Grade 10 Math Exam Review

8/12/2019 Grade 10 Math Exam Review

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Verifying geometric properties Distances from points to lines

Solving Systems of equationsNumerical Problems

1. { How to solve by graphing

i. Change to y=mx+b formii. Graph the y-interceptiii. Find the next points using the slopeiv. Repeat for other lines

2. { 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 variableiii. Substitute that value into the same variable in the other equationiv. Find the value of the other variable and then write: (x,y)

3. { How to solve by elimination

i. Multiply both equations by their factor in the lowest common variableii. Add or subtract one from the other so you eliminate the equal variablesiii. Calculate the value of the remaining variableiv. Substitute the value of that variable to find the value of the otherv. Write as

Word problems

4. There are ninety coins, consisting of quarters and loonies, total $42, how many of each type ofcoin 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)


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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 didshe travel by each mode

Let x represent the amount of time she travelled the amount of time she travelled by carand 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 willearn 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 moneyinvested 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 barbsrunning speed and the wind speed

Let x represent the speed she travelled with the wind and y the wind speed


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{

simplify

{ Then write the statement . /

Polynomials and factoring8. ) Expand then simplify

Write each term within the brackets as multiplied by the terms outside of thosebrackets

4 Then multiply Combine like terms (if any)

9. Factor Extract the greatest common factor In this case, the greatest common factor is Then find two factors of the final term12 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

, -


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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 coefficientsin questions like this, theyre 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 Theres 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 3rd, 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


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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 shorterthan 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


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Solving for X intercepts

1. solve by factoring You have a difference of squares, so the equation can be turned into (3a + 4) (3a4) = 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 3rdterm by the co-efficient of the first, then determine the two numbers

that multiply to that number and add to the 2ndterm

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



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 -2 and -2/3


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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


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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

ParabolaBasic function Can draw table of values, randomly select values of x and see their relation to yParabola

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


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

Y=-x^2

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 FunctionsDomain: 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: , the two stretches the lineCompressed: 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

x Y

-3 -9

-2 -4

-1 -1

0 0

1 -1

2 -4

3 -9


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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


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Analytic geometry Review sheetIn 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 linesThe equation of a circle with centre h,k) and radius r


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Length of a line segmentLine 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 agraph, 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 and the othershould be 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 , h and k being the (x,y)coordinates of the circle centre, and r being the radius


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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


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Midpoint of a line segmentThe 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 and endpoint 2 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:

( ) ( ) () ( )


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Equations of linesCalculating 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 linesParallel lines have the same slope, by different y intercepts.

Perpendicular lines are lines that intersect at a 90 degree angle


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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 altitudesMedianThe 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. Finding the midpoint of the opposite side2. Using its coordinates and the coordinates of the opposite vertex to calculate slope3. Using that slope, and any known point on the line (midpoint or vertex) determine the value of b4. 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


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3. Determine the value of b

4. Write the equation in slope, y-intercept form Or in standard form Right bisectorsA 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. Find the midpoint of the line being bisected2. Calculate the slope of the line being bisected3. Find the negative reciprocal of that line: the slope of the right bisector4. 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


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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

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

angle


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To calculate an altitude

1. Calculate the slope of the line intersected by the altitude2. 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 equationThe 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


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Similar and Congruent trianglesCongruent triangles

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

Observe from the diagram that, ABCand DEF are congruent they

have equal side lengths and equal

angles

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 sidesare equal.

2) Side, Angle, Side: congruence can be demonstrated by showing that two of the sides, and thecontained 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 thetriangle are equal, then the triangles are also congruent.

Note: means is congruent to, in the above example Note: means triangle, literallyNote: means similar to

F

E

D

A

B

C


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Similar triangles

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

Observe from the diagramthat both triangles, ABC and

DEF have identical

correspondent angles, while

they have proportional side

lengths.

Also Each corresponding pair has proportional side lengths, suchas = =

Proving similarities

To prove that two triangles are similar, there are three main theorems that apply

1) Angle, Angle: show that two angles are similar2) Side, Side, Side: show that three sides are proportional3) Side, Angle, Side: show that two sides are proportional and the contained angles are equal

Ratios of similar triangles

B Ec a f d

A 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 g=4cm e

B c D E f=6cm G

A D

B C

E F


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, find the values of e and cSince 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 eSo, e is 8cm and c is 7.5 cm

Finding Areas

AB = 8cm, DE = 12 cmThe area of is 54cm^2SolutionAB/DE = 8/12=2/3

The ratio of the areas of the triangles isor 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

Show why 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,

A 7cm B

5cm 4cm

C

6cm x

D y E


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Find the lengths of x and y

Since , the ratios of the corresponding sides are equalSolution of x

Solution of y

Therefore, x is 7.5 cm and y is 10.5 cm

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

A

B C

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/hypotenuseCosine Cosine, shown as cos on a calculator, is represented by the equationadjacent/hypotenuse

There are names for each of the sides in the triangle, theseare:

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


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Tangent Tangent, shown as tan on a calculator, is represented by the equation opposite/adjacentIn this triangle:

Sine = O/H=15/17Cosine

= A/H=8/17

Tangent=O/A=15/8Note: When you are using a calculator ensure that your calculator is in degree mode

ote for future reference When using inverted side ratios use the nd buttonFinding 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.

Finding sides in right triangles when given an acute angleand another side

Label the three sides, opposite, hypotenuse and adjacent.

Remember the three ratios

Sin=O/HCos=A/HTan=O/AFind X, the opposite side, given that the angle known is and the hypotenuse is 50cm longTo 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

A 15 O B

A 8

17 H C

If Cos= .7071 cos

If Tan tan

If sin sin

B X (O) C

Y (A) 50cm

(H)

A


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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.

Unknown denominators

Sometimes the unknown may end up in the denominator. To solve for this, you will need an extra

step:

Two sides known

Sometimes, two sides of a triangle are known, and the third side length and the other acute angle

must be known.

cos cos

ii) Using trigonometry

i) Using Pythagorean Theorem

P

A

6cm X H

Q R

O

cos

X XX=6.38

As you know, you can find a third side using the Pythagorean

Theorem, or using trigonometry

P

A

X 34cm H

Q R

O 30cm


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B. Find the distance of the boat from the observerProcedure

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 thatthe angle of depression of the boat and sea is 25 degrees

(This step does not necessarily need to be written down)

Step two: Diagram

Step Three: Labeling sides and angles

Notice that the two unknown sides are

Labelled, as well as the Adjacent,

Hypotenuse and Opposite

Step four: select a side to solve for, and solve

Find X (the adjacent)tan xtan

xtanx Step five: find the other side (keep in mind, the Pythagorean Theorem can be used as well)

Find Y (the hypotenuse)sin sin

20 Metre

Cliff

20 Metre y (H)

Cliff (O)

X (A)


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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

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

A E

15

10 x 4

B

C D


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Example

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

In tan

tan In , tantan

D

h

A 25 B x C


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Then substitute

The height of the building is 46.2 metres


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Sine law and its applicationProof of sine law for acute angles

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

s i n s i n s i n s i n )Now, h=h

s i n s i n Divide both sides by sinsin

sinsinsin sinsinsinThen 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 ismore 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

B

c a

A C

D b


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

So,

sin sinSubstitute

sin sinCross 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 sin

Q

r p

P 40 cm q R

B

c=10cm a=22cm

A C

b


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s i n sin sin

Solve for the other angles using sine law

Example problem

The angle of elevation of the top of a building is. If you move 10 metres closer, the angle ofelevation is. Find the height of the building.Consider In triangle

sinsin

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

Proof of cosine law

In

D

h

A 10m B C

B

c a

A x b b-x C

h


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c o s cosTo find other side lengths, re-arrange the equation to b and c

cos

cosTo find angles, the formulas can be manipulated again, bringing Cosby itself on the left sidec os c o s

c o s Example

Find a to the nearest tenth of a centimetre:

cos cos

B

10cm a

A 20cm C


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, Find angle B using cosine law

c o s

c o s c o s cos c o s

Documents

Grade 10 Math Exam Review