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Natasha_Volkova [10]
3 years ago
10

Help me please asap :(

Mathematics
1 answer:
GREYUIT [131]3 years ago
6 0
This is a fairly tedious problem. The only way to solve that I see is to simply list out all the possible cases and go through each one by one. You'll use the triangle inequality theorem to see if the triangles can be formed. The theorem says that a+b > c must be true for all sets of pairs. In other words, take any two sides and add them up. The sum must be greater than the third side. 

Going through all the combos possible (that I could find), this is what I got
Blue9,Green7,Orange4
Blue9,Green7,Purple12
Blue9,Green7,Red3
Blue9,Green7,Yellow5
Blue9,Orange4,Purple12
Blue9,Purple12,Yellow5
Green7,Orange4,Yellow5
Green7,Red3,Yellow5
Orange4,Red3,Yellow5

In all, I count 9 cases. So the answer to problem 1 is 9.

-----------------------------------------------------------

For problem 2, the best way may be to pick two segments at a time instead of one. However, I have a feeling that will take just as long as the first method. I haven't tried it out. Even though going through the rods one at a time takes a while, it's probably the best option so you don't overlook any cases. 


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Step-by-step explanation:

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These are special right triangles, a 45, 45, 90 and a 30, 60, 90 triangle. The formla for these triangles are shown in the picture below.

Now for the first one ]

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For the second one this is easier

We know that the x is (6) becuase its already in the radical number, now to find the hypotnuse, we double that and get 12. Now we know what y is becuase since the 2nd triangle is a 45/45/90, we know that 12 is y, the hypotnuse is just N\sqrt{2} and we just plug in the n with 12 and solve it!

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