Instead of using the slope formula, on a diagram, you can find the slope by doing the rise / run. So make a right angled triangle on the diagram between the two points, then count the number of squares rising and running. It don't matter if the squares are not full.
Also syran, i see where your prediction come from for theorom 19 but i dont think maths will be that predictable. As paper 1 was easy i imagine they might ask an awkward variation of the theorom rather than just it straight.
To find missing sides or angles in a right-angled triangle, we use Sin, Cos or Tan. Follow these steps whilst doing so:
1. Label the sides of the triangle with all the information you're given.
2. Select which formula to use.
3. Substitute values into the formula.
4. Solve the resulting equation to find the missing value.
So for example of a past question,
ABC is a right-angled triangle
|ACB (triangle)| = 50 degrees and |AC| = 10cm
Calculate the length of [AB], correct to two decimal places.
Label the sides first!
We have the hypotenuse and we want the opposite.
Which formula has H and O in it? Soh Cah or Toa? Soh has. So you use Sin's formula.
Sin 50 degrees = |AB| / 10
Bring the ten over and multiply (because it was dividing before you brought it over) with Sin of 50 degrees.
Fill it into your calculator and you should get:
7.66044 = |AB|
Correct to two decimal places, 7.66 = |AB|.
Theorem 19 answer for anybody here
Given: Circle, centre 0, containing points A, B and C
To prove: |BOC (triangle)| = 2|BAC (triangle)|
Construction: Join A to 0 and continue to D
Label angles 1, 2, 3, 4 and 5
Proof: Consider |AOB (triangle)|:
|1| = |2| + |3|
But |2| = |3|
Similarly, |4| = 2|5|
|1| + |4| = 2|2| + 2|5|
Therefore |BOC (triangle)| = 2|BAC (triangle)|
When it asks you for the probability of something, DO NOT answer in words unless they specifically say to. If they ask probability, it has to be a fraction. Like what's the probability of getting a King out of a 52 card pack:
The probability is 4/52 which simplifies to 2/26 which simplifies to 1/13.
Unless they specifically say to, it is not advised to simplify probabilities. You should keep it as the original probability, in this case, 4/52 but there is NO HARM in simplifying it.
I'm lost with some of the volume questions. The way that sometimes they ask you about surface area's or something of shapes but the formula's aren't in the log tables. And how do you learn off the formula's? Thanks.
Volume is always something cubed whether it's cm or m etc.
So if the output is something cubed, then there are 3 inputs that you multiply to make it. This is the clue I do <
Volume (not Cylinder, Sphere & Cone which you're given in the tables book) = Length x Base x Height (which you could also write as lbh or l (cubed)
Surface area (not Cylinder, Sphere & Cone which you're given in the tables book) = All the sides added with Base and Height (Otherwise: 6l (squared) (squared because it's area)
You MUST KNOW these formulae as they are not in the tables book.
Well here's past constructions Danielle:
Bisector of a line segment - 2015, 2013
Constructing a right-angled triangle - 2014, 2012
Drawing line segments, 2014
Maybe constructing a right-angle triangle... but also maybe one of the other 12 constructions..
Given: A right-angled triangle
To prove: a(squared) + b(squared) = c(squared)
Construction: Draw a square PQRS with sides of length a + b.
Draw four congruent right-angled triangles in the square with sides of length a and b and hypotenuse c.
Label angles 1, 2, 3 and 4
Proof: Each of the four inscribed triangles is congruent to the original triangle.
Each side of the inner quadrilateral has length c.
|1| + |2| = 90 degrees (Remaining angles in the triangle)
|1| = |3| (Corresponding angles in congruent triangles)
|2| + |3| = 90 degrees
|4| = 90 degrees (Straight angle)
The inscribed quadrilateral is a square.
Area of square PQRS = 4
(a + b) (squared) = 4(half x ab) + c(squared)
a(squared) + 2ab + 2b(squared) = 2ab + c(squared)
a(squared) + b(squared) = c(squared)
The only acceptable form of the theorem of pythagaros is where you divide a right angled triangle in 2 then use similar triangles to prove the theorem
The one that sryan bruen has posted above isn't acceptable as far as i know as we did that one but after the mocks we were told we had to learn the version project maths wants
The correct version is in this link
Point of intersection came up and I wouldn't have a clue how to do it if it weren't for the person who posted how to do it on this thread. Thank you very much - sorry I cannot see your post right now. Plus, Rocky how? Less Stress More Success and my teacher told me to learn off the theorem that way? And yeah my Less Stress More Success book says PROJECT MATHS on it, not just Maths. And no constructions came up after all that Danielle haha.
It was a combination of the rectangle and triangle constructions. I realised that 5 mins before the end (and then I was frantically rubbing out lol). I just drew it with a ruler the first time but the second time I used the constructions methods