# Thread: Volume, Surface Area, and Lateral Surface Area

1. ## Volume, Surface Area, and Lateral Surface Area

Find the Volume, Surface Area, and Lateral Surface Area of the pyramids and cone. If someone could include their work and how they got their answer that would be great! Thanks.

2. Originally Posted by peachgal
Find the Volume, Surface Area, and Lateral Surface Area of the pyramids and cone. If someone could include their work and how they got their answer that would be great! Thanks.
Let $a_B$ denote the base area, $a_L$ the lateral surface area and H the height of the solid.
Then the volume is calculated by:

$V = \frac13 \cdot a_B \cdot H$

to #a.: a_B is a rectangle (84) and the height is 19. Thus $V = 532$

to #b.: Let s denote the side of the hexagon. Then $a_B = 6 \cdot \frac12 \cdot s \cdot \frac12 \sqrt{3} \cdot s = \frac32 \cdot s^2 \sqrt{3}$
Now use the general formula to calculate the volume.

to #c.: All areas are half rectangles. So I'm sure that you can do the calculations.

to #d.: $a_B$ is a circle with $a_B = \pi r^2$. Use the general formula to calculate the volume ( $256 \pi \approx 804.2477...$)

To calculate the surface area or the lateral surface area is a little bit more tricky. If the solid is a right solid and the base area is a regular polygon or a circle then the lateral surface consists of congruent triangles or a sector of a circle. In this case the lateral surface is calculated by:

$a_L = \frac12 \cdot (perimeter\ of \ a_B) \cdot (height\ of\ triangle)$. This formula can be used at #b. and #d.. In both cases you must calculate the height h of the triangle first.

to #a.: The lateral surface area consists of 4 right triangles.

$a_L=\frac12 \left( 14 \cdot 19+6 \cdot 19 + 6\cdot\sqrt{14^2+19^2} + 14\cdot\sqrt{6^2+19^2} \right)$

The complete surface area consists of $a_B + a_L$

to #d. Let s denote the line segment from the tip of the cone to the edge of the base area then

$a_L = \pi r s$. The complete surface area is: $a = \pi r^2 + \pi r s = \pi r(r+s)$

to #c.: All areas are right triangles (that means half rectangles) and I'm sure that you can do the calculations. Post your results so we can check.