Area of Region

**Vicky1997** · December 13th 2009, 05:39 PM

On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is contructed. The interiors of the square and the 12 triangles have no points in common. Let R be the region formed by the union of the square and all the trianlges, and let S be the smallest convex polygon that contains R. What is the area of the region that is inside S but outside R?

I tried drawing the square and triangles several times but I must be misunderstanding the problem. Could i get some help pls?

Vicky.

**Soroban** · December 13th 2009, 07:32 PM

Hello, Vicky!

On each side of a unit square, an equilateral triangle of side length 1 is constructed.
On each new side of each equilateral triangle, another equilateral triangle of side length 1 is contructed.
The interiors of the square and the 12 triangles have no points in common.
Let R be the region formed by the union of the square and all the trianlges,
and let S be the smallest convex polygon that contains R.
What is the area of the region that is inside S but outside R?

Code:

                     B
    *-------*-------o
     *:::::*:*:::::* \
  *   *:::*:::*:::*   o C
  |::* *:*:::::*:* *::|
  |:::::*-------o:::::|
  |::*::|      A|::*::|
  *:::::|       |:::::*
  |::*::|       |::*::|
  |:::::*-------*:::::|
  |::* *:*:::::*:* *::|
  *   *:::*:::*:::*   *
     *:::::*:*:::::*
    *-------*-------*

The region $R$ is comprised of a unit square and 12 equilateral triangles.

The area of $R$ is: . $A_R \;=\;1 + 12\left(\tfrac{\sqrt{3}}{4}\right) \;=\;1 + 3\sqrt{3}$

$\Delta ABC$ has two sides $AB = AC = 1$ and included angle $\angle BAC - 30^o$
Its area is: . $\tfrac{1}{2}(1^2)\sin30^o \:=\:\tfrac{1}{4}$

The area of $S$ is the area of $R$ plus four of those triangles.
. . $A_S \;=\;(1 + 3\sqrt{3}) + 4\left(\tfrac{1}{4}\right) \:=\:2 + 3\sqrt{3}$

The area inside $S$ and outside $R$ is: . $(2+3\sqrt{3}) - (1 + 3\sqrt{3}) \;=\;1$

**Vicky1997** · December 15th 2009, 08:43 PM

Originally Posted by Soroban

Hello, Vicky!

Code:

                     B
    *-------*-------o
     *:::::*:*:::::* \
  *   *:::*:::*:::*   o C
  |::* *:*:::::*:* *::|
  |:::::*-------o:::::|
  |::*::|      A|::*::|
  *:::::|       |:::::*
  |::*::|       |::*::|
  |:::::*-------*:::::|
  |::* *:*:::::*:* *::|
  *   *:::*:::*:::*   *
     *:::::*:*:::::*
    *-------*-------*

The region $R$ is comprised of a unit square and 12 equilateral triangles.

The area of $R$ is: . $A_R \;=\;1 + 12\left(\tfrac{\sqrt{3}}{4}\right) \;=\;1 + 3\sqrt{3}$

$\Delta ABC$ has two sides $AB = AC = 1$ and included angle $\angle BAC - 30^o$
Its area is: . $\tfrac{1}{2}(1^2)\sin30^o \:=\:\tfrac{1}{4}$

The area of $S$ is the area of $R$ plus four of those triangles.
. . $A_S \;=\;(1 + 3\sqrt{3}) + 4\left(\tfrac{1}{4}\right) \:=\:2 + 3\sqrt{3}$

The area inside $S$ and outside $R$ is: . $(2+3\sqrt{3}) - (1 + 3\sqrt{3}) \;=\;1$

Thanks!!!!!!

Now I clearly understand what I was doing wrong.

Vicky.

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