GEOMETRY AND TRIGONOMETRY /25
Represent, in a variety of ways, two-dimensional shapes and three-dimensional figures arising from
real-world applications, and solve design problems;
Solve problems involving trigonometry in acute triangles using the sine law and the cosine law,
including problems arising from real-world applications.
One of the most important concepts in all of geometry is Pythagorean Theorem. It states that for right angle triangles such as the one shown here, the square of the hypotenuse, c, is equal to the sum of the squares of the other two sides, a and b. As a formula, this is:
Example Find the hypotenuse of the following triangle.
c is always the hypotenuse but it doesn’t matter which sides we pick for a and b, so we will just plug them into the formula and solve.
Therefore the hypotenuse is 5.
Question 1 Find the hypotenuse of the following triangle. 
Two Dimensional Shapes
Question 2 Find the perimeter and area of the following composite shapes. Hint: You will need to use Pythagorean Theorem to find the missing side length in the first shape. There is a formula sheet at the end of this package. 
Three Dimensional Shapes
Example A cylindrical roll of tape is almost used up. Determine the volume of tape that remains.
First we can find the volume of the entire object.
Next we will find the volume of the inside portion that contains no tape. The entire diameter is 26mm, but there is 5mm of tape on either side leaving the diameter of the inside portion to be 16mm. This means that the radius of this portion is 8mm.
Finally we subtract the inside portion from the total.
Therefore there are3298.7 mm^3 of tape.
Question 3 Three tennis balls that are each 8.4cm in diameter are stacked in a cylindrical can.
a) What is the minimum amount of aluminum required to make the can? Round to the nearest tenth. Hint: The minimum amount to make the can means you need to find the surface area. 
b) If the three balls fit perfectly in the can, how much empty space is in the can? Round to the nearest tenth. 
From grade 10, you should know about the primary trig ratios.
Sine Cosine Tangent
sinx=opposite/hypotenuse cosx=adjacent/hypotenuse tanx=opposite/adjacent
Question 4 Determine the primary trig ratios (sinx, cosx, and tanx) for the following triangle. 
Sine Law and Cosine Law
To find angles in non right angle triangles, we can use sine law and cosine law.
For finding angles:sinA/a=sinB/b=sinC/c
For finding sides:a/sinA=b/sinB=c/sinC
When to use Sine Law
When you have the measure of an angle and the corresponding side opposite to that angle, you can use Sine Law to find the remaining sides and angles.
Example Determine angle x and side length y in the following triangle.
Since we have two angles, we can find the third using the property that the interior angles add to 180O.
Now we can use Sine Law to find the missing side length.
Therefore the missing angle is 53O and the missing side is 22.4m.
When to use Cosine Law
If you have all three sides and no angles, or two sides with the angle between them, then you can use cosine law. A good rule is that if you can’t use sine law then you can use cosine law.
Example Determine the missing side length in the following triangle.
We have everything we need for the cosine law.
Therefore the missing side length is 7.8m.
Question 5 Marv and Elvis are each flying a kite when they collide. At the point where the kites collide, Marv’s string is 25 metres and makes a 35O angle with the horizontal, while Elvis’ string is 22 metres and makes a 45O angle with the horizontal. How far apart are Marv and Elvis? 
Question 6 The Leaning Tower of Pisa is 55.9 metres tall and leans 5.50 from the vertical. If its shadow is 90.0 metres long, what is the distance from the top of the tower to the top edge of its shadow? 
MATHEMATICAL MODELS /19
Make connections between the numeric, graphical, and algebraic representations of quadratic relations, and use the connections to solve problems;
Demonstrate an understanding of exponents, and make connections between the numeric, graphical, and algebraic representations of exponential relations;
Describe and represent exponential relations, and solve problems involving exponential relations arising from real-world applications.
Quadratic functions take the form
The curve they form is called a parabola.
Vertex – This is the highest or lowest point on the parabola. The x-value can be found by using the formula x=(-b)/2a.
y-Intercept – This is where the parabola crosses the y-axis. It is the value of c.
Roots – These are the x-intercepts of the parabola. They can be found by factoring, or by using the quadratic formula.
Example Determine the vertex and x-intercepts of the parabola y=3x^2+6x-1
Therefore the vertex is (-1,-4) X-Intercepts
This doesn’t factor so we can use the quadratic formula.
(-6+√48)/6=0.15or (-6-√48)/6= -2.15
Therefore the two roots are 0.15 and -2.15.
Question 1 Determine the vertex and x-intercepts of the parabola y=-2x^2+4x+5. 
Example The height, h, in metres, above the ground of a football t seconds after it is thrown can be modelled by the function h=-4.9t^2+19.6t+2.
a) Determine the maximum height that the football reaches.
For the maximum height, we need to find the h value of the vertex. First we need to find the t value of the vertex.
We can now plug this into the function to determine the height.
This means that the maximum height reached by the football is 21.6 metres.
b) Determine how long the football will be in the air.
For this, we need to find the x intercepts which we can do using the quadratic formula.
Since the ball was thrown at t=0, the ball was in the air for 4.1 seconds.
Question 2 A diver followed a path defined by h=-4.9t^2+3t+10in her dive, where t is the time, in seconds, and h represents her height above the water, in metres.
a) What was the maximum height the diver reached? 
b) For how long was the diver in the air? 
There are several exponent laws that can be used to simplify expressions that have the same base. Here are four common and important ones.
Product Quotient Power to a Power Negative Exponents
Add the exponents 2^7/2^2 =2^(7-2)=2^5
Subtract the exponents (5^4 )^2=5^(4×2)=5^8
Multiply the exponents 6^(-3)=1/6^3
Multiply the exponents
Examples Simplify the following expressions.
Solution (〖4^2/4^7 )〗^(-3)
=(5^3 5^(-6))/5^4 Start by simplifying the top and the bottom separately =(〖4^2/4^7 )〗^(-3) Start by simplifying the inside of the brackets
=5^(3+(-6))/5^4 =5^(-3)/5^4 Apply the product rule =(〖4^(2-7))〗^(-3)=(〖4^(-5))〗^(-3) Apply the quotient rule
=5^(-3-4)=5^(-7) Apply the quotient rule 〖=4〗^(-5×(-3)) 〖=4〗^15 Apply the power rule
=1/5^7 Rewrite as a positive exponent
Question 3 Simplify the following expressions. Leave your answers with positive exponents.  a) 4^(-7)×4^3 b) (9^6/9^3 )^(-2) c)(7^5 )^(-8) (7^(-1))
Consider a rumour at your high school that your principal is replacing the library with a room full of puppies. On the first day, five people are aware of the rumour. On the second day, each of those people tell two additional people, meaning that the number of people that hear the rumour each day doubles. The number of people who hear the rumour continues to double every day after that. An equation that models this situation would take the following form:
This is an example of an exponential growth, where the dependent variable increases by a constant factor while the independent variable increases by a constant amount. In this case, the number of people who hear a rumour increases by a factor of 2 anytime the number of days increases by 1.
In the previous situation, the growth factor was 2 because the number of people were doubling. If three people were told the rumour every day, the growth factor would be 3 and the number of people would be tripling. This would make the equation N=5(3^t).
Question 4 If the growth factor was 4, and initially there were 8 people that knew the rumour, which of the following equations would model this situation? Select one.  N=4(3^t) N=8(4^t) N=5(4^t) N=4(8^t)
Question 5 Bacteria A has an initial population of 500 and doubles every day, while bacteria B has an initial population of 50 and triples every day.
Write an equation for the population of each bacteria. 
Bacteria A: Bacteria B:
After 10 days, which bacteria population is greater? Use your models from part a) to determine the populations of each bacteria. 
The graph of an exponential is shown on the right. Specifically, this is y=2^xand you can see that the height of the graph doubles every for every step taken to the right.
a is the initial amount, and is also the y-intercept on the graph.
b is the growth or decay factor. If b is less than one then it will be a decay and the graph will be decreasing.
Question 6 Which of the following graphs can be represented by the equation y=10(4^x)? 
Graph 1 Graph 2 Graph 3 Graph 4
Question 7 Which of the following graphs can be represented by the equation y=3(〖1/2)〗^x? 
Graph 1 Graph 2 Graph 3 Graph 4
PERSONAL FINANCE /18
Compare simple and compound interest, relate compound interest to exponential growth, and solve problems involving compound interest;
Compare services available from financial institutions, and solve problems involving the cost of making purchases on credit;
Interpret information about owning and operating a vehicle, and solve problems involving the associated costs.
Simple and Compound Interest
There are two types of interest, simple and compound. Simple interest is always calculated based on the principal, or initial amount. With simple interest, the amount is added to the total every year. Compound interest is recalculated based on the most recent amount. Every compounding period, the amount of interest added to the total is larger than the previous amount.
Simple Interest Interest=Prt
Compound Interest Total Amount=P(1+i)^n
P- Principal Amount, r- Annual Interest rate, t- Number of years,
i- Compounding Interest Rate, n- Number of Compounding Periods
Question 1 Which graph below represents simple interest, and which represents compound interest? Explain your answer. 
Example Steve deposits $835 into an account that earns 8% per year. Determine the amount in the account after 5 years if the interest is:
The interest rate is 8% so r is 0.08. The principal amount is $835 and t is 5.
This is just the interest, so to find the total amount in the account, we add the interest to the principal amount.
Therefore there would be $1169 in the account after 5 years.
The interest rate is 8% and the compounding frequency is semi-annual, meaning twice a year. We divide the interest rate by 2 to determine the compounding interest rate, soi is 0.04. In 5 years, there would be 10 compounding periods (n=10).
Unlike simple interest, this is the total amount that is in the account, therefore there would be $1236 in the account after 5 years.
Question 2 Ryder deposits $500 into an account that earns 4.5% per year. Determine the amount in the account after 5 years if the interest is:
Simple Interest  Compounded Monthly 
Credit cards are great for making a variety of purchases, but they can easily accumulate large amounts of interest, making these purchases much more expensive in the long term.
Credit cards charge daily compounded interest. This means that we divide the interest rate by 365 to obtain the daily interest rate. If the bill is paid before the grace period ends, no interest is earned.
Question 3 Refer to the sample statement above.
If the bill is issued on the 8th of each month, when is the minimum payment due? 
If Ahmad’s balance is $85, what will the minimum payment be? 
If Ahmad’s balance is $413, what will the minimum payment be? 
Example Ahmad made a purchase of $462.18 on September 15th. If he makes his payment on November 3rd, how much interest will he owe?
First we need to determine the daily compounded interest rate by dividing the annual rate, 14.9% by 365.
We use the compound interest formula to determine the final amount. If the payment was made on November 3rd, that would be 49 days after the purchase, so n=49.
Therefore he would owe $9.33 in interest.
Question 4 You recently obtained a credit card with an annual interest rate of 14.9%, and you made a purchase of $1200. You paid the entire balance 55 days after you made the purchase. How much interest did you owe? 
Owning a Vehicle
Question 5 Bailey is 19 and owns a seven-year-old mid-sized car. He called several insurance agents and the lowest quote he received was $2620/year. There are two payment options: he can pay the insurance premium in full once a year or he can make monthly payments of $230.
a) Calculate the annual cost if he chooses the monthly instalments 
b) Calculate the difference between the two payment methods 
c) Suggest reasons why Bailey might choose each option.