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Luv2Teach
Use the geometric mean for right triangles here.  Like this [tex] \frac{9}{y}= \frac{y}{7} [/tex].  Cross multiply to get [tex] y^{2}=63 [/tex], and [tex]y= \sqrt{63} [/tex].  That simplifies down to [tex]y= \sqrt{9*7} [/tex].  Pull out the 9 as a perfect square of 3 and you're left with [tex]y=3 \sqrt{7} [/tex], first choice above.
tonb
Another way to do this is to apply pythagoras 3 times, since there are 3 right triangles. Assuming the sides not named here are a and b, you get:

a^2 + b^2 = 16^2
9^2 + y^2 = a^2
y^2 + 7^2 = b^2

You can solve these to y=3√7

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