You have to use the geometric mean of right triangles here. Your ratio is set up like this: [tex] \frac{3}{y} = \frac{y}{12+3} [/tex]. That gives you [tex] y^{2} =45[/tex] and that [tex]y =3 \sqrt{5} [/tex]. To find z in the second part, use the Pythagorean Theorem. [tex] z^{2} =(3 \sqrt{5} ) ^{2} -(3) ^{2} [/tex] and z = 6. In the third part, the student should have had the length of the hypotenuse as the denominator in the second ratio, which is 8. So your answer there is the second one down. In order to do that last one you need to find the hypotenuse in the triangle that has legs of 4 and 10. Using Pythagorean's theorem we find that that length is [tex] \sqrt{116} [/tex]. Now set up your geometric mean: [tex] \frac{4}{ \sqrt{116} }= \frac{ \sqrt{116} }{x+4} [/tex]. Cross multiply to get 4x + 16 = 116 and that 4x = 100. Solve for x to get x = 25
