Area And Volume Revision

Area and Volume Revision –Questions 1 of Paper 2Measurements –Firstly make sure that before you put any measurements into a formula they are of thesame unit. If the height is in mm and the radius is in cm then you will not have the correctanswer.Be careful here1cm 10 mm BUTIf you are dealing with length give your answer in cm1 cm 3 10 mm3If you are dealing with area give your answer in cm 21cm 3 1cm 1cm 1cmIf you are dealing with volume give your answer in cm 3 10mm 10mm 10mm 1000mm322Remember that is equal to 3.14 or7If you are asked to give your answer in terms of though do NOT put in 3.14 just leave in the answer.For example a circle with radius 3 would have an area of r 2 which we would write as (3) 2 9 cm 2If they do not ask to express in terms of then we multiply in 3.14 (3) 2 (3.14)(9) 28.26 cm 2Pythagoras –In a right angle triangle the square of the hypotenuse is equal to the sum of the squares ofthe other two sides.Basically the long side squared is equal to the sum of the other sides squared.This can be asked a number of ways but one of the most important and least obvious iswhen dealing with cones.Example – Calculate the length of the missing side in the following example.x43x 2 4 2 32x 2 16 9x 2 25x 51

If we are given a cone we are sometimes given the radius and length but NOT the height.The volume of a cone formula requires the height not the length and students often makethe mistake of putting the length in instead.In this case we use Pythagoras to find the height and then put the height into the formulafor h. LENGTH IS NOT THE SAME AS HEIGHT.Example –Calculate the height of the following cone.135Use Pythagoras132 5 2 x 2169 25 x 2169 25 x 2144 x 2x 122

Area and Perimeter of Basic Shapes –Examples – Find the (i) area and (ii) perimeter of the following shapes.Area4cmPerimeter l b 7 4 28cm 2 2l 2b 2(7) 2(4) 14 8 22cm7cm1bh21 (8)(6) 24cm 22Area of Triangle 683598Area of Shape Area of Triangle Area of Rectangle1 bh l b21 (9)(3) (9 5)21 (9)(3) (9 5) 13.5 45 58.5cm 22Perimeter (Circumference) 2 r 2(3.14)(8) 50.24cmArea of Circle r 2 (3.14)(8) 2 3.14(64) 201cm 214cmArea of Shaded region Area of Square – Area of CircleArea of Square l b 14 14 196cm 222Area of Circle r 2 ()(7) 2 154cm 272Area of Shaded region 196cm 154cm 2 42cm 23

Volume and Surface Area –Questions regarding volumeand surface area require theuse of formula that are foundin the tables. The first thing todo in these questions is towrite down the relevantformula then put in anyknown numbers.Example – Find the (i) volume and (ii) total surface area of the rectangular solid below.6cm2cm12cmVolume l w h 12 2 6 144cm 3Total Surface Area 2lw 2lh 2wh 2(12)(2) 2(12)(6) 2(2)(6) 48 144 24 216cm 2Example – A rectangular solid of length 3cm and height 4 cm has a volume of 84cm.Calculate its width.Volume l w h 843 w 4 8412w 8484w 12w 7cmWrite down formula and let it 84Fill in known valuesSimplifyDivide across by 12Width of the solidExample – Calculate the volume of a prism that has a perpendicular height of 5cm, abase of 10cm and a length of 15cm.1b h l21 10 5 152 375cm 3Volume 510154

Be careful when doing questionsinvolving cones. The surface areaformulae require the length of theside of the cone whereas the volumeformula requires the height.If you are given either the height orthe length and you are looking to findthe other use Pythagoras theorem.Example – Express to two decimal places the volume of a cone of radius 7cm and height10 cm.Volume of cone 1 r 2 h31(3.14)(7) 2 (10)31(3.14)(49)(10)31(1538.6)3512.87cm 2Write down the formulaInsert known valuesSimplifySimplifyExpressed to 2 decimalsExample – Express in terms of the volume of a cone of radius 5cm and height 12cm.Volume of cone 1 r 2 h31 (5) 2 (12)31 (25)(12)3 100 Write down the formulaInsert known valuesSimplifyExpressed in Example – The volume of a cone of radius 3cm is 132 cm 2 . Find the (i) cone’s heightand hence (ii) the cone’s total surface area.(i) Volume of cone 1 r 2 h 13231(3.14)(3) 2 h 13231(3.14)(9)h 13239.42h 132132h 14cm9.42Write down the formula and let equal 132Insert known valuesSimplifySimplifyDivide across by 9.42 and simplify5

(ii) Before we can use the Total Surface Area formula we must find the slant length ofthe cone.Use PythagorasIf the height is 14 and the radius is 3 then:x -slant heightl 2 14 2 32214l 196 9l 2 2053l 14.32Total Surface Area rl r 2 (3)(14.32) (3) 2 43 9 52 Write down the formulaInsert known valuesSimplifyExpressed in terms of Example – Express to two decimal places the volume of a cylinder of radius 14cm andheight 10 cm.Volume of cylinder r 2 h (3.14)(14) 2 (10) (3.14)(196)(10)Write down the formulaInsert known valuesSimplify 6154.40 cm 2Expressed to 2 decimalsExample – Express in terms of the (i) volume and (ii) total surface area of a cylinderof radius 5cm and height 12cm.(i) Volume of Cylinder r 2 h(ii) Total Surface Area 2 rh 2 r 2 (5) 2 (12) (25)(12) 300 2 (5)(12) 2 (5) 2 120 50 170 Insert valuesSimplifyExpressed in Insert valuesSimplifyExpressed in Example – The volume of a cylinder of height 7cm is 88 cm 2 .Find the cylinder’s radius.(i) Volume of cylinder r 2 h 8822 2r (7) 88722r 2 8888r2 22r2 4Write down formula and let equal 88r 2cmGet the square rootInsert known valuesSimplifyDivide across by 22Simplify6

The formulae for hemispheresare not in the logs tables butcan be arrived at using theformulae for the sphere.A hemisphere is basically justhalf a sphere so for thevolume and curved surfacejust get the sphere formulaand divide by 2.Example – Express to two decimal places the volume of a sphere of radius 6cm.Volume of sphere 4 r 334(3.14)(6) 334 (3.14)(216)3 904.32cm 3 Write down the formulaInsert known valuesSimplifyExpressed to two decimalsExample – Express in terms of the curved surface area of a sphere of radius 14cm.Volume of sphere 4 r 222 4()(14) 2722 4()(196)717248 7 2464cm 3Write down the formulaInsert known valuesSimplifySimplifyExpressed in Example – The volume of a hemisphere is 57 cm 2 . Find the hemisphere’s radius.Volume of hemisphere 2(3.14)r 3 5732.09r 3 5757r3 2.093r 27r 32 r 3 573Write down the formula and let 57Insert known valuesSimplifyDivide across by 2.09SimplifyGet the cubed root7

More complex problems –Part (c) will involve more complex thinking, taking the formula from the tables andapplying them to a two or three part question.The first thing to always do is write down the shapes involved and their formulae. Alsodraw a rough sketch to help you visualise what is being asked.It is also useful to leave and not put in 3.14 as a on each side of an equation willcancel each other out.The following are the types of questions asked in section (c) and examples of each followon the subsequent pages.1. Recasting – one object is melted down and made into another2. Displacement of water – one object placed into a water, the amount the waterrises is equal to the volume of the object.3. Shapes within a shape – calculate the area remaining when a number of smallobjects are placed in a large object.4. Two shapes in one – find the volume of a shape that is made from two othershapes.5. Flow of liquid – calculate the time taken for liquid to flow through a cylinder.Recasting –For this question the volumes of the two shapes will be equal so begin by writing downthe relevant formulae and letting them equal one another.Fill in all the known numbers and then solve for whatever is left.Example – A cylinder and a sphere have equal volumes. If the radius of the sphere is6cm and the radius of the cylinder is 8cm, calculate the height of the cylinder.Volume of cylinder r 2 h4Volume of sphere r 334 r 2 h r 334 (8) 2 h (6)33464h ( 216)364h 288288h 64 4.5cmWrite down formulaWrite down formulaLet formula equal each otherInsert known valuesCancel on both sides and simplifySimplifyDivide across by 64Height of cylinder8

Displacement of water –An object or group of objects is placed into water. The volume of the water that rises isequal to the volume of the object or objects placed in the water.Example – When a solid sphere is dropped into a cylinder partly filled with water, thelevel of the water rises by h cm. If the diameter of the cylinder is 16cm, calculate h.Draw a rough sketchWrite down the volume of the sphere and the volume of water displaced.Put down the formulae and enter known numbers.444Volume of Sphere r 3 (6) 3 216 288 (sphere)333Volume of Water displaced r 2 h (8) 2 h 64 h(cylinder)(Remember if diameter 16cm then the radius is 8cm)Volume of Water displaced Volume of SphereWater displaced volume of object64 h 288 We can cancel on both sides64h 288288Divide across by 64 and simplifyh 4.5cm64Shapes within a shape –Small objects are placed in a larger object and we are asked to find the area or volumeremaining.Example – Three tennis balls of diameter 7cm are placed inside a cylindrical container.The diameter of the cylinder is equal to the diameter of the tennis ball and the height issuch that the tennis balls cannot move about. What fraction of space is occupied by thetennis balls?Draw a rough sketchWrite down the volume of the large object, in this case a cylinder. Thediameter is 14cm so the radius is 7cm.The height is 3 tennis balls therefore 3 x 14 42 cm.Volume of cylinder r 2 h (7) 2 42 2058 Next find the volume of one tennis ball, radius 7cm.441Volume of sphere (7) 3 (343) 457 3331Total volume of 3 tennis balls 3 457 1372 321372 1372 Space occupied by tennis balls 32058 20589

Two shapes in one –Sometimes we will be asked to find the volume of a shape that consists of a combinationof cones, cylinders, spheres etc.In this case just add the volume of each shape separately to find the total volume of thefigure.Example – Calculate the volume of the shapes below.8410Shape Cone Cylinder1 r 2 h r 2 h31 (4) 2 (8) (4) 21031 (16)(8) (16)1032 42 160 32 202 cm 33Shape Cone Hemisphere14 r 2 h r 33314 (4) 2 (10) ( 4) 33314 (16)(10) (64)3311 53 85 332 138 cm 33Shape Prism Rectangular Solid1 l w h l w h21 (12 8 2) (12 8 5)2 96 480 576cm 310

Flow of liquid –Example – Water flows through a circular pipe of radius 2cm at the rate of 7cm/ sec intoa rectangular tank that measures 1.2m long by 1.1m wide by 30cm high. How long will ittake to fill the tank?ANSWER –Time taken to fill the tank Size of the TANKRate of flowWe need to find out two things first.1. The size of the tank2. The rate of flowFirst change all our measurements into cm1.2m 120cm1.1m 110cm30cm 30cmSize of tank Rectangular Solid l w h 120 110 30 396000cm 3To find the rate of flow we need to find the volume of the cylinder.Speed of 7cm/ s means that the height of the cylinder is 7cmVolume of cylinder r 2 h22 ()(2) 2 (7) 88cm 37Therefore 88 cubic cm passes through into the tank per second.Size of the TANK 396000 4500 seconds88Rate of flow4500 75mins6075 1.25hrs601 hour 15 minsDivide across by 60 to get hours (60 secs in minute)Divide across by 60 to get hours (60 mins in hour)The time in hours and minutes11

4 Volume and Surface Area – Example – Find the (i) volume and (ii) total surface area of the rectangular solid below. Volume Example – A rectangular solid of length 3cm and height 4 cm has a volume of 84cm. Calculate its width. Volume luwuh 84 Write down formula and let it