Perimeter

The perimeter of a two-dimensional figure is the length of the boundary of the figure. If the figure is a polygon such as a triangle, square, rectangle, pentagon, etc., then the perimeter is the sum of the edge lengths of the polygon. For example, a square with side length 3 has perimeter \(3+3+3+3 = 12.\)

Any closed figure has a perimeter, such as the irregular polygon to the right with perimeter \(1+1+2+ 3= 7.\) Some shapes do not have a finite number of sides, so calculating the perimeter can be less straighforward. For example, the perimeter (or circumference) of a circle is \(2\pi r,\) where \(r\) is the radius of the circle. Fascinatingly, some shapes (such as certain fractals) have infinite perimeter, despite having a finite area.

What is the perimeter of a square with side length 6?

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Since a square has four sides of equal length and the perimeter of a square is the sum of its side lengths, the perimeter of the square is

\[ 6 + 6 + 6 + 6 = 24. \ _\square \]

You can also do \(6 \times 4\) on the count that it is a square.

What is the perimeter of a regular pentagon whose side lengths are all 6?

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Since a regular pentagon has five sides and all its side lengths are \(6\), the perimeter of the pentagon is

\[ 6 + 6 + 6 + 6 + 6= 5 \cdot 6 = 30 . \ _\square \]

What is the perimeter of a rectangle with width \(4\) and height \(7?\)

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Since a rectangle has two sides with equal width and two sides with equal height, and the perimeter of a rectangle is the sum of its side lengths, the perimeter of the rectangle is

\[ 4 + 4 + 7 + 7 = 2(4) + 2(7) = 8 + 14 = 22 . \ _\square \]

Find the perimeter of a triangle \(ABC\) with \(AB= 3\text{ cm}, BC=4\text{ cm},\) and \(\angle B= 90^{\circ}.\)

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First, use the Pythagorean theorem to find the length of \(AC:\)

\[AC^{2}=AB^{2}+BC^{2}=3^{2}+4^{2} \implies AC=5.\]

Then add \(AB,BC,CA\) to get the perimeter, which is

\[3+4+5 =12\text{ (cm)}.\ _\square\]

Suppose an ant walks along the boundary of a triangle with side lengths \(4, 6,\) and \(7\). If the ant starts at one vertex of the triangle and makes \(3\) complete trips around the triangle boundary to return to its starting point, what is the total distance traveled by the ant?

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Since the perimeter is the sum of edge lengths, the perimeter of the triangle is \(4 + 6 + 7 = 17\). Since the ant makes \(3\) complete trips around the triangle boundary, the total distance traveled by the ant is

\[ 17 \times 3 = 51. \ _\square \]

Find the perimeter of the given figure.

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We have

\[\begin{align}
\text{Perimeter} & = \text{Sum of side lengths}\\ & = 9 + 8 + 7 + 3 + 4 + 5\\ & = 36.\ _\square \end{align}\]

The perimeter of a circle with radius \(r\) is \(2\pi r\). The perimeter of an arc of a circle subtending angle \(\theta\) (in degrees) at the center is \(2\pi r \times \frac {\theta}{360}\).

What is the perimeter (circumference) of a circle of diameter \(14\text{ cm}?\)

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We have

\[\begin{align}
\text{Radius} & = \dfrac{\text{Diameter}}{2}\\ & = \dfrac{14}{2}\\ & = 7\text{ (cm)}\\ \\ \text{Circumference} & = 2× \pi × r\\ & = 2 × \dfrac{22}{7} × 7\\ & = 44\text{ (cm)}.\ _\square \end{align}\]

Find the perimeter of a semicircle of radius \(21\) cm.

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We have

\[\begin{align}
\text{Perimeter of a semicircle} & = \left( \dfrac{1}{2} × 2 × \pi × r \right) + 2r\\ & = \left( \dfrac{1}{2} × 2× \dfrac{22}{7} × 21\right) + 2×21\\ & = 66 + 42\\ & = 106\text{ (cm)}.\ _\square \end{align}\]

We can calculate the perimeter of an enclosed figure if its graph on a Cartesian plane is a function in polar coordinates. Specifically, suppose the figure has the equation \(r = f(\theta)\) such that \(f\) is a continuous, nonnegative function for \(\theta \in [0, 2\pi]\) and \(f(0) = f(2\pi)\). Then its perimeter is given by the following formula:

\[P = \int_{0}^{2\pi} \sqrt{r^2 + \left(\dfrac{dr}{d\theta}\right)^2} d\theta.\]

Suppose we want to calculate the circumference of a circle with radius \(r\). Then \(r\) is a constant, and its derivative is \(0\). So the formula gives us

\[P = \int_{0}^{2\pi} \sqrt{r^2 + 0^2} d\theta = \int_{0}^{2\pi} r d\theta = r(2\pi-0) = 2\pi r.\]

Therefore the circumference is \(2\pi r.\)

Unfortunately, for other shapes, this formula usually leads to complicated integrals which often do not have a simple closed form.

In this section, we will learn about applications of the concept of perimeter in real life. They are mostly of fencing-type problems. Here is an example:

Mr. Antony has a rectangular garden of area \(630\text{ m}^2\) and length \(42\text{ m}.\) To save his garden from domestic animals grazing them, he wants to fence his whole garden. Find much Mr. Antony has to spend for fencing his field at the rate of $3 per meter?

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Given, the area of the rectangle \(A = 630\text{ m}^2\) and its length \(l = 42\text{ m},\) we know that

\[A = l \times b \implies b = \dfrac{A}{l} = \dfrac{630}{42} = 15\ (\text{m}).\]

Now, let's find perimeter: \(P = 2(l + b) = 2(42 + 15) = 114\ (\text{m}).\)

Since the cost of fencing per meter is $3, the total cost of fencing the whole feild is \( 3 \times P = 3 \times 114 = 342\) dollars. \(_\square\)