10 circular pieces of paper each of radius 1 cm have been cut out from a piece of paper having a shape of an equilateral triangle. What should be the minimum area of the equilateral triangle?

2 Answers
Cesareo R.
Jun 8, 2016

A = 6(3+2sqrt(3))["cm"^2]

Explanation:

The circular pieces of paper aligned in rows, like

1 over
2 over
3 over
4

configure the skeleton of an equilateral triangle.

The external equilateral triangle must have a minimum side length of 6r plus an addition of

2 xx r/(tan(30^@)) = 2xx sqrt(3)xx r.

So each side must have a length of l=(6+2 xx sqrt(3))xxr or l = lambda r. For the equilateral triangle the height is h = sqrt(3)/2 l so the area is given by

A = 1/2 l xx h = 1/2 sqrt(3)/2 (lambda r)^2 = 6(3+2sqrt(3))r^2

now if r = 1["cm"] then A = 6(3+2sqrt(3))["cm"^2]

P dilip_k
Jun 10, 2016

=6(3+2sqrt3)cm^2

Explanation:

After cutting out 10 circular pieces the triangular paper will have structure as shown below
enter image source here

r =1 cm->"Radius of each circular piece of paper"

BT = "projection of BO"=x

/_OBT=1/2/_ABC=1/2xx60^@=30^@

x/r=cot30^@=sqrt3

=>x=sqrt3r

"Each side of the " DeltaABC=6r+2x=6r+2sqrt3r =(6+2sqrt3) cm

"Area " Delta ABC=sqrt3/4*"side"^2=sqrt3/4(6+2sqrt3)^2 cm^2

=sqrt3(3+sqrt3)^2cm^2

=sqrt3(12+6sqrt3)cm^2

=(18+12sqrt3)cm^2

=6(3+2sqrt3)cm^2

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