Rectangle $ABCD$ is the base of pyramid $PABCD$. If $AB = 8$, $BC = 4$, $\overline{PA}\perp \overline{AD}$, $\overline{PA}\perp \overline{AB}$, and $PB = 17$, then what is the volume of $PABCD$?
To find the volume of a pyramid, I need to multiply the area of the base by the height and divide by 3. The base is a rectangle, so its area is just the product of its length and width, which are 8 and 4, respectively. So the area of the base is 8 times 4, which is 32. The height of the pyramid is the perpendicular distance from the vertex $P$ to the base. Since $\overline{PA}$ is perpendicular to two adjacent edges of the base, it must also be perpendicular to the plane containing the base, and therefore it is the height of the pyramid. So I just need to find the length of $\overline{PA}$. I notice that triangle $PAB$ is a right triangle, so I can use the Pythagorean theorem to find the length of $\overline{PA}$. By the Pythagorean theorem, $PA^2 + AB^2 = PB^2$. Plugging in the values given in the problem, I get $PA^2 + 8^2 = 17^2$. Simplifying, I get $PA^2 + 64 = 289$. Subtracting 64 from both sides, I get $PA^2 = 225$. Taking the square root of both sides, I get $PA = 15$. So the height of the pyramid is 15. Now I can find the volume by multiplying the area of the base by the height and dividing by 3. The volume is 32 times 15 divided by 3, which is 160. 