Question

the base of a solid oblique pyramid is an equilateral triangle with a base edge length of 18 inches. what is the height of the triangular base of the pyramid? 9√2 in. 9√3 in. 18√2 in. 18√3 in.

Explanation:

Step1: Divide the equilateral triangle

Divide the equilateral triangle into two right - angled triangles. The base of each right - angled triangle is $\frac{18}{2}=9$ inches, and the hypotenuse is 18 inches.

Step2: Apply the Pythagorean theorem

Let the height of the equilateral triangle (which is the height of the base of the pyramid) be $h$. According to the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = 18$ (hypotenuse) and $a = 9$ (one of the legs). So $h=\sqrt{18^{2}-9^{2}}=\sqrt{(18 + 9)(18 - 9)}=\sqrt{27\times9}=\sqrt{9\times3\times9}=9\sqrt{3}$

Answer:

$9\sqrt{3}$ in.