Given:
• AB = 6 cm
,• SM = √15 cm
Let's solve for the following:
• 1) the base elevation AM.
Given that we have a regular triangular pyramid, the length of the three bases are equal.
AB = BC = AC
BM = BC/2 = 6/2 = 3 cm
To solve for AM, which is the height of the base, apply Pythagorean Theorem:
[tex]\begin{gathered} AM=\sqrt{AB^2-BM^2} \\ \\ AM=\sqrt{6^2-3^2} \\ \\ AM=\sqrt{36-9} \\ \\ AM=\sqrt{27} \\ \\ AM=5.2\text{ cm} \end{gathered}[/tex]The base elevation of the pyramid is 5.2 cm.
• (2)., The elevation SO.
To find the elevation of the pyramid, apply Pythagorean Theorem:
[tex]SO=\sqrt{SM^2-MO^2}[/tex]Where:
SM = √15 cm
MO = AM/2 = 5.2/2 = 2.6 cm
Thus, we have:
[tex]\begin{gathered} SO=\sqrt{(\sqrt{15})^2-2.6^2} \\ \\ SO=\sqrt{15-6.76} \\ \\ SO=2.9\text{ cm} \end{gathered}[/tex]Length of SO = 2.9 cm
• (3). Area of the base:
To find the area of the triangular base, apply the formula:
[tex]A=\frac{1}{2}*BC*AM[/tex]Thus, we have:
[tex]\begin{gathered} A=\frac{1}{2}*6^*5.2 \\ \\ A=15.6\text{ cm}^2 \end{gathered}[/tex]The area of the base is 15.6 square cm.
• (4). Area of the side surface.
Apply the formula:
[tex]SA=\frac{1}{2}*p*h[/tex]Where:
p is the perimeter
h is the slant height, SM = √15 cm
Thus, we have:
[tex]\begin{gathered} A=\frac{1}{2}*(6*3)*\sqrt{15} \\ \\ A=34.86\text{ cm}^2 \end{gathered}[/tex]• (5). Total surface area:
To find the total surface area, apply the formula:
[tex]TSA=base\text{ area + area of side surface}[/tex]Where:
Area of base = 15.6 cm²
Area of side surface = 34.86 cm²
TSA = 15.6 + 34.86 = 50.46 cm²
The total surface area is 50.46 cm²
• (6). Volume:
To find the volume, apply the formula:
[tex]V=\frac{1}{3}*area\text{ of base *height}[/tex]Where:
Area of base = 15.6 cm²
Height, SO = 2.9 cm
Thus, we have:
[tex]\begin{gathered} V=\frac{1}{3}*15.6*2.9 \\ \\ V=15.08\text{ cm}^3 \end{gathered}[/tex]The volume is 15.08 cm³.
ANSWER:
• 1.) 5.2 cm
,• 2.) 2.9 cm
,• 3.) 15.6 cm²
,• 4.) 34.86 cm²
,• (5). 50.46 cm²
,• 6). 15.08 cm³.
