Question

the base of a solid oblique pyramid is an equilateral triangle with a base edge length of 14 units. what is bc, the height of the pyramid? 7 units 7√2 units 14 units 14√2 units

Explanation:

Step1: Find the distance from a vertex of the base - equilateral triangle to the center of the base.

For an equilateral triangle with side - length \(a = 14\) units, the distance from a vertex to the center of the base (\(AC\)) can be found using the property of an equilateral triangle. The distance from a vertex of an equilateral triangle to its centroid (center) is \(\frac{2}{3}\) of the altitude of the equilateral triangle. The altitude \(h\) of an equilateral triangle with side - length \(a\) is \(h=\frac{\sqrt{3}}{2}a\). So the distance from a vertex to the center of the base \(AC=\frac{2}{3}\times\frac{\sqrt{3}}{2}a\). Substituting \(a = 14\), we get \(AC=\frac{2}{3}\times\frac{\sqrt{3}}{2}\times14=\frac{14\sqrt{3}}{3}\) units. In the right - triangle \(ABC\), assume the slant - height (not given, but we can also use another approach). Consider the right - triangle formed by half of the side of the base and the height of the pyramid. If we consider the right - triangle \(ABC\) where \(\angle BAC = 45^{\circ}\) and the side of the base of the equilateral triangle is \(14\) units. The distance from the center of the base to a side of the base is \(\frac{1}{3}\) of the altitude of the equilateral triangle. The altitude of the equilateral triangle \(h=\frac{\sqrt{3}}{2}\times14 = 7\sqrt{3}\) units. The distance from the center of the base to a side of the base is \(\frac{7\sqrt{3}}{3}\) units.
Since \(\triangle ABC\) is a right - triangle and \(\angle BAC=45^{\circ}\), \(\tan45^{\circ}=\frac{BC}{AC}\). Also, if we consider the fact that in right - triangle \(ABC\), and assume the base of the right - triangle (related to the base of the equilateral triangle) and using the property of \(45 - 45-90\) triangle. The base of the right - triangle (adjacent to \(\angle BAC\)) is half of the side of the equilateral triangle, i.e., \(AC = 7\) units.

Step2: Use the property of a 45 - 45-90 triangle.

In a \(45 - 45-90\) triangle, if the length of one of the legs (adjacent to the \(45^{\circ}\) angle) is \(x\), and the other leg (opposite to the \(45^{\circ}\) angle) is \(y\), and the hypotenuse is \(z\), then \(x = y\) and \(z=\sqrt{2}x\). Here, in right - triangle \(ABC\) with \(\angle BAC = 45^{\circ}\) and \(\angle ACB=90^{\circ}\), if \(AC = 7\) units, then \(BC = 7\) units because \(\tan45^{\circ}=1=\frac{BC}{AC}\), so \(BC = AC\).

Answer:

7 units