Revision Topic 12: Area and Volume Area of simple shapes

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Revision Topic : Area and Volume Area of simple sapes You need to learn ALL of te following area formulae: Rectangle Triangle W L b Area = lengt widt Area = base eigt = ½ b Parallelogram Trapezium a b Area = base eigt = b Area = sum b of parallel sides eigt = ( a b ) You also need to know ow to find te area of a kite: You can find te area of a kite by splitting it into triangles. Alternatively, you can find te area of a kite by multiplying te two diagonals togeter and ten dividing by. Diagonal Area of a kite = product of diagonals Diagonal Dr Duncombe February 004

Example : Find te area of te following trapezium. 7cm 4.5cm cm To find te area, you add te parallel sides 7 + = 8 you multiply by te eigt 8 4.5 = 8 you divide by 8 = 40.5 cm Alternatively, you can use te formula: Area = ( a b ) ( 7 ) 4.5 8 4.5 40. 5 cm Note: In all area questions it is important to sow your metod and to give units. Example : If te area of tis triangle is 4cm, find x. x cm Te formula for te area of a triangle is A b Here te triangle as been rotated te eigt is cm and we need to find te base, x. So A b 4 = ½ x Double bot sides 48 = x So x = 4 cm.. Circles You need to learn te following: Circumference of a circle = πd Area of a circle = πr were D is te diameter of te circle and r is te radius (D = r). Dr Duncombe February 004

Example 3: A circle as a circumference of 50cm. Calculate te radius of te circle. Solution Te formula for te circumference is So. Terefore So C = πd 50 = π D D = 50 π D = 5.95 cm Te radius of te circle is 5.95 = 7.96 cm Example 4: Calculate te percentage of te diagram below tat is saded..5cm 4.6cm Te area of te larger circle is r 4.6 66. 48 cm Te area of te smaller circle is r.5 7. 07 cm So te area saded is 66.48 7.07 = 59.4 cm 59.4 Terefore te percentage saded is 00 89 % (to te nearest wole number). Examination Question 66.48.5 m.86 m A mat is made in te sape of a rectangle wit a semicircle added at one end. Te widt of te mat is.5 metres. Te lengt of te mat is.86 metres. Calculate te area of te mat, giving your answer correct to decimal places. Dr Duncombe February 004 3

Examination Question 4cm cm Te diagram sows a sape, made from a semi-circle and a rectangle. Te diameter of te semi-circle is cm. Te lengt of te rectangle is 4cm. Calculate te perimeter of te sape. Give your answer to 3 significant figures. Examination style question 3 7.4cm 6cm 9.6cm Calculate te saded area in te above diagram. Dr Duncombe February 004 4

Volume of prisms A prism is a tree dimensional sape wit a cross-section tat is te same all te way troug te sape. Tese are all examples of prisms: Cylinder Cuboid Triangular prism (circular cross-section) (rectangular cross-section) Prism wit a trapezium-saped cross-section Prism wit a cross-saped cross-section At grade B/C level, you need to be able to work out te volume of any prism. Te formula is Volume of prism = cross-sectional area lengt Cuboids and cylinders ave teir own formulae, wic are special cases of te general formula above: eigt Volume of a cuboid = lengt widt eigt lengt widt eigt, Volume of a cylinder = r radius, r Dr Duncombe February 004 5

Worked examination question : A cylinder is 3 cm ig. Te diameter of te base is.76 m. Calculate te volume in cm 3 of te 3 cm cylinder. Give your answer correct to 3 significant figures..76 metres Te formula for te volume of a cylinder is r (remember tis!!). Te diameter is.76 m or 76 cm (we want lengts in cm as volume must be given in cm 3 ). So te radius is 76 = 88 cm. As te eigt is 3 cm, te volume must be 88 3 3670 cm 3 or 36000 cm 3 (to 3 sf). Density Some questions on volume relate to density. You need to know te formula: Density = mass volume Worked examination question : or mass = density volume 5cm 6cm 0 cm Te diagram sows a prism. Te cross-section of te prism is a trapezium. Te lengts of te parallel sides of te trapezium are 8 cm and 6 cm. Te distance between te parallel sides of te trapezium is 5 cm. Te lengt of te prism is 0 cm. 8cm (a) Work out te volume of te prism. (b) Te prism is made out of gold. Gold as a density of 9.3 grams per cm 3. Work out te mass of te prism. Give your answer in kilograms. (a) To find te volume, we must first find te cross-sectional area (i.e te area of te trapezium). Area of trapezium = ( a b ) ( 6 8 ) 5 4 5 35 cm. Volume of prism = cross-sectional area lengt = 35 0 = 700 cm 3. (b) Mass = density volume = 9.3 700 = 350 grams. Te answer must be in kilograms, so 350g = 3.5 kg. Dr Duncombe February 004 6

Examination Question : COPPER NICKEL Tese two metal blocks eac ave a volume of 0.5m 3. Te density of te copper block is 8900 kg per m 3. Te density of te nickel block is 8800kg per m 3. Calculate te difference in te masses of te blocks. Examination Question : On a farm, weat grain is stored in a cylindrical tank. Te cylindrical tank as an internal diameter of 6 metres and a eigt of 9 metres. a) Calculate te volume, in m 3, of te tank. Give your answer correct to decimal places. 9m m 3 of weat grain weigts 0.766 tonnes. b) Calculate te weigt, in tonnes, of te weigt grain in te storage tank wen it is full. 6 m Examination Question 3: A gold bar is in te sape of a prism of lengt 0cm. It as a cross-section in te sape of tis trapezium. Calculate te volume of te bar..5 cm.4cm 3.5 cm Dr Duncombe February 004 7