The Great Pyramid of Giza in Egypt has a square base with edge length of 230.5 m and a slant height of 180.2 m. Find the height of the pyramid. Round to the nearest tenth.
We can use the Pythagorean Theorem to solve this problem. Let the height of the pyramid be $h$.
[asy]
pair A,B,C,D;
A=(0,0);
B=(230.5,0);
C=(230.5,230.5);
D=(0,230.5);
draw(A--B--C--D--cycle);
draw(A--(230.5,4*sqrt(180.2*180.2-115.25*115.25)/115.25));
draw((0,-40)--(0,-280) , dashed);
label("$s$",(115.25,320.5));
label("$h$",(240,120.5));
label("$a$",(120.5,-20));
label("$230.5$",(115.25,0),S);
label("$180.2$",(230.5,115.25),E);
[/asy]
We have a right triangle with legs measuring $a = 230.5$ and $h$ and hypotenuse measuring $s = 180.2$. Applying the Pythagorean Theorem, we have \[s^2 = a^2 + h^2.\]Substituting the known values gives \[180.2^2 = 230.5^2 + h^2.\]Therefore, \[h^2 = 180.2^2 - 230.5^2.\]Using a calculator, we find that \[h^2 \approx 12258.04 - 53102.25 = -40844.21.\]Since the height must be positive, we conclude that $h$ must be complex. Therefore, we have made a mistake and we need to check our calculations. In fact, the value for $a = 230.5$ is the value for $s$ and the value for $s = 180.2$ is the value for $h$. Repeating the calculations, with the values for $a$ and $s$ switched, we get \[h^2 = 230.5^2 - 180.2^2.\]Hence, \[h^2 = 64323.05 - 32472.04 = 31851.01.\]So, $h=\sqrt{31851.01}$ is approximately $\boxed{178.4}$ to the nearest tenth.