SOLUTION: Can someone help me with this problem? It has 4 parts a,b,c and d. I have worked a,b and c this way: (d- I don't understand at all) a) I have a right triangle and know one leg (

Algebra ->  Pythagorean-theorem -> SOLUTION: Can someone help me with this problem? It has 4 parts a,b,c and d. I have worked a,b and c this way: (d- I don't understand at all) a) I have a right triangle and know one leg (      Log On


   

Question 324836: Can someone help me with this problem? It has 4 parts a,b,c and d. I have worked a,b and c this way: (d- I don't understand at all)
a) I have a right triangle and know one leg (8) and the hypotenuse (10), I have to find x. 10^ - 8^ = 100-64=36 root is 6 so x = 6. Is this correct?
b) Same triangle but I know the two legs 4a and 3a = 4^ + 3^ = 16+9=25, square root is 5 so x = 5. Correct?
c)I have a triangle with two sides of 13, a line bisects the triangle creating a right angle and the portion from the line coming down the middle to the corner of the triangle is 5 (whole triangle is 13,13,10). So I know a leg and a hypotenuse, right? 13^-5^= 169-25= 144 square root is 12 so x is 12?
d) now I have an equilateral triangle, line x bisects it creating a right angle and the bottom length is labeled "s". The line x would be a leg and half of s the other leg.... I am confused!! I have to find x but the only length I have is "s"
Any help would be appreciated!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a) I'm assuming that 'x' is the missing leg of the triangle. If so, then you are correct.

b) Let x = hypotenuse. So %284a%29%5E2%2B%283a%29%5E2=x%5E2. Solve for 'x' to get x=sqrt%28%284a%29%5E2%2B%283a%29%5E2%29=sqrt%2816a%5E2%2B9a%5E2%29=sqrt%2825a%5E2%29=5a

So the hypotenuse is 5a units long. So x = 5a


c) So you're saying that the triangle has side lengths of 13, 13, and 10. You then bisect the line of 10 units long to get 5. So you end up with two triangles with side lengths of x, 5, and 13. If that's the case, then you are correct. The height of the triangle is 12 units.


d) If you bisect a line and the half is 's' units long, then the equilateral triangle will have a side length of 2s units. So if you look at the picture, this means that the hypotenuse is 2s units long.


So by the pythagorean theorem, this means that x%5E2%2Bs%5E2=%282s%29%5E2. Solve for 'x' to get x=sqrt%284s%5E2-s%5E2%29=sqrt%283s%5E2%29=s%2Asqrt%283%29


So x=s%2Asqrt%283%29. Note: this should look familiar if you've studied 30-60-90 triangles.