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?

Jun 8, 2016

$A = 6 \left(3 + 2 \sqrt{3}\right)$[${\text{cm}}^{2}$]

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 $6 r$ plus an addition of

$2 \times \frac{r}{\tan \left({30}^{\circ}\right)} = 2 \times \sqrt{3} \times r$.

So each side must have a length of $l = \left(6 + 2 \times \sqrt{3}\right) \times r$ or $l = \lambda r$. For the equilateral triangle the height is $h = \frac{\sqrt{3}}{2} l$ so the area is given by

$A = \frac{1}{2} l \times h = \frac{1}{2} \frac{\sqrt{3}}{2} {\left(\lambda r\right)}^{2} = 6 \left(3 + 2 \sqrt{3}\right) {r}^{2}$

now if $r = 1$[$\text{cm}$] then $A = 6 \left(3 + 2 \sqrt{3}\right)$[${\text{cm}}^{2}$]

Jun 10, 2016

$= 6 \left(3 + 2 \sqrt{3}\right) c {m}^{2}$

Explanation:

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

$r = 1 c m \to \text{Radius of each circular piece of paper}$

$B T = \text{projection of BO} = x$

$\angle O B T = \frac{1}{2} \angle A B C = \frac{1}{2} \times {60}^{\circ} = {30}^{\circ}$

$\frac{x}{r} = \cot {30}^{\circ} = \sqrt{3}$

$\implies x = \sqrt{3} r$

$\text{Each side of the } \Delta A B C = 6 r + 2 x = 6 r + 2 \sqrt{3} r = \left(6 + 2 \sqrt{3}\right) c m$

${\text{Area " Delta ABC=sqrt3/4*"side}}^{2} = \frac{\sqrt{3}}{4} {\left(6 + 2 \sqrt{3}\right)}^{2} c {m}^{2}$

$= \sqrt{3} {\left(3 + \sqrt{3}\right)}^{2} c {m}^{2}$

$= \sqrt{3} \left(12 + 6 \sqrt{3}\right) c {m}^{2}$

$= \left(18 + 12 \sqrt{3}\right) c {m}^{2}$

$= 6 \left(3 + 2 \sqrt{3}\right) c {m}^{2}$