You're working with two right triangles here.
The big one is the whole drawing. It's lying flat on its hypotenuse.
One side of the big triangle ... the left side ... has no label.
We'll be working with that side. Let's call it ' S ' .
The smaller triangle is the left piece of the big one.
Its height is 'y', and its base is 4. ' S ' is its hypotenuse.
Go to your toolbox and pull out the little box marked "Pythagoras".
Inside that box is the equation you need in order to work on this
problem, or any problem with right triangles. The equation is ...
(hypotenuse)² = (length of one leg)² + (length of the other leg)² .
In the big triangle, 'x' is one leg, and the hypotenuse is 16 .
(16)² = x² + S²
256 = x² + S²
Subtract x² from each side: S² = 256 - x²
Take the square root of each side: S = √(256 - x²)
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Now look at the small right triangle.
One leg is 4. The other leg is 'y'. The hypotenuse is ' S '.
S² = (4)² + y²
Subtract 16 from each side: y² = S² - 16
Writer in the value of S² that
we worked out before: y² = (256 - x²) - 16 .
y² = 240 - x²
Take the square root
of each side: y = √(240 - x²) .
That's what you're supposed to find. So we're done.
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Oh phooey ! Sometimes I am so stupid.
Forget almost everything I did up above, and
let's back up several steps:
==> Cut the picture along the line marked 'y', and throw away
the left piece, with the '4' on the bottom. We don't need it.
-- Now you have a right triangle that I didn't even notice before.
One leg is 12. The other leg is 'y'. The hypotenuse is 'x'.
Pythagoras says: (x²) = (12²) + (y²)
Subtract 12² from each side: y² = x² - 144
Take the square root of each side: y = √(x² - 144) .