A pyramid is formed on a $6\times 8$ rectangular base. The four edges joining the apex to the corners of the rectangular base each have length $13$. What is the volume of the pyramid?
I know that the volume of a pyramid is one-third times the base area times the height. To find the base area, I can multiply the length and width of the rectangle, which are $6$ and $8$. So the base area is $6\times 8=48$. To find the height, I need to use the Pythagorean theorem, since the height is the perpendicular distance from the apex to the base. If I draw a line from the apex to the center of the base, it will form a right triangle with the height as one leg and half the diagonal of the base as the other leg. The diagonal of the base is the hypotenuse of a right triangle with the length and width of the base as the legs. So the diagonal is $\sqrt{6^2+8^2}=\sqrt{100}=10$. Half the diagonal is $10/2=5$. The hypotenuse of the triangle with the height and half the diagonal as the legs is one of the edges of the pyramid, which is $13$. So I can use the Pythagorean theorem to find the height: $h^2+5^2=13^2$. Simplifying, I get $h^2=169-25=144$. Taking the positive square root, I get $h=\sqrt{144}=12$. So the height of the pyramid is $12$. Now I can find the volume by plugging in the base area and the height into the formula: $V=\frac{1}{3}\times 48\times 12$. Simplifying, I get $V=\frac{1}{3}\times 576=192$. 