A triangular pyramid has a surface area of 100, the base equals 6, and the slant height equals 8. What is the height of the base to the nearest tenth?

Answer 1

To start, we know that the surface area of a triangular pyramid can be found using the equation $A = \frac{1}{2}Pl$, where $A$ is the surface area, $P$ is the perimeter of the base, and $l$ is the slant height. In this problem, we are given that $A = 100$ and $l = 8$. Thus, our equation becomes $100 = \frac{P}{2}(8)$. Solving for $P$ gives us $P = \frac{200}{8} = 25$.

Next, we can use the fact that the perimeter of a triangle can be found using the equation $P = s_1 + s_2 + s_3$, where $s_1, s_2,$ and $s_3$ are the lengths of the sides of the triangle. Since the base of the triangular pyramid has a length of 6, we can rewrite the equation as $25 = 6 + s_2 + s_3$. Solving for $s_2 + s_3$ gives us $s_2 + s_3 = 19$.

Finally, we can find the height of the base of the triangular pyramid by using the Pythagorean Theorem. The height forms a right triangle with one of the sides being the slant height (8) and the other side being the height of the base of the pyramid. Thus, we can set up the equation $h^2 + 6^2 = 8^2$ and solve for $h$. Simplifying, we get $h^2 + 36 = 64$, so $h^2 = 28$. Taking the square root of both sides gives us $h = \sqrt{28}$. To the nearest tenth, the height of the base is approximately $\boxed{5.3}$.