Semicircle how to calculate perimeter, area, centroid, exercises

The semicircle is a plane figure delimited by a diameter of the circumference and one of the two flat circular arcs determined by said diameter.

In this way, a semicircle is bordered by a semicircumference, which consists of a flat circular arc and a straight segment that joins the ends of the flat circular arc. The semicircle encompasses the semicircle and all points inside it..

We can see this in figure 1, which shows a semicircle of radius R, whose measure is half that of diameter AB. Note that unlike a circle, in which there are infinite diameters, in the semicircle there is only one diameter.

The semicircle is a geometric figure with many uses in architecture and design, as we see in the following image:

Article index

• 1 Elements and measures of a semicircle
  • 1.1 Perimeter of a semicircle
  • 1.2 Area of ​​a semicircle
  • 1.3 Centroid of a semicircle
  • 1.4 Moment of inertia of a semicircle
  • 1.5 Inscribed angle
• 2 Solved exercises
  • 2.1 Exercise 1 
  • 2.2 Exercise 2
  • 2.3 Exercise 3
  • 2.4 Exercise 4
  • 2.5 Exercise 5
• 3 References

Elements and measures of a semicircle

The elements of a semicircle are:

1.- The flat circular arc A⌒B

2.- The segment [AB] 

3.- The points inside the semicircle composed of the arc A⌒B and the segment [AB].

Perimeter of a semicircle

The perimeter is the sum of the contour of the arc plus that of the straight segment, therefore:

Perimeter = arc length A⌒B + segment length [AB]

In the case of a semicircle of radius R, its perimeter P will be given by the formula:

P = π⋅R + 2⋅R = (π + 2) ⋅R

The first term is half the perimeter of a circle of radius R, while the second is the length of the diameter, which is twice the radius..

Area of ​​a semicircle

Since a semicircle is one of the plane angular sectors that remain when drawing a diameter through the circumference, its area A will be half the area of ​​the circle that contains the semicircle of radius R:

A = (π⋅R^two) / 2 = ½ π⋅R^two

Centroid of a semicircle

The centroid of a semicircle is on its axis of symmetry at a height measured from its diameter of 4 / (3π) times the radius R.

This corresponds to approximately 0.424⋅R, measured from the center of the semicircle and on its axis of symmetry, as shown in figure 3.

Moment of inertia of a semicircle

The moment of inertia of a plane figure with respect to an axis, for example x-axis, is defined as:

The integral of the square of the distance of the points belonging to the figure to the axis, the integration differential being an infinitesimal element of area, taken at the position of each point. 

Figure 4 shows the definition of the moment of inertia I_x of the semicircle of radius R, with respect to the X axis that passes through its diagonal:

The moment of inertia about the x axis is given by:

I_x = (π⋅R⁴) / 8

And the moment of inertia with respect to the axis of symmetry y is:

Iy = (π⋅R⁴) / 8

It is noted that both moments of inertia coincide in their formula, but it is important to note that they refer to different axes.

Inscribed angle

The angle inscribed in the semicircle is always 90º. Regardless of where the point is taken on the arc, the angle formed between sides AB and BC of the figure is always right..

Solved exercises

Exercise 1 

Determine the perimeter of a semicircle of radius 10 cm.

Solution

Remember that the perimeter as a function of the radius is given by the formula that we saw earlier:

P = (2 + π) ⋅R

P = (2 + 3.14) ⋅ 10 cm = 5.14 ⋅ 10 cm = 51.4 cm.

Exercise 2

Find the area of ​​a semicircle with radius 10 cm.

Solution

The formula for the area of ​​a semicircle is:

A = ½ π⋅R^two = ½ π⋅ (10cm)^two = 50π cm^two = 50 x 3.14 cm^two = 157 cm^two.

Exercise 3

Determine the height h of the centroid of a semicircle of radius R = 10 cm measured from its base, the diameter of the semicircle being the same. 

Solution

The centroid is the equilibrium point of the semicircle and its position is on the axis of symmetry at a height h from the base (diameter of the semicircle):

h = (4⋅R) / (3π) = (4⋅10 cm) / (3 x 3.14) = 4.246 cm

Exercise 4

Find the moment of inertia of a semicircle with respect to the axis that coincides with its diameter, knowing that the semicircle is made of a thin sheet. Its radius is 10 cm and its mass is 100 grams.

Solution

The formula that gives the moment of inertia of the semicircle is:

I_x = (π⋅R⁴) / 8

But since the problem tells us that it is a material semicircle, then the previous relation must be multiplied by the surface density of mass of the semicircle, which will be denoted by σ.

I_x = σ (π⋅R⁴) / 8

We then proceed to determine σ, which is nothing other than the mass of the semicircle divided by its area.

The area was determined in exercise 2 and the result was 157 cm^two. Then the surface density of this semicircle will be:

σ = 100 grams / 157 cm^two = 0.637 g / cm^two

Then the moment of inertia with respect to the diameter will be calculated like this:

I_x = (0.637 g / cm^two) [3.1416 ⋅ (10cm)⁴] / 8

Resulting:

I_x = 2502 g⋅cm^two

Exercise 5

Determine the moment of inertia of a semicircle of radius 10 cm made of a material sheet with a surface density of 0.637 g / cm^two along an axis that passes through its centroid and is parallel to its diameter.

Solution

To solve this exercise, it is necessary to remember Steiner's theorem on moments of inertia of parallel axes, which says:

The moment of inertia I with respect to an axis that is at a distance h from the centroid is equal to the sum of the moment of inertia I_c with respect to an axis that passes through the centroid and is parallel to the first one plus the product of the mass times the square of the separation of the two axes.

I = I_c + M h^two

In our case, I is known which is the moment of inertia with respect to the diameter, which was already calculated in exercise 4. The separation h between the diameter and the centroid is also known, which was calculated in exercise 3.

We only have to clear Ic:

I_c = I - M h^two

I_c = 2502 g⋅cm^two - 100g ⋅ (4.246 cm)^two giving as a result that the moment of inertia through an axis parallel to the diameter and passing through the centroid is: 

I_c = 699.15 g⋅cm^two

References

1. Alexander, D. 2013. Geometry. 5th. Edition. Cengage Learning.
2. Math Open Reference. Semicircle. Recovered from: mathopenref.com.
3. Universe Formulas Semicircle. Recovered from: universoformulas.com.
4. Universe Formulas. Area of ​​a Semicircle. Recovered from: universoformulas.com.
5. Wikipedia. Semicircle. Recovered from: en.wikipedia.com.