Question

type the correct answer in the box. use numerals instead of words.
a right square pyramid has an altitude of 10 and each side of the base is 6. to the nearest tenth of a centimeter, what is the distance from the apex, or top of the pyramid, to each vertex of the base?

x = units

Explanation:

Step1: Find base diagonal half-length

The base is a square with side 6. The diagonal of the square is $6\sqrt{2}$, so half the diagonal is $\frac{6\sqrt{2}}{2} = 3\sqrt{2} \approx 4.2426$.

Step2: Apply Pythagorean theorem

The altitude (10), half the base diagonal, and $x$ form a right triangle. Use $x = \sqrt{(3\sqrt{2})^2 + 10^2}$.
First calculate $(3\sqrt{2})^2 = 9\times2 = 18$, then $18 + 10^2 = 18 + 100 = 118$. So $x = \sqrt{118} \approx 10.8628$.

Step3: Round to nearest tenth

Round $\sqrt{118}$ to one decimal place.

Answer:

10.9