# Grade 10 Math Exam Review

August 13, 2017 | Author: BrendanJHiggins | Category: Sine, Triangle, Trigonometric Functions, Quadratic Equation, Trigonometry

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Full and comprehensive math review for grade 10 created for ontario curriculum information in this document was studen...

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Grade 10 Math Exam Review Concepts Solving systems of equations    

Solving by graphing Solving by substitution Solving by elimination Word problems

Polynomials and factoring    

Expansion and simplification Factoring Special Quadratics Special Products

Solving by graphing Solving by factoring The quadratic formula

Quadratic functions    

Graphing functions Sketching parabolas Parabolic properties Word problems

Trigonometry      

Trigonometric ratios Solving right triangles Problems involving two right triangles Sine law Cosine law Word problems

Analytic geometry  

Line segment length Midpoints of line segments

 

Verifying geometric properties Distances from points to lines

Solving Systems of equations Numerical Problems 1. { 

How to solve by graphing i. Change to y=mx+b form ii. Graph the y-intercept iii. Find the next points using the slope iv. Repeat for other lines

How to solve by substitution i. Isolate one of the variables (one with the lowest co-efficient usually works best) ii. Divide both sides by the co-efficient of the isolated variable iii. Substitute that value into the same variable in the other equation iv. Find the value of the other variable and then write: (x,y)

How to solve by elimination i. Multiply both equations by their factor in the lowest common variable ii. Add or subtract one from the other so you eliminate the equal variables iii. Calculate the value of the remaining variable iv. Substitute the value of that variable to find the value of the other ) v. Write as (

2. {

3. {

Word problems 4. There are ninety coins, consisting of quarters and loonies, total \$42, how many of each type of coin are there?  Begin with the “let… then… statement”  In this case, let x represent the number of quarters, and y represents the number of loonies. Then, { 

Then simplify both equations, in this case, only the second can be simplified {

Then you solve the system of equation, by whatever means easiest (or required)

   Then substitute the value of “y” (in this case) into the other equation    Then you write your statement:  There are 24 Loonies and 66 quarters 5. Amy travels 1860km in 6.5h, partly by car at 80km/h and the rest by plane at 750km/h, how did she travel by each mode  Let x represent the amount of time she travelled the amount of time she travelled by car and y the amount of time she travelled by plane. Then, { 

{

( )      Then you substitute the value of y into the first equation   Then you write the statement  Amy travelled 4.5 hours by car and 2 hours by plane 6. Carol invests \$25000, partly in low-risk bonds at 4% and the rest in a stock that she thinks will earn 8%. If she expects to earn \$1280 interest, how much did she invest at each rate?  Let x represent the amount of money invested in bonds and y the amount of money invested in stocks. Then, { 

(

)

(

)

{

( )   100000-4y+8y=128000     ( )  Then substitute the value of y into the first equation   Then write the statement:  Carol invested 18000 in bonds and 7000 in stocks 7. Barb ran 100 meters with the wind in 20 seconds and against the wind in25 seconds, find barb’s running speed and the wind speed  Let x represent the speed she travelled with the wind and y the wind speed

{

{

{

(

simplify

)

    

Then write the statement

. /

Polynomials and factoring 8.

9.

( ) Expand then simplify  Write each term within the brackets as multiplied by the terms outside of those brackets ) ( )  4 ( ( )  Then multiply   Combine like terms (if any) Factor  Extract the greatest common factor  In this case, the greatest common factor is ( )   Then find two factors of the final term – 12 that add up to the coefficient of the middle term, -8. -2 and -6 work, so substitute them into the expression in place of -8a    

( – – ) Then group the terms , ( (

– ) – ( – ))( – )

Factor difference of squares 

Take the square roots of the two terms. In brackets, add them together, and in another set of brackets, subtract the square root of the negative term ( ) from the positive one ( ) ( )( – ) –

Factor

Group the terms, as appropriate. 12vw can be grouped with -8w, the greatest common factor being 4w, and 15v with -10, the greatest common factor being 5.

 

( – ) ( – ) Take the two coefficients – in questions like this, they’re the terms outside the brackets – and put them in a bracket together ( )( )

 –

Factor

 

Find two factors of -40 that add up to -6 -10 and 4 work, so substitute those into the expression in place of -6d

 

– – Group the terms: (

(

)( –

)–

(

)

) Factor (Difference of squares)

  

Factor out ( ) There’s a difference of squares in the bracket (

)( –

) Factor (Trinomial factoring)

 

Multiply the coefficient of the first term, 6, by the coefficient of the last term, -4, which gives -24. Note that you always multiply the coefficient of the first term by the 3’rd, but often the coefficient is 1 and in that case it makes no difference. Find two factors of -24 that add up to 5. -3 and 8 work, so substitute those in place of +5ab. 5ab

 

– Group

 

( – ) ( – )(

( )

– )

Factor out the greatest common factor, ab

 

( – ) Find two terms that add to 1 and multiply to -12. -3 and 4 work, so substitute them in

  

( – , ( – ) ( – )(

– ) ( – )) Factor (difference of squares)

 

Factor out the greatest common factor, 11, to get: ( ) You have a difference of squares in the brackets, which can be simplified:

)(

)(

– )

)

Factor 

You have a difference of squares, which can be simplified: ( –

(

Factor (Trinomial factoring)

Multiply the coefficients of the first and last terms to get -48, then solve normally; find two factors that multiply to -48 and add to 22. -24 and 2 work out, so substitute them in, then group.

 –  ( ) ( )  ( )( ) 10. A box has a length of x centimeters, find the volume and surface area if width is 4 cm shorter than the length and the height is 3 times the length  First calculate for the volume  ( )   ( )( )  ( ) 11. Find the area between the two rectangles with dimensions  To calculate this, you must subtract the area OF smaller rectangle from the area of the larger rectangle  ( )( ) ( )( )    Then write a statement  The area between the two rectangles is

Solving for X intercepts 1.       

          

solve by factoring You have a difference of squares, so the equation can be turned into (3a + 4) (3a – 4) = 0 To solve by factoring, you must determine the two possible answers One of those factors has to equal 0 If 3a + 4 = 0, then 3a = - 4, and a = -4/3 If 3a -4 = 0 , then 3a = 4, and a = 4/3 The two possible answers are a = -4/3 and a = 4/3

Trinomial factoring Multiply the 3rd term by the co-efficient of the first, then determine the two numbers that multiply to that number and add to the 2nd term In this case ( ) You would re-write the equation as: ) ( ) Then separate the first two terms and the last two terms ( This then becomes ( )( )=0 To solve by factoring, you must determine the two possible answers One of those factors has to equal 0 If , then If , then The two possible answers are 8 and 5

 Trinomial factoring  Multiply the 3rd term by the co-efficient of the first, then determine the two numbers that multiply to that number and add to the 2nd term )  In this case (  You would re write the equation as: ) ( )  Then separate the first two and last two terms as ( )( )  This becomes (  To solve by factoring, you must determine the two possible answers  One of those factors has to equal 0  If , then 

If

The two possible answers are -2 and -2/3

, then

 Trinomial factoring  Multiply the 3rd term by the co-efficient of the first, then determine the two numbers that multiply to that number and add to the 2nd term )  In this case (  You would re write the equation as: ) ( )  Then separate the first two and last two terms as ( )( )  This becomes (  To solve by factoring, you must determine the two possible answers  One of those factors has to equal 0  If , then  If , then  The two possible answers are 8 and -2

 You have a difference of squares, factor out the GCF in this case (x)  ( )( )=0  To solve by factoring, you must determine the two possible answers  One of those factors has to equal 0  would be obvious  If

, then

 The two possible answers are 0 and

The Quadratic formula √  A quadratic equation is written as  If you notice, the value of the variable x is the negative of the co-efficient of the second term plus or minus the square root of three squared minus four times the co-efficient of the first term minus the last divided by twice the co-efficient of the first term 

and

( ) ( )

Graphing functions A function is a set of ordered pairs for each value of x there is only one for y To find a function one would conduct a vertical line test which will only intersect one point on the line A relation will have multiple points, A quadratic function will have a degree of , or Parabola Basic function Can draw table of values, randomly select values of x and see their relation to y Parabola

X

Y

-3

18

-2

8

-1

2

0

0

1

2

2

8

3

18

Y=ax^2 If a is more than one the parabola will be stretched vertically What if we then have a fraction for a (much smaller number?) x

Y

-3

3

-2

1.3

-1

0.3

0

0

1

0.3

2

1.3

3

3

Y=ax^2, if a is less than one the parabola will be compressed vertically. Y=-x^2 x Y -3

-9

-2

-4

-1

-1

0

0

1

-1

2

-4

3

-9

Domain (X values) and range (Y values) The X value can only happen once, if it happens multiple times, it is not a function. The vertical line test, Take a ruler and make sure it never hits the line more than one time ( )

Functions Domain: the domain states what x can be, and is written as: *

+

This says; the domain of x is any real number, and must be equal to or less than two Range: the range states what y is and can be written as:

Either of these can display the value of x as *

}

The vertex: The point where both “sides” of the function come to a point and is written (_, _) Opening: where the function appears to open to, can be up or down. Stretched: when a function has an integer co-efficient of x, ex: Compressed: when a function has a fractional co-efficient of x, ex: The scale factor: the co-efficient of x, can be an integer or a fraction

, the two stretches the line

The y-intercept: pretty self-explanatory, the point where the line intersects the y axis, usually takes the form of the variable K. The x intercepts: the points where the line intersects the x axis, if they intersect the y axis at a positive point, this will be non-existent (written as none), is written as an ordered pair.

Types of equations you may come across The equations on the test should come across as:

Practice Equation

vertex

opening

Stretched compressed

Scale factor

y-intercept

X-intercept

range

Domain

Analytic geometry Review sheet In grade 10 analytic geometry, the few concepts in the course and their formulas are: Finding lengths of line segments

)

√(

(

)

Midpoints of line segments

(

) (

)

Slope of a line segment

Slope and y-intercept form of the equation of a line:

Standard Form

Distances from points to lines The equation of a circle with centre (h,k) and radius r

(

)

(

)

Length of a line segment Line segments are a given set of points, often two endpoints of a graphed line. To find the length of one of these lines, one can attempt to count how many spaces they cover on a graph, or they can calculate the length using the Cartesian coordinates (x,y), (x,y) As shown above, the formula for length of a line segment is:

)

√(

(

)

And use the coordinates provided, (x,y), (x,y), but note one should be ( should be ( )

) and the other

Example Given points A(2,1) and B(3,5) determine the length of the line segment to the nearest tenth A good way to dissect equations is to use GRASS, given, required, analysis, solution statement Given: A(2,1), B(3,5) Required: l

√(

Analysis:

)

(

)

Solution (substitute) )

√( √( )

(

)

( )

√ √

Note: this is the exact solution; you do not need to round any (If asked for one tenth of a unit only)

The length of the line segment is roughly 4.1 units

Equations of circles with radius r and centre (h,k) Use GRASS to determine the value of the unknown variable The formula for this type of equation is (

) ( ) coordinates of the circle centre, and r being the radius

, h and k being the (x,y)

Example Determine the radius a circle with centre (0,0) and point (2,7). Round to the nearest tenth, if necessary Given: the centre of the circle (h,k)=(0,0), therefore h=0 and k=0, a known point is (2,7), x=2, y=7 Required: value of r Analysis: (

)

(

)

Solution ( ( )

)

(

)

( )

Statement: (

)

(

)

Note: if centre is not (0,0) substitute it into (h,k) You may also be asked to find the equation of a circle with a given radius and centre. To find this, substitute known values, and isolate the variables

Example Write an equation for a circle with Centre (5,2) and radius 4 Given: h=5, k=2 and r=4

( ( )

)

(

) ( )

Midpoint of a line segment The midpoint of a line segment is the point on the line that is the equal distance from both endpoints To find the midpoint, you can calculate using endpoints, endpoint 1 (

(

)

As shown above the formula for finding midpoints is: (

) (

)

Example Determine the midpoint of the following lines given their endpoints: a) (5,7) and (3,9,) b) (-2, -4) and (-2, 8) a) Given:

=5,

,

,

Required: mp Analysis: (

) (

)

Solution: ( ( ) (

) ( )

)

) and endpoint 2

Equations of lines Calculating equations of lines, both in slope- y- intercept form and standard form is an essential skill. You will be asked to find the equation of lines using endpoints, and use these lines to calculate intersections. The formula for calculating slope is:

You can use the slope (once found) and another point on the line, such as an endpoint to calculate the y-intercept.

Example Find the slope and y-intercept of the line that passes through (1,2) and (-2,5)

Now, use one of the endpoints to find the equation of the line

(

)

Parallel and perpendicular lines Parallel lines have the same slope, by different y intercepts.

Perpendicular lines are lines that intersect at a 90 degree angle

The slope of a perpendicular line is the negative reciprocal of the line it is perpendicular to, meaning to “flip” its fractional value and multiply it by -1, for example 3/2 becomes -2/3.

Medians, right bisectors and altitudes Median The median of a triangle is a line segment that joins a vertex to the midpoint of the opposite side. In grade 10, the objective is to find the equation of its line. The equation of a median is found by: 1. 2. 3. 4.

Finding the midpoint of the opposite side Using its coordinates and the coordinates of the opposite vertex to calculate slope Using that slope, and any known point on the line (midpoint or vertex) determine the value of b Write the equation of the line in y=mx+b form

The place where three medians within a triangle intersect is called the Centroid

Example has verticies A(3,4), B(-5,2), C(1,-4) Find the equation for the median from C to AB: CD 1. Find the midpoint of the opposite side, AB

(

) (

(

) (

(

) ( )

) )

2. Calculate the slope using the midpoint and C

3. Determine the value of b

4. Write the equation in slope, y-intercept form

Or in standard form

Right bisectors A right bisector is a line that divides a line in two parts at 90 degrees. In grade 10 mathematics, the objective is to determine the equation of the line To find the equation of a right bisector: 1. 2. 3. 4.

Find the midpoint of the line being bisected Calculate the slope of the line being bisected Find the negative reciprocal of that line: the slope of the right bisector Use the slope of the right bisector and the midpoint (where it bisects) to determine the equation of the right bisector

The place where three right bisectors intersect within a triangle is called the circumcentre

Example has verticies A(3,4), B(-5,2), C(1,-4) Find and equation for GH, the right bisector of AB

1. Find the midpoint of the line being bisected

(

) (

(

) (

(

) ( )

) )

2. Calculate the slope of the line being bisected

3. Find the negative reciprocal; the slope of the right bisector

4. Use the slope of the right bisector and the midpoint to find the y intercept

Or in standard form

Altitudes An altitude of a triangle is a segment from a vertex to an opposite side, which it intersects at a right angle

To calculate an altitude 1. Calculate the slope of the line intersected by the altitude 2. Use the negative reciprocal of that slope, as the slope of the altitudes (because it intersects the line at a 90 degree angle.) 3. Use the opposite vertex and the slope of the altitude to find the altitude equation The place where the three altitudes intersect is called the orthocentre

Example has verticies A(3,4), B(-5,2), C(1,-4) Find an equation for CE, the altitude from C to AB 1. Calculate the slope of the line intersected by the altitude

2. Find the negative reciprocal; the slope of the altitude

3. Use the opposite vertex and the slope of the altitude to find the altitude equation

Or in standard form

Similar and Congruent triangles Congruent triangles A congruent triangle is a triangle with where corresponding side lengths and angles are equal A

F

B have angles

E

C

Observe from the diagram that, ΔABC and ΔDEF are congruent they equal side lengths and equal

D

Proving congruence Although one may be told that these triangles are congruent, to prove that they are, there are three mainly used theorems (congruence postulates) 1) Side, Side, Side: congruence can be demonstrated by showing that all corresponding sides are equal. 2) Side, Angle, Side: congruence can be demonstrated by showing that two of the sides, and the contained angles (angle of the two corresponding sides intersection) are equal, the triangles are congruent 3) Angle, Side, Angle: if one corresponding side, and any two corresponding angles within the triangle are equal, then the triangles are also congruent. Note:

means is congruent to, in the above example

Note: Δ

means triangle, literally

Note:

means similar to

Similar triangles The corresponding angles of similar triangles are equal, while their side lengths are proportional A

D

B

Observe from the diagram that both triangles, ABC and DEF have identical correspondent angles, while they have proportional side lengths.

C E

F

Also

Each corresponding pair has proportional side lengths, such

as ½ = ½ = ½

Proving similarities To prove that two triangles are similar, there are three main theorems that apply 1) Angle, Angle: show that two angles are similar 2) Side, Side, Side: show that three sides are proportional 3) Side, Angle, Side: show that two sides are proportional and the contained angles are equal

Ratios of similar triangles B c

E a

A

f

d

C b

D

F e

If two triangles are similar, the ratio of their heights is equal to the ratio of corresponding sides,

The ratio of their areas is equal to

=

Finding unknown side lengths C

F

d=5cm

b=10cm B

c

D

g=4cm E

e f=6cm

G

, find the values of e and c Since the triangles are similar, the ratios of the corresponding sides are equal or Substitute the know values Take the first part of the ratio, and use it to calculate the values of c and e So, e is 8cm and c is 7.5 cm

Finding Areas

AB = 8cm, DE = 12 cm The area of

is 54cm^2

Solution AB/DE = 8/12=2/3 The ratio of the areas of the triangles is

or 4/9

= Let the area of ABC be x, x/54 = 4/9 Then calculate using the cross product rule: 9*x=54*4, 9x=216, x=24 Therefore ABC= 24cm^2

Showing and using similarity

A

7cm

B

5cm

Show why

4cm C

Since AB is parallel to DE angle A = angle E

Alternate angles

angle b = angle d

Alternate angles

angle ABC = angle ECD

Opposite angles

since the corresponding pairs of angles are equal,

6cm D

x y

E

Find the lengths of x and y Since

, the ratios of the corresponding sides are equal

Solution of x

Solution of y

Therefore, x is 7.5 cm and y is 10.5 cm

Trigonometry Trigonometry is the study of the relationship between angles and sides in triangles. A

B

C

There are names for each of the sides in the triangle, these are: AC- The Hypotenuse: generally the longest side, the hypotenuse is located opposite the right angle AB- Opposite: This angle is the one always opposite from the chosen angle BC-Adjacent: The third side of a triangle

Theta:

Side ratios There are three important ratios of sides in right triangles. These functions are all functions of the given angle, theta. These sides are: Sine

Sine, shown as sin on a calculator, the equation to calculate sine is opposite/hypotenuse

Cosine Cosine, shown as cos on a calculator, is represented by the equation adjacent/hypotenuse

Tangent A

Tangent, shown as tan on a calculator, is represented by the equation opposite/adjacent 15 O

In this triangle: Sine = O/H=15/17

B A

17 H

Cosine = A/H=8/17

8

Tangent =O/A=15/8

C

Note: When you are using a calculator ensure that your calculator is in degree mode Note for future reference When using inverted side ratios use the “

nd”

button

Finding angles If you know a trig ratio of an angle, you can find the angle using a calculator. You can use inverted functions to find the angles from the trig ratio. Cos 𝜃= .7071

If

‘’

𝜃

cos

𝜃

𝑜

(

If )

Tan 𝜃

sin 𝜃

If

𝜃

tan (

𝜃

𝑜

)

𝜃

sin (

𝜃

𝑜

)

Finding sides in right triangles when given an acute angle and another side B

Label the three sides, opposite, hypotenuse and adjacent.

X (O)

C

Remember the three ratios Sin =O/H

Y (A)

50cm

Cos =A/H

(H)

Tan =O/A A Find X, the opposite side, given that the angle known is

𝑜

and the hypotenuse is 50cm long

To find X, use the angle ratio that has X and one of the known sides, in this case, the best ratio to use is Sine, O/H, because we know the length of the hypotenuse

(

)

Now that two sides of the right angled triangle are known, the length of the other side can be found using either the Pythagorean Theorem or using trigonometry. Generally, using trigonometry is more common and also easier. ‘ i)

Using Pythagorean Theorem

𝑦

ii)

𝑥

𝑦

Using trigonometry

cos

𝑜

cos

𝑜

𝑦 𝑦

( 𝑐𝑚

𝑦

𝑦 (

) )

𝑦 𝑦

𝑐𝑚

Unknown denominators Sometimes the unknown may end up in the denominator. To solve for this, you will need an extra step: P

𝐴 𝐻

𝑜

cos

A

𝑋

6cm

X

H

X X

Q

R

X=6.38

O

Two sides known Sometimes, two sides of a triangle are known, and the third side length and the other acute angle must be known. As you know, you can find a third side using the Pythagorean Theorem, or using trigonometry

P A X

𝑋

34cm H

𝑋 Q

R O

30cm

𝑋

√ 𝑐𝑚

To determine angles, first label the angles and then select an angle to determine For example: Solve for the angle of P sin sin sin sin (

)

To solve for the last angle, remember that all angles add up to 180-90-61.9=28.1

Word problems Some terms you will need to know The angle of depression: The angle of depression is the line of sight and the horizontal, but only above the below.

The angle of elevation: The angle of elevation is the angle between the line of sight and the horizontal, but only above the horizontal. The angle of elevation is also known as the angle of inclination

Example word problems solved with trigonometry From the top of a vertical cliff 20 metres high and is vertical, the angle of depression of a boat at sea is A. Find the distance of the boat from the base of the cliff

B. Find the distance of the boat from the observer

Procedure Step one: Determine and apply what is given in the problem We know that the cliff is 20 meters high, and that it is at 90 degrees with the sea, we also know that the angle of depression of the boat and sea is 25 degrees (This step does not necessarily need to be written down) Step two: Diagram

20 Metre Cliff Step Three: Labeling sides and angles 𝑜 Notice that the two unknown sides are Labelled, as well as the Adjacent, Hypotenuse and Opposite

20 Metre Cliff (O)

y (H) 𝑜

X (A)

Step four: select a side to solve for, and solve Find X (the adjacent) tan x tan ( ) x tan x Step five: find the other side (keep in mind, the Pythagorean Theorem can be used as well) Find Y (the hypotenuse) sin sin ( )

sin

Step six: statement The distance from the boat to the base of the cliff is 42.9 metres and the distance from the boat to the observer is about 47.3 m

Problems involving right triangles A

E

15 10

x

4

B C In Find angle ACB sin sin

CBA=180-90-41.8=48.2 Find CE, (x) ECD=180-90-48.2=41.8 sin

sin

D

Example From a point on the ground, the angle of elevation of the top of a building is . If one moves 25 metres closer to that building, the angle of elevation will be . Find the height of the building.

D

h 𝑜

𝑜

A

25

B

In tan (

)(tan

(

)(

In

)

, tan (tan

)

)

x

C

Then substitute (

)

The height of the building is 46.2 metres

Sine law and its application Proof of sine law for acute angles B c

a

A

C D

b

Draw a perpendicular line from B to AC at D, this line is BD, Let BD=h sin (sin ) sin (sin ) Now, h=h sin

sin

Divide both sides by sin sin sin sin sin

sin sin sin

Then cancel out terms, creating Together we have Sine law can be used with two of the above terms (or three, but if that is possible, another method is more practical.)

Sine law application When we are given any 2 angles and 1 opposite side, sine law can be used to find other sides. (If two angles are known, the third can be calculated, because all angles in a triangle must add to

Q r

p

𝑜

P

𝑜

40 cm q

R

In this equation, we know all three angles, because we can equate Q = and one side, (opposite to Q).

, we know all three angles,

So, sin

sin

Substitute sin

sin

Cross multiply sin

sin (

)

Now solve for side r

Solving with 2 sides and 1 uncontained angle Use sine law to find another angle (and subsequently, the other angles) Try to find the acute angles first. (Only use sine law to find acute angles) Remember:

, the reciprocals of the fractions are also equal sin

B

sin c=10cm

a=22cm

A

C 𝑜

b

sin

sin sin

Solve for the other angles using sine law Example problem The angle of elevation of the top of a building is elevation is . Find the height of the building.

. If you move 10 metres closer, the angle of D

Consider (

)

(

)

𝑜

h

In triangle sin

𝑜

(sin

) A

10m

B

𝑜

Cosine Law Cosine law can be used when Two sides, and a contained angle are given or when three sides are given.

Proof of cosine law B c

a h

A

x

b

In (

)

b-x

C

C

(

)

(

(

)

cos ) cos

To find other side lengths, re-arrange the equation to b and c cos cos To find angles, the formulas can be manipulated again, bringing Cos by itself on the left side cos

cos

cos

Example

B

Find a to the nearest tenth of a centimetre:

10cm

cos ( (

a

𝑜

)(

) cos )

A

20cm

C

, Find angle B using cosine law cos cos

(

)(

cos cos cos

(

)

)