Shsat Ninth Grade - Triangle Inequalities


Based on this problem were asked, what is the smallest integer exports? The sides of a triangle can be 2, X, 4 X, plus 10 and X plus 20. This is actually my second attempt at this issue, I realized things to a viewer comment that I had made a mistake. So let's take a look at this and explain the premise of it and then solve it. The premise is what's called the triangle inequality. And the basic idea is that if you have a triangle, any triangle, and you look at the three sides of the triangle, let's say, we.

Have side a, and we have side B, and we have side see what the triangle. Inequality says is that. Okay, no matter what trying really of have and no matter how you write it. If you add up any two sides, it'll, be greater than the third side. So we can quickly write and say that. Okay, if this thing here is a triangle, then a plus B has to be greater than a C. And every other combination has to be greater than in the other side.

So for example, if I add a plus C that has to be greater than what well a plus C has. To be greater than B and then B, plus C, oops, B, plus C has to be greater than a. And this is actually not something I think crazy or mysterious property. If you were to grab three sticks and connect those three sticks to form a triangle.

Well, if you've got to connect these sticks at their endpoints, right and stick a and B were not longer that someone's not longer than the length of stick C, you could not create a triangle. It wouldn't work. There would be a gap somewhere. So you can try this and test it. Out, but what does this mean with this equation?

And this problem here? Well, it means I get to look at this triangle and the size aren't, a B and C right? The sides are 2, X, 4, X, plus 10 and X, plus 20. We have to look at every combination and find out what's the smallest integer. So that the sum of any two sides is greater than the third.

So let's set this up in the first scenario we have let's say, 2x, plus 4, X, plus 10 that has to be great, an X plus 20. So now we could just solve for x. If I combine X's here. Get 6x plus 10 is greater than X, plus 20, I'm going to subtract X from both sides and subtract 10 from both sides, kind of combining two things at once X minus X is zero 20. Minus 10 is 10. This is zero 6x.

Minus X is 5x and 5x is greater than 10 divide both sides by 5, all right inverse operations here and X is greater than 10 divided by 5 or X is greater than 2. What does that mean? Well, that means already, you know that X has to be some integer greater than 2 right? So, for example, 3 would work, but.

That's greater than 2, it's, an integer greater than 2 where the sum of these two sides is greater than this side. We'd also check other combinations. We might need a bigger, a bigger result. So let's try that so let's, try 4 X, plus 10, plus X, plus 20 has to be greater than 2. X right?

I could have I mean, person error, added 2, X, plus 4 X, plus 10, I didn't want to just reverse that order I know, I'm going to get the same result. So I'm trying that to other sides all together and adding them here when we. Simplify we get 5x, plus 30 is greater than 2x. And now what I'm going to do is divide, or I could divide both sides by 2, sorry, but what makes nothing more sense, subtract 5x on both sides, right?

This cancels out we get 30 is greater than negative. 3X. Remember, here we're going to divide by a negative values that means it will reverse everything and make inequality. And this is going to become X is greater than negative. 10? Remember here, you're, you're reversing inequalities. The reason you're doing that is for.

You're dividing or multiplying by a negative dividing once upon by negative reverses everything, of course, because if you think with a number line, the negative numbers, we've you have a huge positive number like 100, if you multiply that or divide it by negative 1 or any negative number, what ends up happening well-hunter, it becomes negative 100. Maybe it was a tiny. Whereas a small number like 1 becomes something like negative 1, which is small, but not nearly as small as negative 100. So. Yeah, there's idea reversing values when you're multiplying divided by negatives. Anyway, we can go to that much for the detail, but let's keep going ahead.

So now we know X tends to also be greater than negative 10. To be greater than 2 and greater than negative 10, which means so far, if X was 3, it would work where in both cases exit of being 3 that's being a negative and bigger than 2. Now the last scenario we'll, look at blast, combination we're going to check it just to make sure that 3 would.

Also work for this last scenario. So here we check would check out 2x, plus X, plus 20 that needs to be greater than 4x plus 10. So here we combine AXA to get 3x, plus 20 is greater than 4x, plus 10 I'm going to subtract 10 from both sides and subtract 3x from both sides, cancels out all right cancels out 20. Minus 10 is 10 for X. Minus 3 X is X.

So X could be anything less than positive 10. So in the last video, I think I chose 1, but clear, that's a mistake because our first our prerequisite here is that X. Has to be a value greater than 2. Remember, it has to work on all three combinations of whether a plus B is greater than C or a plus C is greater than B or so forth. Whatever you decide to work in all three scenarios.

So really you're asking what's. The smallest integer that satisfies X is greater than 2 greater than negative 10 and less than 10. Because if you decide an integer like 3, which is the answer here that works it's greater than 2, it's greater than negative, 10 and less than 10 it's. The. Smallest one available, of course, also equal work, right? But that that's, the smallest integer that's, the largest integer on this list that will work 10 would not work because you know, excellent would be less than 10 in our last scenario. Okay.

So I hope this helped, and thanks for the feedback.

Dated : 18-Apr-2022

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