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 $\displaystyle a_B$ denote the base area, $\displaystyle a_L$ the lateral surface area and H the height of the solid.
Then the volume is calculated by:
$\displaystyle V = \frac13 \cdot a_B \cdot H$
to #a.: a_B is a rectangle (84) and the height is 19. Thus $\displaystyle V = 532$
to #b.: Let s denote the side of the hexagon. Then $\displaystyle a_B = 6 \cdot \frac12 \cdot s \cdot \frac12 \sqrt{3} \cdot s = \frac32 \cdot s^2 \sqrt{3}$
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.: $\displaystyle a_B$ is a circle with $\displaystyle a_B = \pi r^2$. Use the general formula to calculate the volume ($\displaystyle 256 \pi \approx 804.2477...$)
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:
$\displaystyle a_L = \frac12 \cdot (perimeter\ of \ a_B) \cdot (height\ of\ triangle)$. 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.
$\displaystyle a_L=\frac12 \left( 14 \cdot 19+6 \cdot 19 + 6\cdot\sqrt{14^2+19^2} + 14\cdot\sqrt{6^2+19^2} \right)$
The complete surface area consists of $\displaystyle a_B + a_L$
to #d. Let s denote the line segment from the tip of the cone to the edge of the base area then
$\displaystyle a_L = \pi r s$. The complete surface area is: $\displaystyle a = \pi r^2 + \pi r s = \pi r(r+s)$
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.