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?

Question

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?

Theodore Abad · Accepted Answer

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

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#]

Maverick Smith · Answer

#=6(3+2sqrt3)cm^2#

After cutting out 10 circular pieces the triangular paper will have structure as shown below

# 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#

Evelyn Arboleda · Answer

undefined The minimum area of the equilateral triangle would be such that it can accommodate the 10 circular pieces of paper without overlapping.

Each circular piece of paper has a diameter of 2 cm, which means it needs a minimum distance of 2 cm from the center of one circle to the center of the adjacent circle to avoid overlapping.

For an equilateral triangle, the centers of the circles would be arranged in a hexagonal pattern, where the distance between adjacent centers would be the same as the diameter of the circle, which is 2 cm.

Considering that the circles are arranged in this hexagonal pattern, the minimum distance between two sides of the equilateral triangle (which corresponds to the diameter of the circle) is 2 cm.

Therefore, the minimum side length of the equilateral triangle would be 2 cm. Using the formula for the area of an equilateral triangle, $ \frac{\sqrt{3}}{4} \times \text{side length}^2 $, the minimum area of the equilateral triangle would be $ \frac{\sqrt{3}}{4} \times (2 \, \text{cm})^2 = \sqrt{3} \, \text{cm}^2 $.