Surface Area and Volume of a Cone

Surface Area and Volume of a Cone Goal: To learn how to calculate the surface area and volume of a cone.Draw the net of a cone and determine the surface area. LA=Lateral Area The ratio of the areas of the cone and the circle is the same as the ratioof their circumferences. 𝐿𝐴 𝑐𝑜𝑛𝑒 𝐿𝐴 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑐𝑜𝑛𝑒 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝐿𝐴 𝑐𝑜𝑛𝑒 π𝑠 2 = 2π𝑟 2 π𝑠 π𝑠 2 𝐿𝐴 𝑐𝑜𝑛𝑒 π𝑠 2 ( ) = π𝑠 2 2π𝑟 2 π𝑠 ( ) 𝐿𝐴 𝑐𝑜𝑛𝑒 = π𝑟 𝑠 𝑆. 𝐴. 𝑐𝑜𝑛𝑒 = 𝐿𝐴 𝑐𝑜𝑛𝑒 + 𝐴 𝑏𝑎𝑠𝑒 = π𝑟 𝑠 + π𝑟 𝑠 Example 1: Determine the Surface Area of this cone. Solution: 𝑆. 𝐴. = π 𝑟 𝑠 + π𝑟 2 = π 1 ( ) 2 ( ) + π 1 ( ) 2

= 2π + π = 3π = 9. 42 The surface area of this cone is approx. 9. 42 𝑚 2 Example 2 : The lateral area of a cone with radius 4 cm is . 60 𝑐𝑚 2 a) Determine the slant height of the cone, to the nearest centimetre.b) Determine the height of the cone, to the nearest centimetre. Solution: a) 𝐿𝐴 = 60 𝑐𝑚 2 𝑟 = 4 𝑐𝑚𝑆 =?𝐿𝐴 = π𝑟 𝑠60 = π 4 ( )𝑠 60 = 4 π𝑠 60 4 π = 4 π𝑠 4 π 𝑆 = 4. 77𝑆 = 5 S is approx. 5 cm b)Use Pythagorean Theorem to find h. ℎ 2 + 𝑟 2 = 𝑠 2 ℎ 2 + 4 2 = 5 2 ℎ 2 + 16 = 25 ℎ 2 = 25 − 16 ℎ 2 = 9 ℎ = ±3 So, h=3 and h=-3h is 3 cm Explain how you can use the General Volume formula to develop the formula for a cone. V= 13 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 ( )× ℎ𝑒𝑖𝑔ℎ𝑡 ( ) 𝑉 = 13 π𝑟 2 ℎ 𝑉 𝑐𝑜𝑛𝑒 = π𝑟 2 ℎ 3

The base of a cone is the circle. Note that 𝑉 𝑐𝑜𝑛𝑒 = 13 𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 3 𝑉 𝑐𝑜𝑛𝑒 Example 3 : Determine the volume of this cone. Solution: 𝑉 = 13 π𝑟 2 ℎ We need to find the height, h of the cone (See example 2b) h=2.7 m. r=1 m S=2m 𝑉 = 13 π 1 ( ) 2 2. 7 ( ) = 2.7 π 3 = 2. 8 The volume of the cone is approx. 2. 8 𝑐𝑚 3 Example 4 : A cone just fits inside a cylinder with volume 300 𝑐𝑚 2 . What is the volume of the cone? Solution: 𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 300 𝑐𝑚 3 𝑉 𝑐𝑜𝑛𝑒 = 13 𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟

= 13 ( ) 300 ( ) = 100 The volume of the cone is . 100 𝑐𝑚 3 Example 5 : A cone has a height of 4 cm and a radius of 3 cm. Another cone h as a height of 3 cm and a base radius of 4 cm. Can you predict which cone has the greater volume? Solution:Cone_1: h=4 cm 𝑟 = 3 𝑐𝑚 𝑉 𝑐𝑜𝑛𝑒 = 13 π𝑟 2 ℎ = 1 3 π 3 ( ) 2 4 ( ) [ ] = 36 π 3 = 12π Cone_2: ℎ = 3 𝑐𝑚𝑟 = 4 𝑐𝑚 𝑉 𝑐𝑜𝑛𝑒 2 = 13 π𝑟 2 ℎ = 13 π 4 ( ) 2 3 [ ] = 48 3 π = 16 π The cone with the bigger base has a greater volume.