I.G.C.S.E. Volume & Surface Area. You can access the solutions from the end of each question

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I.G.C.S.E. Volume & Surface Area Index: Please click on the question number you want Question 1 Question Question Question 4 Question 5 Question 6 Question 7 Question 8 You can access the solutions from the end of each question

Question 1 Find the volume of the following prisms. All lengths are in cm. a. b. 7 4. 5.7 9.4 4 c. d. 6.4 8 4.8. 7.7 16 Click here to read the solution to this question

Solution to question 1 a. b. 7 4. 5.7 9.4 4 6.4 V = l w h = 4 7 = 56cm V = area of cross section length = area of trapezium length 1 = ( 6.4 + 4. ) 5.7 = 0.1 = 0.cm c. d. 8 4.8. 7.7 16 V = area of cross section length = area of triangle length = 1 base height 1 = 4.8. 7.7 = 60.984 = 61.0cm V = area of cross section length = area of semicircle length 1 = π r length 1 = π 4 16 = 18π = 40cm Click here to read the question again

Question. Find the surface area of the following rectangular prism. All lengths are in cm. 7 4 Click here to read the solution to this question

Solution to question 7 Drawing the net we can see 4 4 4 7 The surface area is the sum of the area of each of the six rectangles. ( 4 7 7 4) Surface area = + + = 100cm Click here to read the question again

Question a. Find the volume in litres of the following cylinder. ( 1L = 1000cm ). 15 cm 6 cm b. Calculate the surface area in cm. Click here to read the solution to this question

Solution to question a. Note that r 1 1 = d = 6 = cm 15 cm 6 cm V = area of base = πr h = π ( ) ( 15) = 15π = 44cm height b. Drawing the net r π r r h From the net we can see that the surface area is the sum of the area of the two circles and the rectangle. SA = πr + πrh ( ) π ( )( 5) = π + 1 = 18π + 90π = 108π = 9cm Click here to read the question again

Question 4 A solid cylinder of radius 5 cm and height 9 cm is melted down and recast into a solid cube. Find the side of the cube. Click here to read the solution to this question

Solution to question 4 9 cm 5 cm s The cylinder and the cube have the same volume. The volume of the cylinder is given = πr h = π ( 5) ( 9) = 5π cm Now the volume of the cylinder = the volume of the cube Let s be the length of the side of the cube The volume of the cube = s 5π = s s = 5π = 8.91cm Click here to read the question again

Question 5 a. Find the volume of the following cone, with radius 5 cm and vertical height 1 cm. 1 cm l 5 cm b. The cone has a slant height of l cm. Find the value of l. c. Find the curved surface area of the cone. Click here to read the solution to this question

Solution to question 5 1 cm l = Slant height 5 cm 1 a. The volume of a cone = πr h 1 = π 5 1 = 100π b. = 14cm ( ) ( ) 1 cm l By Pythagoras l = 5 + 1 l = 169 = 1cm 5 cm c. The curved surface area = πrl = π = 65π ( 5)( 1) = 04cm Click here to read the question again

Question 6 Find the volume and curved surfaced area of a sphere radius 4 cm. Click here to read the solution to this question

Solution to question 6 r 4 Volume of a sphere = π r 4 = π ( 4) 56 = π = 68cm Curved surface area = 4πr ( ) = 4π 4 = 64π = 01cm Click here to read the question again

Question 7 Find the height of a squared based pyramid of volume 40 9m Click here to read the solution to this question m and base area

Solution to question 7 h The volume of a pyramid 1 = base area height height volume = base area 40 = 9 = 1.cm Click here to read the question again

Question 8 A small pencil consists of a cylinder of radius 6 mm, which is sandwiched between a hemisphere and cone of the same radius. The height of the 50 mm. Find the total volume of the pencil. Diagram not to scale Click here to read the solution to this question

Solution to question 8 6 mm 50 mm Diagram not to scale 6 mm 0 mm Considering each shape separately, leaving our answers in terms of π, we have Volume of hemisphere 1 4 = πr 1 4 6 = π = 144π Volume of cylinder = π = π r h ( 6) ( 50) = 1800π Volume of cone 1 = 1 = π = 60π π r h ( 6) ( 0) Total volume of pencil = 144π + 1800π + 60π = 04π = 78 = 740mm Click here to read the question again