Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
• Status: Solved
• Priority: Medium
• Security: Public
• Views: 436

# area of a snooker triangle

Hi

Another question:

A triangle frame is made to enclose 2 identical spheres. They are stacked in 3 rows (3,2,1). Each sphere has a radius of 2cm. Find the lengths of the sides of the frame.

I'm sure this is obivous, but again I can't see it :)

thanks
0
andieje
• 6
• 4
• 2
• +3
1 Solution

Commented:
May be 6 spheres?
You have 3 spheres on each side, it's 6R.
The distance from the center of corner sphere to triangle vertix is 2R, so the distance from vertix to projection of center to side can be calculated as sqrt(4R^2-R^2)=sqrt(3)R
So the side is (2sqrt(3)+6)R
0

Commented:
Do you mean 6  identical spheres?
Can you find the lengths of the sides of a triangle enclosing one sphere?
0

Commented:
Sorry, it's (2sqrt(3)+4)R
0

Commented:
You can get very close by just calculating the area of an equilateral triangle. The problem is that snooker triangles are rounded in the corners, thus cutting off a little of the area.

Given the radius of 2cm and 3 balls/side each side is 12cm

Plug it into this equation: Area= length of side squared x (square root of 3 / 4)

Area = 62.35
0

Commented:
62.25 is too small for area. 12 cm is too small for side.
yuk99 has a very nice answer.
The tricky part is
"The distance from the center of corner sphere to triangle vertix is 2R"
but a good diagram will help derive it.
0

Commented:
To the author.
It's unclear do you need to calculate the area (in the title) or side length (in the question).
Is it ok to neglect roundness of real snooker triangle's corners?
Do we assume correctly that the question is about 6 spheres?
0

Author Commented:
hi,
i wrote the question verbatim. There is a picture with angluar corners and not round corners. There are 6 spheres. The answer is 14.9 if that helps you reverse engineer the question!
0

Commented:
If you substitute R by 2 in my answer, you will get 14.928..., as you need.
0

Author Commented:
Sorry, you are right. I can see my typo. It is 6 spheres
0

Author Commented:
I don't understand the answer. Not because it is poorly explained but because my geometry is poor. If you are willing to talk me through it, it is muh appreciated. Thanks
0

Commented:
You have already the complete answer in yuk99's response 10:35 and 10:38
0

Commented:
1. Make a plot. You will see the side can be calculated as a sum of 4 sphere's radius and 2 equal segments from triangle corner to projection of center of sphere to the side. Is it clear.
2. How to calculate this segment? Consider the corner sphere as enclosed in a equilateral triangle. There is a theorem, that  divides intersection of heights in equilateral trianglehe height as 2:1. Try to prove it, it's easy. Smaller part of height is R, then larger part is 2R, the distance from sphere center to triangle corner.
3. Apply Pifagor theorem to calculate the projection: sqrt((2R)^2-R^2)=sqrt(4R^2-R^2)=sqrt(3)R
4. Finally just sum all segments.
0

Commented:
Mistyped: intersection of heights in equilateral triangle divides the height as 2:1
0

Commented:
Are you allowed to use trig?

Along one side of the triangle, drop perpendiculars from the centers of the first and third circles.  This will divide the side into three segments.

The middle segment will have a length of four radii (one each from the first and third circles and two from the center circle.  Hence, its length will be 4r = 8.

The outer segments will each have a length x = r / Tan(30 = pi/6) = 2 / 0.577350269189625 = 3.46410161513776.  This is true because we know from yesterday that a line from the center of the corner circle to the corner of the triangle bisects the corner angle (60).  Then, using trig, we see that Tan(30) = opposite side / adjacent side = r /x. or x = r/Tan(30).

The total length will be 8 + 2x = 14.9282032302755.
0

Author Commented:
thanks
0

## Featured Post

• 6
• 4
• 2
• +3
Tackle projects and never again get stuck behind a technical roadblock.