# Thread: Need help Solid Geometry

1. ## Need help Solid Geometry

1. Given right circular cylinder circumscribed about the regular hexagonal pyramid with base a regular hexagon of side S= 6m, height of cylinder 10m
compute:
a. area of the base of the cylinder
b. lateral surface area of the cylinder
c. total surface area of the cylinder
d. area of the base of the pyramid
e. surface area if the pyramid
f. total area of the pyramid

basically it says we have a hexagonal pyramid (pyramid with a hexagon for it's base, my classmates just used S for radius r...and I'm very sure that's wrong...unless someone explains to me how did S = r,)

I know everything else to solve this but...I need to know:
a. how to solve for the area of a hexagon? (or for any other polygon for future reference since our instructor said we'd be facing more problems like this)
b. how do I derive the radius r from the sides S = 6?

2.Given: Oblique circular cylinder with base circumscribed about the square of a side 10cm, edge 20cm inclined at 45 degrees
compute
a. area of the base of the cylinder
b. lateral surface area of the cylinder
c. total surface area if the cylinder
d. surface area of prism
e. total area of prism

similar to 1st question but instead of a hexagon we get a prism, and it is inclined at a 45 degree angle. I can calculate for the height which is
h= 20cos 45 degrees = 14.14cm (I think?)

Off topic...engineering day in the Philippines in September 13 right?

2. ## Re: Need help Solid Geometry

Originally Posted by keosan22

1. Given right circular cylinder circumscribed about the regular hexagonal pyramid with base a regular hexagon of side S= 6m, height of cylinder 10m
compute:
a. area of the base of the cylinder
...

basically it says we have a hexagonal pyramid (pyramid with a hexagon for it's base, my classmates just used S for radius r...and I'm very sure that's wrong...unless someone explains to me how did S = r,)
• a regular hexagon consists of 6 equilateral triangles and therefore the radius of the circumscribed circle of the hexagon has the same length as the side of the hexagon.
• The area of a regular hexagon is determined by: $a_{6-gon} = 6 \cdot \overbrace{ \frac12 \cdot s \cdot \underbrace{\frac12 \cdot s \cdot \sqrt{3}}}_{\text{height of triangle}}^{\text{area of one triangle}}$

I know everything else to solve this but...I need to know:
a. how to solve for the area of a hexagon? (or for any other polygon for future reference since our instructor said we'd be facing more problems like this)
...
If you have a regular n-gon then this shape consists of n isosceles triangles with the radius of the circumscribed circle as legs and the side of the n-gon as base. The half of such an isosceles triangle is a right triangle. Assuming that n and r are known then

$s = 2 \cdot r \cdot \sin\left(\frac{2 \pi}{2n} \right)$

The area of one triangle is consequently: $a_\Delta = \frac12 \cdot 2 \cdot r \cdot \sin\left(\frac{2 \pi}{2n} \right) \cdot r \cdot \cos\left(\frac{2 \pi}{2n} \right) = r^2 \cdot \sin\left(\frac{\pi}{n} \right)\cdot \cos\left(\frac{\pi}{n} \right)$

Then the area of a regular n-gon is:

$a_{n-gon} = n \cdot a_\Delta$

3. ## Re: Need help Solid Geometry

for me, i think the best way to solve for the area of any polygon is to draw or make triangles on them,and make sure that you are very familiar with the sine law or cosine law so that you could easily solve for the sides of the triangle,the area of the triangle multiply iby the number of triangles you had drawn is equal to the area of the polygon.
in "how S=r?"
simply because we know that in one revolution there is 360 degrees,so knowing that hexagon has 6 sides, the angle in every triangle would be 360/6 = 60,since it is a regular hexagon, the two remaining angles in a single triangle would be (180-60)/2 = 60.
therefore the triangle is said to be an equilateral triangle which has equal sides,
therefore S=r.
I hope I helped somehow.

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# oblique pentagonal prism with edge 10cm lateral edge e=10 cm is inclined 30° with the base

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