Area of extended triangle

Question

I have three points

$$(0,0),\ (1,1),\ (2,0)$$

and $k$, where $k$ is a number, in this task $k = 2$.

I need to calculate the area of ​​the figure extending it points less than or equal to $k$.

(In other words, the desired figure is the boundary of the set of points whose distance from the triangle is less than or equal to $k$.)

I'd figure , but I have no idea what it looks like.

Answer 1

Use the same approach as in the other question. You're still going to end up with three sectors that will add up to a complete circle, the area of which will be $\pi\cdot k^2$; plus a rectangle comprising $(0,0)-(1,1)$ and the line $k$ units away to the north-west, whose area is $\sqrt 2\cdot k$, plus two more rectangles off the lines $(0,0)-(2,0)$ and $(1,1)-(2,0)$.

If you still can't picture the resulting shape, imagine drawing the shape out on the ground, then walk round the shape clockwise holding a stick out to your left. The outer edge of the shape is the line taken by the far end of the stick.

Answer 2

First some general theorem so that you don't need to ask same type of questions again.

Let $T$ be any convex body in the plane with area $A$ and perimeter $P$.
For any $r > 0$, let $T_r$ be the set $$T_r = \bigg\{ x \in \mathbb{R^2} : \inf\{ |x - y| : y \in T \} \le r \bigg\}$$

The area of $T_r$ will be equal to $$\verb/Area/(T_r) = A + P r + \pi r^2$$

For the special case of your triangle,

$$\begin{cases} A &= 1,\\ P &= 2\sqrt{2} + 2 \end{cases} \implies \verb/Area/(T_2) = 1 + (2\sqrt{2}+2)2 + \pi\cdot 2^2 = 4\pi + 4\sqrt{2} + 5$$

As an illustration of the geometric content of this formula, here is a picture of the figure you wish to find the area.

As you can see, the final figure is composed of seven pieces.

1. The original triangle in red.
2. Three rectangles in green, one for each side of the triangle. The width of each rectangle is the length of corresponding side. The heights of all triangles are $r$.
3. Three circular sectors, all of them has radius $r$. If you look carefully, you will discover you can translate the three pieces and combine them into a single circle of radius $r$.

Their contribution to $\verb/Area/(T_2)$ are $A$, $Pr$ and $\pi r^2$ respectively.

For other convex polygons, like the square in your other question, you can verify the formula in essentially the same way.

This theorem is not limited to polygons. As long as the starting shape is convex and regular enough so that $A$ and $P$ make sense, you can apply the same formula (e.g. circles, ellipses, ... ).

This sort of formula can be generalized to higher dimensions. For example, if you start from a convex body $K$ (say a dodecahedron or an ellipsoid) in $\mathbb{R}^3$ and construct the set ( this sort of construction is known as the Minkowski sum, see wiki for more details):

$$K_r = \bigg\{ x \in \mathbb{R^3} : \inf\{ |x - y| : y \in K \} \le r \bigg\} = \bigg\{ x + y : x \in K, |y| \le r \bigg\}$$

You will find the volume of $K_r$ has a very simple dependence on $r$

$$\verb/Vol/(K_r) = V + A r + 2\pi \ell r^2 + \frac{4\pi}{3}r^3$$

The $V$ and $A$ are volume and area of $K$. $\ell$ is some constant depends on $K$ you don't need to know.

Answer 3

Here is a diagram for your problem.

Each dashed green line has a length of $k=2$. Each point on the red figure is a distance of $k=2$ from the closest point of the triangle, which is a vertex or a point on a side of the triangle. This seems to be what you mean by "extending" the triangle by the amount $k$.

You can find the area of the red figure by finding the area of the triangle, each of the three rectangles that extend from a side of the triangle, and the three circular sectors that extend from the vertices of the triangle. The area of the triangle is trivial. The height of each rectangle is $k=2$, and the base is the length of the corresponding side of the triangle. The radius of each circular sector is $k=2$. To find the area of each sector, find the area of the complete circle with the same radius, find the proportion of the complete circle that the sector takes, and multiply the circle area with the proportion.

Add all those seven areas, and you are done.

I'll leave the remaining work to you.

By the way, this diagram helps you to understand the theorem stated by @achille hui in his answer. The first term of his final answer is the combined areas of the three circular sectors, the second term is the combined area of the upper two rectangles, and the third term is the combined area of the lower rectangle and the triangle polygon. The general formula he gives the combined total of (polygon)+(rectangles)+(circular sectors): the same thing with slightly different combinations.