## yanni Group Title i neeeed serious help..... one year ago one year ago

1. Chelsea04 Group Title

what do you need help with?

2. yanni Group Title

In the figure below, find the exact value of y. (Do not approximate your answer.)

3. yanni Group Title

i attached the problem

4. JuanitaM Group Title

find the side of the larger triange, using pyth theorem

5. EmmaCahoon Group Title

Okay. It is a 45 45 90 triangle. So we know the hypotenuse is five. The hypotenuse in a 45 45 90 triangle is x(sqrt2). So sqrt2 is 1.41. 5/1.41 is 3.546. So x=3.546. Since it equals x.

6. mathstudent55 Group Title

The three triangles are similar, so the lengths of their sides are proportional. 6/5 = 5/y

7. mathstudent55 Group Title

You don't know it's a 45-45-90 triangle.

8. JuanitaM Group Title

then the side that you found is the hypothenuse of 45-45-90 triangle. using that hypotheneuse is twice as large as legs, find the size of the legs

9. mathstudent55 Group Title

If the altitude is drawn to the hypotenuse of a right triangle, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.

10. JuanitaM Group Title

simple - use 45-45-90 logic that y -y - ysqrt2 5 = ysqrt2 solve for y

11. mathstudent55 Group Title

@JuanitaM Why do you keep mentioning a 45-45-90 trianlge when we don't have one here?

12. JuanitaM Group Title

yes an altitude was droppened to form 90 degree at base labeled 6

13. mathstudent55 Group Title

The large triangle has hypotenuse of length 6 and a leg of length 5. The other leg cannot possibly measure 5, so it's not a 45-45-90 triangle. Since all triangles are similar, none of them are 45-45-90 triangles.

14. Chelsea04 Group Title

Find the 3rd side of the triangle, label this x. $x=\sqrt{6^2-5^2}=\sqrt{11}$ Then find the line going straight down the triangle, label this z. You can do this two ways (you'll need both ways). $z=\sqrt{\sqrt{11}^2-(6-y)^2}=\sqrt{11-(6-y)^2}$ $z=\sqrt{5^2-y^2}$ If you make them equal to each other, you'll be able to solve for y. $\sqrt{11-(6-y)^2}=\sqrt{5^2-y^2}$ $11-(6-y)^2=25-y^2$ $11-(36-12y+y^2)=25-y^2$ $2y^2-12y-50=0$ $\therefore y=\sqrt{34}+3$

15. Chelsea04 Group Title

Is this too confusing?

16. Chelsea04 Group Title

|dw:1360214471836:dw| Maybe that will make it more understandable :)

17. mathstudent55 Group Title

@Chelsea04 Can you try to calculate your answer for y as a number rounded off to the nearest tenth?

18. mathstudent55 Group Title

19. agent0smith Group Title

Chelsea, your answer cannot be correct. Your value of y is greater than the 6 at the base of the triangle!

20. agent0smith Group Title

Chelsea, this is your mistake: $11−(36−12y+y^2)=25−y^2$$12y−50=0$ mathstudent55 has the correct answer: The three triangles are similar, so the lengths of their sides are proportional. 6/5 = 5/y You can confirm it by using cosine.

21. mathstudent55 Group Title

@agent0smith Finally a voice of reason. Thanks!

22. agent0smith Group Title

haha no prob. I forgot to make the drawing:

23. agent0smith Group Title

I'll call that angle x |dw:1360216763172:dw| First, look at the bigger/outer triangle $\cos x = \frac{ 5 }{ 6 }$ Now, look at the inner triangle: $\cos x = \frac{ y }{ 5 }$ |dw:1360216959551:dw|

24. Chelsea04 Group Title

Right, oops!!! That seems like a much faster way!!!

25. agent0smith Group Title

once you get cosx = 5/6 and cos x = y/5, then you can equate 5/6 = y/5

26. Chelsea04 Group Title

Either way is correct, just mine was more algebraic (and a little wrong!) than trignometric.

27. agent0smith Group Title

Yes, your way works (other than your mistake) but will probably take longer.