1. The volume of the object below is 186 cm 3. Calculate the Length of x. (a) 3.1 cm (b) 2.5 cm (c) 1.75 cm (d) 1.25 cm

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1 Volume and Surface Area On the provincial exam students will need to use the formulas for volume and surface area of geometric solids to solve problems. These problems will not simply ask, Find the volume of a sphere. They will require a student to preform multiple steps and sometimes use multiple formulas to find solutions. The three problems below illustrate these multistep problems and give ideas for problem solving strategies. The most important tip for solving these problems is to draw a picture (or on the picture given) to sort out the information given and find the path to the solution you seek. 1. The volume of the object below is 186 cm 3. Calculate the Length of x. (a) 3.1 cm (b) 2.5 cm (c) 1.75 cm (d) 1.25 cm Solution: In order to find x we must use the volume of this figure to determine what value of side length x would produce such a volume. The easiest way to find the volume is to divide the object into multiple rectangles, find the volume of each rectangle and then add those volumes together. In the drawing below I have divided this figure into three rectangular solids. There are many ways this object could have been divided. I chose to divide it up in a way that gave me the most figures with which I could find the numerical volume without knowing x. 1

2 3cm 3cm x x 6cm 3cm 6-x 3cm 4cm 4cm Of the three rectangular solids you see above two are identical. They have a width of 3 cm a length of 4 cm and a height of 6 cm. I can easily compute the volume for these figures. V = 3 cm 4 cm 6 cm V = 72 cm 3 My goal is to create the following equation so that I can solve for x: V = 186 cm 3 = Volume of three Rectangular Solid. V = 186 cm 3 = 72 cm cm 3 + Volume of final solid To find the volume of the final solid I notice that the width of the figure is 3 cm and the length is 4 cm, as with the other figure. I also see that the height is 6 cm minus the value of x, 6-x. The volume of that figure is therefore: V = 3 cm 4 cm (6 x) cm V = 12(6 x) cm 3 V = 72 cm 3 12x cm 3 Now I ll add this to the equation above. I will drop the cm 3 units to make our calculations easier. 186 = x 186 = x Now subtract 216 from both sides. 2

3 = 12x 30 = 12x Next divide both sides by = x 2.5 = x Thus our answer is b. 2. The slant height of the pyramid below is 45cm. Calculate its volume. (a) 10,062 cm 3 (b) 12,728 cm 3 (c) 13,500 cm 3 (d) 40,500 cm 3 Solution: The volume of a pyramid is given by the following formula. V = 1 3 (area of base) h The area of the base (which you will notice is a square with 30cm long sides) is: 30 cm 30 cm = 900 cm 2 Therefore we only need to find the height of this pyramid. If we take a cross section cut of the pyramid right through the middle we would see something like this. 3

4 45cm h 15cm 30cm Using this diagram we can use the Pythagorean Theorem which says in a right triangle the following relationship holds: a 2 + b 2 = c 2, where a, b and c are the lengths of the sides of the triangle, and c is the length of the longest side, the side opposite the right angle. Therefore we have: (15cm) 2 + h 2 = (45cm) cm 2 + h 2 = 2025 cm 2 h 2 = 2025 cm cm 2 h 2 = 1800 cm 2 h 2 = 1800 cm 2 h = 42.4 cm Now we have all the information we need to find the volume of this figure. V = 1 3 (area of base) h V = cm cm V = , 160 cm3 V = 12, 720 cm 3 While this is not exactly the same as any of the solutions it is closest to b. My result is a little different due to the way I rounded 42.2cm. The answer therefore is b. 4

5 3. A cone shaped water tank has a volume of 1000 litres. Which diagram best represents the 250L, 500L and 750L marks on the side of the tank? Solution: We will not even ask what situation made it necessary to have a cone shaped water tank, though I am convinced this is not the most efficient use of space or materials. The key to this problem is to consider how does the volume of liquid I put into the water tank change as I move up the cone. After a bit of contemplation it is easy to see that the bottom of the cone will only hold a little water. (If this is confusing think about an ice cream cone. Once you have eaten off the big scoop on top and got down to cone level where is the most ice cream located in your cone? In the top half! By the time you get down to those last two or three bites you are mostly tasting cone.) If we want to put lines on a cone to divide it into 4 equal sections and we know that the bottom of the cone holds the smallest volume we can then deduce that the first mark (the 250L mark) is going to be quite high up the side of the cone. This allows me to rule out C and D as possible solutions. B on closer analysis has 4 marks evenly spaced along the side of the tank, which is not consistent with our observations of an ice cream cone. Therefore A is the logical answer. If you are concerned about remember the many formulas for surface area, area, and volume of geometric figures rest assured. The following tables and charts will be available for you to use during the provincial exam. 5

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