuts makes things easier! But of course you have to remember them.ExampleA cone has radius 5cm and slant height 8cm. Calculate its surface area.

¿𝜋 ∙52+𝜋 ∙8 ∙5¿25𝜋+40𝜋

¿65𝜋 𝑐𝑚2

Answers - EasyS.AS.A

S.A

S.A

S.A

S.A

Tall and thin.

Answers - HardS.A

Both start at the same place and reach the centre but the slant height must reach a higher point too.

S No. (The reasoning is the same as in Question 2.)

𝐶=2𝜋 𝑟 ⇒𝑟=𝐶2𝜋⇒𝐴=𝜋𝑟2⇒𝐴=𝜋 ∙( 𝐶2𝜋 )

2

=C2

4 𝜋

S.A ⇒− 𝜋 𝑙𝑟=𝜋𝑟 2−𝑆 . 𝐴⇒𝑙= 𝑆. 𝐴− 𝜋𝑟2𝜋𝑟

S.A = C S.A = ¿

𝐶2

4 𝜋 +𝐶𝑙2