﻿ Volume of Solids

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## Volume of Solids

LESSON OBJECTIVE

#### By the end of the lesson the you should be able to find the volume of a frustrum.Introduction Volume of a cone To find the volume of the above cone you proceed as follows Example 1 Find the volume of a cone whose slant height is 6cm and radius is 4cm.We use pythagoras theorem to find height of the cone i.e. = 36-16= 20 h = 4.4721cm h = 4.472cm Therefore the area of cone = base Area x height  =74.94cm3             volume of a pyramid     The volume of the pyramid is worked our as follows V = Therefore, if the height is 12cm, the base length 5cm and width 4cm Volume = =128cm3  Introduction Volume of a cone To find the volume of the above cone you proceed as follows Example 1 Find the volume of a cone whose slant height is 6cm and radius is 4cm.We use pythagoras theorem to find height of the cone i.e. = 36-16= 20 h = 4.4721cm h = 4.472cm Therefore the area of cone = base Area x height  =74.94cm3             volume of a pyramid     The volume of the pyramid is worked our as follows V = Therefore, if the height is 12cm, the base length 5cm and width 4cm Volume = =128cm3  Volume of a Cone To find the volume of the above cone you proceed as follows Example 1 Find the volume of a cone whose slant height is 6cm and radius is 4cm. We use Pythagoras theorem to find height of the cone i.e. = 36-16= 20 h = 4.4721cm h = 4.472cm Therefore the area of cone = base Area x height =74.94cm3 Volume of a frustum A frustum is obtained ,when a cone or pyramid is cut along a plane parallel to the base shown below: Example  Find the volume of the given frustum whose top radius is 6cm,and bottom Radius is 2cm and the height is 10cmSolution To obtain the volume of the above frustum ABCD, we have to get the vertical ,height h of the smaller cone DCV. We know that the smaller cone DCV is similar to the bigger cone ABV hence the ratio of their corresponding sides are equal. Type equation here. 6h=2(10+h) 6h =20 + 2h 4h =20 H= = 5cm Volume of cone = h Volume of bigger cone ABV = x 3.142 x 6 x 6 x 15 =565.56cm3 Volume of smaller cone DCV x 3.142 x2 x 2 x5 =20.95 cm3 Therefore the volume of frustum =volume of bigger cone ' volume of smaller cone. =(565.56 -20.95) cm3 =544.61cm3  Exercise 1. find the volume of the lampshade the shape of a frustrum whose bottom radius 12cm,top radius 6cm and slant height is 10cmDiag(a) Solution Diag(b)   Complete the missing part of the cone. Then find L the slant height of the small cone using the ratio of the corresponding of sides of the similar cones i.e. = = = 12L= 6(10 +c) 12L=60 + 6L 6L = 60 L =10 STEP2 Using the information derived find H-the vertical; height of the bigger cone.  H= = = =16 STEP 3. Find the vertical height of the small cone Using the corresponding radios = = = 12h =96 h= 8 step4: volume of the bigger cone = cm3 Volume of smaller cone = = cm3 Volume of the frustum = volume of bigger cone ' volume of smaller cone. = 2413.056 - 301.632cm3 = 2111.424 cm3 2. Find the volume of the given frustum which has a square base of 6cm and a top has Sides 2cm.the height of the frustum is 12cm.  Solution To find the volume of the frustum using dotted lines and the vertec v, then find the volume of big and small pyramids. (use different colour for dotted height)  Diag(d) 6h =2(12 +h) 6h =24 + 2h 4h =24 h =6 Therefore, height of the bigger pyramid =12 +6=18 volume of smaller pyramid= = = 216 cm3 Volume of smaller pyramid = = = 24 cm3 Volume of the frustum = volume of larger pyramid ' volume of small pyramid =(216 ' 24) cm3 =192cm3

Volume of Solids

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Volume of Solids

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Volume of Solids

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