Answer:
134.88 cm³
Step-by-step explanation:
In a regular-based pyramid, the base is a regular polygon, meaning that all of its sides are equal in length, and the lateral faces are congruent isosceles triangles. The slant height is the altitude of one of its triangular faces, and the height of the pyramid is the perpendicular distance from the apex to the center of the base.
The volume of any pyramid can be found by dividing the product of the area of its base and its perpendicular height by 3.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Pyramid}}\\\\V = \dfrac{1}{3}Bh\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$B$ is the area of the base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the perpendicular height.}\end{array}}[/tex]
[tex]\dotfill[/tex]
Since the base of the given pyramid is an equilateral triangle with side length 15 cm, we can find its area by using the formula for the area of an equilateral triangle:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Area of an equilateral triangle}}\\\\A=\dfrac{\sqrt{3}}{4}s^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A$ is the area.}\\ \phantom{ww}\bullet\;\textsf{$s$ is the side length.}\end{array}}[/tex]
Substitute s = 15 into the formula:
[tex]\textsf{Base Area}=\dfrac{\sqrt{3}}{4}\cdot 15^2\\\\\\\textsf{Base Area}=\dfrac{225\sqrt{3}}{4}\; \sf cm^2\\\\\\[/tex]
[tex]\dotfill[/tex]
In a regular pyramid, its apex lies directly above the center of its base. Therefore, the height of the given pyramid is the perpendicular distance from the center of the equilateral triangular base to its apex.
The perpendicular height of the given pyramid is one of the legs of a right triangle. In this triangle, the hypotenuse represents the slant height of the pyramid, while the other leg corresponds to the distance from the midpoint of one of the base edges to the centroid of the equilateral triangular base.
The distance from the midpoint of one of the base edges to the centroid of the equilateral triangular base is the shortest leg of a 30-60-90 triangle, where its longest leg is equal to half the side length of the equilateral triangular base. Therefore, this distance is s / 2√3, where s is the base edge.
To find the perpendicular height (h) of the triangular pyramid given its slant height (l), we can use the Pythagorean Theorem:
[tex]\left(\dfrac{s}{2\sqrt{3}}\right)^2+h^2=l^2\\\\\\\\\left(\dfrac{s}{2\sqrt{3}}\right)^2+h^2=l^2\\\\\\\\\dfrac{s^2}{12}+h^2=l^2\\\\\\\\h^2=l^2-\dfrac{s^2}{12}\\\\\\\\h=\sqrt{l^2-\dfrac{s^2}{12}}[/tex]
[tex]h= \sqrt{l^2 - \dfrac{s^2}{12}}[/tex]
Given that the s = 15 cm and l = 6 cm, then:
[tex]h= \sqrt{6^2 - \dfrac{15^2}{12}}\\\\\\h= \sqrt{36 - \dfrac{225}{12}}\\\\\\h= \sqrt{ \dfrac{69}{4}}\\\\\\h=\dfrac{\sqrt{69}}{2}\; \sf cm[/tex]
So the perpendicular height of the given pyramid is exactly √(69) / 2 cm.
[tex]\dotfill[/tex]
Finally, to find the volume, substitute the area of the base and the perpendicular height into the volume formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Pyramid}}\\\\V = \dfrac{1}{3}Bh\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$B$ is the area of the base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the perpendicular height.}\end{array}}[/tex]
Therefore:
[tex]V=\dfrac{1}{3}\cdot \dfrac{225\sqrt{3}}{4}\cdot \dfrac{\sqrt{69}}{2}\\\\\\\\V= \dfrac{225\sqrt{207}}{24}\\\\\\\\V= \dfrac{225\cdot 3\sqrt{23}}{24}\\\\\\\\V= \dfrac{225\sqrt{23}}{8}\\\\\\\\V=134.88\; \sf cm^3[/tex]
So, the volume of the given pyramid is:
[tex]\Large\boxed{\boxed{134.88\; \sf cm^3}}[/tex]
