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
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#]

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#