Question

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.

Answer 1

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.