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Grade 10s are working with similar triangles and trigonometry. 

We started with a 90 degree angle and the sides 10cm and 9cm in one triangle, and 27cm and 30cm in the other.  Using our knowledge of the pythagorean theorem we calculated the hypotenuse.  We also used our learning from Monday to calculate the value of the angles using our trig tables.  We had the option to use the ratio for sine, cosine or tangent since we knew all 3 sides.  

We noticed that the angles are the same in both triangles, and the side lengths are proportional.  This is always the case!  We have 2 similar triangles.  We also looked at the areas and compare them.  Our big triangle has sides that are 3 times the length of the small triangle sides.  The are of the big triangle is 9 times the area of the small one. Thats 3^2 (three squared).  This relationship holds true for similar triangles.  If the proportionality constant is “k” between the sides of the triangle, it will mean that the areas have a proportionality constant of “k^2”.

We looked at a situation where we can use similar triangles.

We can determine the height of an object by using shadows.  

We’ve been working a lot with lines since coming back from the break.  We already looked at parallel lines a while back.  We know they don’t cross each other, and they run like train tracks always separated by the same distance.  We now can identify that they are increasing or decreasing at the same rate, or they have the same slope (“même pente”)….in fact we have been doing lots of chanting out loud… when someone say “parallèle” the response is “même pente”, and most of us have this word association stuck in our minds.

We know that the slope is the coefficient of x when the equation is written with y isolated.  This is the y=mx+b form.  The b value is the constant, the initial value, the y intercept (“ordonnée à l’origine”).  

We learned that perpendicular lines intersect, and always form a 90 degree angle.  If one slope is positive the other is negative, if one slope is 0 the other is undefined (indéfinie).  If one slope is steep, the other is not.  That’s the only way they’ll have a 90 degree angle between them.

The slopes are related.  This example has one slope of 1/2 and the other of -2/1.   The fraction is inverted, and one is positive and the other is negative.  We determined a process for finding a perpendicular slope is to invert the fraction and to multiply by negative 1.  Our chanting and word association continued, and when someone says “perpendiculaire” the response is “inverse négative”, the rhythm is a bit catchy.  

We see the slopes as the x coefficients again, when y is isolated.

Today we started to work with equations (they are like two expressions on either side of an equal sign). We used algebra and our visual representations help us solve the equation.  We do the same things on both sides, and our goal is to isolate the variable.

We worked through several different types of questions.  Some of us had different approaches to solving…. like here we divided every term by 2 (the coefficient) first.

Some groups tried to add 2 to both sides to start….but at the end it didn’t verify.  (We couldn’t substitute our answer into the variable and have both sides remain equal).  Their next step was to restart with a new first step.  In this case they divided by 3 on both sides, and then have f-2=6 so f=8.

We are learning lots, and getting more confident with our new skills.

The next step was to manage equations with variables  on each side of the equal sign.  We need to choose a side, and use algebra (opposite operations) to make sure that the variables are on one side, and constants on the other.

Our final type of equation had fractions.  We can eliminate fractions first by multiplying by the denominator on both sides.  This is one strategy of many that we used to manage fractions.

We’re practicing a lot this week, and making good progress.

Grade 10s are working on solving problems involving quadratics.  This example shows how we can determine dimensions of a rectangle given the area and perimeter.  We can create equations, substitute, expand to make a trinomial, use algebra to get a trinomial that equals zero, and then use the quadratic formula to solve for the roots (if they exist).

We also saw that sometimes there are no roots.  In this case on desmos we can see that there are no intersection points between our two equations, so we have no possible dimensions that will work.  It corresponds to the fact that the discriminant found is a negative number.

Our work this week is mostly based on modelling (creating equations) and using them to solve problems.  

Some good review questions



In Grade 10 today we learned all about the power of the discriminant (b^2-4ac) that is under the square root of the quadratic formula. 

We looked at multiple examples and practiced solving for the value of x from factored form, vertex form and standard form equations.  We can now find the intersection between 2 parabolas using substitution.

Today grade 9s learned how to represent multiplying in a few different ways.

Some saw this as 3 groups of 2x-4

Others saw it as 3 groups of 2x and 3 groups of -4, which is 6x-12

Some could see it as a rectangle, with the area representing the multiplication of length and width.

In each case, it is the same answer, just represented in a slightly different configuration.  This might not seem to be a big deal, but when we make the question a little more challenging….

We no longer can group things as easily, since we don’t know the value of x.  The area model continues to be useful though, and we can build a rectangle with the dimensions of x and 3x+1

Here’s a representation of 2x(x-4) the rectangle dimensions are placed up on the ledge and the rectangle itself is on the lower level.  The results of the multiplication are 2x^2-4x.  

Look at these hard working grade 9s!  They are collaborating and checking their understanding with some very complex exponent law simplification questions.

We’re using all sorts of surfaces to work on.  Many are choosing the boards, some are writing on desks, others on small whiteboards, and some on paper.

We’re learning about using our textbooks and how to check the answers in the back.

Some of us are expressing our exponents in interesting ways, such as making up our own questions….here’s one that is particularly creative.



We worked on this pattern today

We saw things in so many ways!

We made a lot of equations which all simplified to the same thing.

The next step was to make a graph.  The trick was it’s not a factorable equation.  We needed a new step to find the vertex (le sommet).

We looked at some perfect squares (we’ve done lots of practice with these).  We know they have one root (racine) and it is also the vertex (sommet).  What happens if there is a constant at the end, after the perfect square.  It turns out that if the bracket turns to zero, the final number will be the minimum value of the parabola (part of the vertex).

We got good at that, so we tried to make a perfect square out of our expression

We used algebra tiles, splitting the 8x up into 2 groups of 4x and then we needed 16 squares to complete the square.  We already had 4, but we needed 12 more.  We added 12 zero pairs (blue and red squares together) so now we have a full square, and 12 blue squares left over 

Our equation is y=(x+4)^2-12 and from this we know that the vertex (sommet) is (-4,-12).

We then looked at calculating the roots (racines) by setting y to 0 and then solving for x by using opposite operations.  We’ll do more practice with this next week.

Grade 9s are working on evaluating expressions with exponents.  Today we looked at bases that are integers, and fractions.

We look each time to see what the base is, and can show the repeated multiplication.  In the following examples, if there are parentheses surrounding a negative number which is raised to an exponent, that negative number is what is repeatedly multiplied.  If there are no parentheses, the exponent does not affect the negative.  The base is the positive number the exponent touches.  That number is then repeatedly multiplied, and the negative is applied after.

We noticed that if the base is negative, and the power is an even number, the answer will always be positive.  

We looked also today at the pattern that we see when we have the same base (in this case 2) raised to many different powers.  We made a table, and looked for the patterns.  We made the exponents go all the way into the negative numbers.

We explored to see if this is a linear or non-linear pattern, positive or negative, direct or partial.  It turns out to be pretty interesting on a graph.

It is positive, non-linear, and partial.  The y intercept (ordonnée à l’origine) is 1 and it will never cross the x axis.  It gets super close to it, but will never reach it.  That means there is an asymptote (something to learn more about in grade 11).

We looked a bit at what an exponent of 0 means.  We arrive at it by creating a division with the same base and same power in numerator and denominator.

We know that anything that is the same on top and bottom will divide to make 1.  We also know that with our exponent laws we can subtract the powers in a division question with the same bases.  Using 2 perspectives we get 2 answers which are then equal.

We used similar logic to explain negative exponents.