A triangular pyramid has a surface area of 100^2, the base equals 6, and the slant height equals 8. What is the height of the base to the nearest tenth?
1 answer
We let $B$ be the base edge length and $L$ be the slant height. The lateral side length is $\sqrt{L^2-\left(\frac{B}{2}\right)^2}=\sqrt{8^2-\left(\frac{6}{2}\right)^2}=\sqrt{64-9}=\sqrt{55}$. We can break the base into two 30-60-90 triangles, so the height of the triangle is $\sqrt{55}\sqrt{3}/2$. In a lateral face of the pyramid, we can make a right triangle that has the height we just found as the hypotenuse, $B/2$ as one leg, and the lateral side length as another leg. Call the new leg $y$. Writing the Pythagorean Theorem, we have $y^2+48=y^2+27.5$, which implies that $B/2=\sqrt{21.5}$. Call the radius $r$ of the circumscribed circle of the base of the pyramid. The area of this circle is simply $\pi r^2=\frac{\pi B^2}{4}=\frac{36\pi}{4}=9\pi$. Using $B\cdot \frac{3\sqrt{3}}{2}=9\pi$, we get that $B=\frac{12\pi}{\sqrt{3}}=6\sqrt{3}$, so $B/2=3\sqrt{3}$. Writing the Pythagorean Theorem on a right triangle with hypotenuse $r$ as the slant height and legs $3\sqrt{3}$ and $6\sqrt{3}$, we find that $r=\frac{3\sqrt{3}}{2}\cdot 2=\sqrt{27}=3\sqrt{3}$. Writing the Pythagorean Theorem on a right triangle with hypotenuse $h$ as the slant height of the pyramid and legs $\sqrt{55}/2$ and $3\sqrt{3}$, we find that $h=\left(\frac{\sqrt{55}}{2}\right)^2+27=\frac{55}{4}+27=\frac{163}{4}\approx \boxed{40.8}$.