Question 149702
First, let's draw the picture. Be sure you label every length you can



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/solution5.png" alt="Photobucket - Video and Image Hosting">



So we can see that the hypotenuse of the largest triangle is 26 units. Using pythagoreans theorem, we get:


{{{x^2+y^2=26^2}}}



Now looking at the left most smaller triangle, we see that the legs are 8 and "h" while the hypotenuse is "x". So once again, with pythagoreans theorem, we can say:


{{{8^2+h^2=x^2}}}



Now solving for {{{h^2}}}, we get {{{h^2=x^2-8^2}}}



Finally, the right most smaller triangle has a base of 18 and "h" and a hypotenuse of y. So this gives us:


{{{h^2+18^2=y^2}}}


---------



{{{x^2+y^2=26^2}}} Start with the first equation.



{{{x^2+(h^2+18^2)=26^2}}} Plug in {{{h^2+18^2=y^2}}}. In other words, replace {{{y^2}}} with {{{h^2+18^2}}}



{{{x^2+h^2+324=676}}} Square 18 to get 324. Square 26 to get 676.



{{{x^2+h^2+324=676}}} Square 18 to get 324. Square 26 to get 676.



{{{x^2+x^2-8^2+324=676}}} Plug in {{{h^2=x^2-8^2}}}



{{{x^2+x^2-64+324=676}}} Square 8 to get 64



{{{2x^2+260=676}}} Combine like terms.



{{{2x^2=416}}} Subtract 260 from both sides.



{{{x^2=208}}} Divide both sides by 2.



{{{x=sqrt(208)}}} Take the square root of both sides. Note: only the positive square root is considered.



So the length of x is {{{x=sqrt(208)}}} which is about 14.422 units (this isn't important).



{{{h^2=x^2-8^2}}} Go back to the second equation



{{{h^2=(sqrt(208))^2-8^2}}} Plug in {{{x=sqrt(208)}}}



{{{h^2=208-8^2}}} Square the square root to eliminate it.



{{{h^2=208-64}}} Square 8 to get 64



{{{h^2=144}}} Combine like terms.



{{{h=12}}} Take the square root of both sides. Once again, only the positive square root is considered.


So the height is 12 units which means that altitude is 12 units.