SOLUTION: The problem is solve for x y and z. The diagram is a right triangle with the altitude drawn to the hypotenuse. The altitude divides the hypotenuse into two segements labeled x an

Question 127636This question is from textbook Geometry
: The problem is solve for x y and z. The diagram is a right triangle with the altitude drawn to the hypotenuse. The altitude divides the hypotenuse into two segements labeled x and x+1. Adjacent to x is the leg labeled x+2. Adjacent to x+1 is the leg labeled z.
I am trying to solve first for x and have tried the following, but I know it's not right:
x over x+2 = x+2 over 2x+1
x squared + 4x+4 = 2x squared +x
(subtract x squared from both sides)
4x + 4 = x squared + x
(subtract x from both sides)
3x + 4 = x squared
the square root of (3x + 4) =x
to find y and z, i need to isolate x
Thanks! This question is from textbook Geometry

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The problem is solve for x y and z. The diagram is a right triangle with the altitude drawn to the hypotenuse. The altitude divides the hypotenuse into two segements labeled x and x+1. Adjacent to x is the leg labeled x+2. Adjacent to x+1 is the leg labeled z.
I am trying to solve first for x and have tried the following, but I know it's not right:
x over x+2 = x+2 over 2x+1
x squared + 4x+4 = 2x squared +x
(subtract x squared from both sides)
4x + 4 = x squared + x
(subtract x from both sides)
3x + 4 = x squared
the square root of (3x + 4) =x
to find y and z, i need to isolate x
-----------------------------------------
The hypotenuse is 2x+1
One leg is x+2
Solve for z: z = sqrt[(2x+1)^2 - (x+2)^2]
z = sqrt[4x^2+4x+1 -x^2-4x-4]
z = sqrt[3x^2-3]
-------------
You say you want to isolate "x":
3x^2-2 = z^2
x^2 = (z^2+2)/3
x = sqrt[(z^2+2/3)]
=======================
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

You didn't specify the segment labeled y, but I just assumed by process of elimination. I'm pretty sure this is the picture you described, and I'm going to base my solution on this assumption.

What we are going to do is use Pythagoras' Theorem repeatedly to get two expressions for in terms of x that we can then equate and solve for x.

The hypotenuse of the large triangle is , so we can say:

Similarly, using the smallest triangle:

Now that we have an expression for we can develop the second expression for using the middle-sized triangle:

Now set these two expressions for to be equal:

So or . We can exclude the negative root because we are looking for a measure of length, hence .

From a previous result we have:

And

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