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Year 11 maths similar triangles
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first, I am not a mathematician. However, my idea is this.
In exercise a) you know that you can use similar triangles. In exercise b) I would rely on pythagoras. You don't know triangles that are 100% reliable comparable. (i could be wrong).
Besides that, y is not sqrt(101.25) but y=sqrt(1010.25)-2
using pythagoras a^2+b^2=c^2
a= 9, b=4.5 , c= (2+y)
81+20.25=(2+y)^2
sqrt(101.25) =2+y
Y= sqrt(101.25)-2
If your educational system is anything like ours, do not try to be intelligent or clever but do it they way they tell you to.
You're on the right track with the Pythagorean theorem, although note that y is not the entire length of the hypotenuse of the bigger triangle. I don't remember how to do similar triangles because they're pretty much useless.
Yeah I meant -2 on the end. I know the answer was around 10.(weird decimal) and then take away 2 makes 8.(weird decimal). But the point is using similar triangles I get x = 1 and y = 16, whilst using pythagoras' theorem i get x = sqrt(12) and y = 8.(weird decimal) equal to sqrt(101.25)-2. Why don't I get the same answer? Is the question messed up?
And to those that do not understand how similar triangles work, it is really quite simple. First you slip the triangle so that all of the angles correspond with the angles on the other triangle. The size will be different. Then you see what sides correspond with what. In this case, the 2 corresponds with the 9, the x corresponds with the 4.5 and the 4 corresponds with the y+2, going from the smaller triangle to the bigger triangle.
Once you know that, you simply multiply the ratio to figure out the rest. The ratio of small to big = 2 to 9, since the 2 in the smaller triangle corresponds with the 9 in the bigger triangle.
To find x, it corresponds with 4.5, so set up an equation where x/4.5 = 2/9. Then let x/4.5 = 1/4.5 and x = 1.
To find y, you know that y+2 corresponds with 4, but since you are going from big to small, you reverse the formula. So you multiply (y+2)/4 by 9/2 (which is the same as dividing by 2/9). That will calculate to be y = 16.
Deranged wrote:
Yeah I meant -2 on the end. I know the answer was around 10.(weird decimal) and then take away 2 makes 8.(weird decimal). But the point is using similar triangles I get x = 1 and y = 16, whilst using pythagoras' theorem i get x = sqrt(12) and y = 8.(weird decimal) equal to sqrt(101.25)-2. Why don't I get the same answer? Is the question messed up?
I don't think that you know that the triangles are similar, and I think that they aren't. Finding the hypotenuse of the bigger triangle, you know that it is not two times greater than any side of the same triangle, whereas in the other triangle, 2 times 2 = 4. So, if they were really similar, then at least one side of the bigger triangle would be the hyotenuse divided by 2. So, obviously, they weren't similar.
cxue2010 wrote:
Deranged wrote:
Yeah I meant -2 on the end. I know the answer was around 10.(weird decimal) and then take away 2 makes 8.(weird decimal). But the point is using similar triangles I get x = 1 and y = 16, whilst using pythagoras' theorem i get x = sqrt(12) and y = 8.(weird decimal) equal to sqrt(101.25)-2. Why don't I get the same answer? Is the question messed up?
I don't think that you know that the triangles are similar, and I think that they aren't. Finding the hypotenuse of the bigger triangle, you know that it is not two times greater than any side of the same triangle, whereas in the other triangle, 2 times 2 = 4. So, if they were really similar, then at least one side of the bigger triangle would be the hyotenuse divided by 2. So, obviously, they weren't similar.
The test was called "Ratios and Similar Triangles". I think he intended for the triangles to be similar, but NOT for them to be right-angled triangles.
As cxue2010 pointed out, the triangles are not similar. In the larger triangle, it is the non-hypotenuse sides which are in a 2:1 ratio. In the smaller triangle, it is the hypotenuse and one of the other sides which is in the ratio. The part of the problem that deals with similar triangles is in realizing that they're not. After all, part of truly understanding a mathematical procedure is in knowing when you can't use it. Just because you have two different right triangles with a 2:1 ratio on two of the sides doesn't mean that the triangles are similar, and this part of the problem is designed to make sure you understand that. Stick with the Pythagorean solutions for part b.
Ignore question 4a. I already know that x = 7.5m. Only focus on 4b.
My year 11 maths class had a test on similar triangles, but one of the questions looked like it could be solved using Pythagoras' Theorem. Would I get full marks if I solved this question using Pythagoras' Theorem, or do I have to use similar triangles?
Also, I solved it with Pythagoras' Theorem and I got x = sqrt(12)m, then I tried it again using similar triangles and I got x = 1m. I checked it and it all seemed to work out, but I got 2 different answers by using different methods. I then solved for y, and using similar triangles, I got y = 16m, but using Pythagoras' Theorem, I got y = sqrt(101.25)-2m. Why did I get different answers????? What is the real answer?