Based on the Diagram, Can Point D Be the Centroid of Triangle Acf? Explain.

Mean ("boilerplate") position of all the points in a shape

In mathematics and physics, the centroid or geometric eye of a plane figure is the arithmetic mean position of all the points in the figure. Informally, information technology is the indicate at which a cutout of the shape (with uniformly distributed mass) could be perfectly balanced on the tip of a pin.^[1] The same definition extends to any object in northward-dimensional space.^[2]

While in geometry, the give-and-take barycenter is a synonym for centroid, in astrophysics and astronomy, the barycenter is the center of mass of two or more bodies that orbit 1 some other. In physics, the center of mass is the arithmetic mean of all points weighted by the local density or specific weight. If a physical object has uniform density, and then its heart of mass is the same as the centroid of its shape.

In geography, the centroid of a radial projection of a region of the Earth'south surface to ocean level is the region'due south geographical eye.

History [edit]

The term "centroid" is of recent coinage (1814).^{[ commendation needed ]} It is used as a substitute for the older terms "heart of gravity," and "heart of mass", when the purely geometrical aspects of that bespeak are to exist emphasized. The term is peculiar to the English. The French employ " centre de gravité " on most occasions, and others use terms of similar pregnant.

The center of gravity, as the name indicates, is a notion that arose in mechanics, most likely in connection with edifice activities. When, where, and by whom it was invented is not known, as information technology is a concept that likely occurred to many people individually with minor differences.

While Archimedes does non state that proposition explicitly, he makes indirect references to it, suggesting he was familiar with information technology. However, Jean-Étienne Montucla (1725–1799), the writer of the first history of mathematics (1758), declares categorically (vol. I, p. 463) that the center of gravity of solids is a subject Archimedes did not affect.

In 1802 Charles Bossut (1730–1813) published a two-volume Essai sur l'histoire générale des mathématiques. This volume was highly esteemed by his contemporaries, judging from the fact that inside two years after its publication it was already bachelor in translation in Italian (1802–03), English (1803), and German language (1804). Bossut credits Archimedes with having found the centroid of aeroplane figures, but has nothing to say about solids.^[iii]

While it is possible Euclid was still active in Alexandria during the childhood of Archimedes (287–212 BCE), it is sure that when Archimedes visited Alexandria, Euclid was no longer there. Thus Archimedes could not take learned the theorem that the medians of a triangle meet in a point—the center of gravity of the triangle—direct from Euclid, as this proposition is not in Euclid's Elements. The first explicit argument of this proffer is due to Heron of Alexandria (peradventure the get-go century CE) and occurs in his Mechanics. It may be added, in passing, that the proposition did not become common in the textbooks on airplane geometry until the nineteenth century.

Properties [edit]

The geometric centroid of a convex object e'er lies in the object. A non-convex object might have a centroid that is outside the figure itself. The centroid of a ring or a basin, for example, lies in the object's primal void.

If the centroid is divers, it is a fixed indicate of all isometries in its symmetry group. In particular, the geometric centroid of an object lies in the intersection of all its hyperplanes of symmetry. The centroid of many figures (regular polygon, regular polyhedron, cylinder, rectangle, rhombus, circle, sphere, ellipse, ellipsoid, superellipse, superellipsoid, etc.) tin be determined by this principle lonely.

In detail, the centroid of a parallelogram is the meeting point of its two diagonals. This is not true of other quadrilaterals.

For the same reason, the centroid of an object with translational symmetry is undefined (or lies exterior the enclosing space), considering a translation has no stock-still bespeak.

Examples [edit]

The centroid of a triangle is the intersection of the three medians of the triangle (each median connecting a vertex with the midpoint of the opposite side).^[4]

For other properties of a triangle'southward centroid, meet beneath.

Locating [edit]

Plumb line method [edit]

The centroid of a uniformly dense planar lamina, such as in effigy (a) below, may be determined experimentally by using a plumbline and a pin to observe the collocated middle of mass of a thin trunk of uniform density having the same shape. The body is held past the pivot, inserted at a point, off the presumed centroid in such a way that it can freely rotate around the pin; the plumb line is and then dropped from the pin (figure b). The position of the plumbline is traced on the surface, and the procedure is repeated with the pin inserted at whatsoever unlike point (or a number of points) off the centroid of the object. The unique intersection betoken of these lines will be the centroid (figure c). Provided that the body is of uniform density, all lines made this manner will include the centroid, and all lines will cross at exactly the same place.

This method tin can be extended (in theory) to concave shapes where the centroid may lie outside the shape, and virtually to solids (again, of uniform density), where the centroid may prevarication within the body. The (virtual) positions of the plumb lines need to be recorded by ways other than by cartoon them along the shape.

Balancing method [edit]

For convex two-dimensional shapes, the centroid tin can be found by balancing the shape on a smaller shape, such as the top of a narrow cylinder. The centroid occurs somewhere inside the range of contact between the two shapes (and exactly at the point where the shape would residue on a pin). In principle, progressively narrower cylinders can be used to notice the centroid to capricious precision. In practice air currents make this infeasible. All the same, by marking the overlap range from multiple balances, one can achieve a considerable level of accuracy.

Of a finite fix of points [edit]

The centroid of a finite set of ${\displaystyle one thousand}$ points ${\displaystyle \mathbf {10} _{ane},\mathbf {x} _{2},\ldots ,\mathbf {x} _{k}}$ in ${\displaystyle \mathbb {R} ^{due north}}$ is^[2]

$${\displaystyle \mathbf {C} ={\frac {\mathbf {x} _{i}+\mathbf {10} _{2}+\cdots +\mathbf {ten} _{one thousand}}{k}}.}$$

This betoken minimizes the sum of squared Euclidean distances between itself and each point in the set.

By geometric decomposition [edit]

The centroid of a airplane figure ${\displaystyle X}$ can exist computed by dividing it into a finite number of simpler figures ${\displaystyle X_{1},X_{2},\dots ,X_{n}}$ , calculating the centroid ${\displaystyle C_{i}}$ and area ${\displaystyle A_{i}}$ of each part, and and so computing

$${\displaystyle C_{x}={\frac {\sum _{i}{C_{i}}_{ten}A_{i}}{\sum _{i}A_{i}}},C_{y}={\frac {\sum _{i}{C_{i}}_{y}A_{i}}{\sum _{i}A_{i}}}}$$

Holes in the figure ${\displaystyle Ten}$ , overlaps betwixt the parts, or parts that extend outside the figure tin all exist handled using negative areas ${\displaystyle A_{i}}$ . Namely, the measures ${\displaystyle A_{i}}$ should exist taken with positive and negative signs in such a way that the sum of the signs of ${\displaystyle A_{i}}$ for all parts that enclose a given signal ${\displaystyle p}$ is 1 if ${\displaystyle p}$ belongs to ${\displaystyle Ten}$ , and 0 otherwise.

For example, the figure below (a) is hands divided into a square and a triangle, both with positive surface area; and a circular hole, with negative area (b).

(a) 2D Object

(b) Object described using simpler elements

(c) Centroids of elements of the object

The centroid of each function can be constitute in any list of centroids of uncomplicated shapes (c). Then the centroid of the figure is the weighted average of the 3 points. The horizontal position of the centroid, from the left edge of the effigy is

$${\displaystyle x={\frac {5\times 10^{two}+thirteen.33\times {\frac {1}{ii}}10^{2}-3\times \pi 2.5^{ii}}{x^{two}+{\frac {ane}{2}}10^{2}-\pi ii.5^{2}}}\approx 8.5{\text{ units}}.}$$

The vertical position of the centroid is found in the same mode.

The same formula holds for whatever 3-dimensional objects, except that each ${\displaystyle A_{i}}$ should be the volume of ${\displaystyle X_{i}}$ , rather than its surface area. It likewise holds for any subset of ${\displaystyle \mathbb {R} ^{d}}$ , for any dimension ${\displaystyle d}$ , with the areas replaced past the ${\displaystyle d}$ -dimensional measures of the parts.

By integral formula [edit]

The centroid of a subset X of ${\displaystyle \mathbb {R} ^{n}}$ can also be computed by the integral

$${\displaystyle C={\frac {\int xg(x)\;dx}{\int k(x)\;dx}}}$$

where the integrals are taken over the whole space ${\displaystyle \mathbb {R} ^{n}}$ , and k is the characteristic part of the subset, which is 1 inside 10 and 0 outside it.^[five] Note that the denominator is merely the measure of the set X. This formula cannot be applied if the fix X has zero measure, or if either integral diverges.

Another formula for the centroid is

$${\displaystyle C_{k}={\frac {\int zS_{k}(z)\;dz}{\int g(x)\;dx}}}$$

where C _k is the thouth coordinate of C, and S _k (z) is the measure of the intersection of X with the hyperplane divers past the equation ten _m = z. Over again, the denominator is simply the measure out of X.

For a plane figure, in item, the barycenter coordinates are

$${\displaystyle C_{\mathrm {x} }={\frac {\int xS_{\mathrm {y} }(10)\;dx}{A}}}$$

$${\displaystyle C_{\mathrm {y} }={\frac {\int yS_{\mathrm {10} }(y)\;dy}{A}}}$$

where A is the surface area of the figure X; Southward _y(ten) is the length of the intersection of X with the vertical line at abscissa x; and S _x(y) is the analogous quantity for the swapped axes.

Of a divisional region [edit]

The centroid ${\displaystyle ({\bar {x}},\;{\bar {y}})}$ of a region bounded by the graphs of the continuous functions ${\displaystyle f}$ and ${\displaystyle chiliad}$ such that ${\displaystyle f(10)\geq grand(x)}$ on the interval ${\displaystyle [a,b]}$ , ${\displaystyle a\leq x\leq b}$ , is given by^{[v] [6]}

$${\displaystyle {\bar {x}}={\frac {one}{A}}\int _{a}^{b}x[f(10)-1000(10)]\;dx}$$

$${\displaystyle {\bar {y}}={\frac {ane}{A}}\int _{a}^{b}\left[{\frac {f(10)+thousand(x)}{2}}\correct][f(x)-m(ten)]\;dx,}$$

where ${\displaystyle A}$ is the area of the region (given by ${\textstyle \int _{a}^{b}\left[f(x)-1000(10)\right]dx}$ ).^{[7] [8]}

Of an L-shaped object [edit]

This is a method of determining the centroid of an L-shaped object.

1. Divide the shape into two rectangles, as shown in fig 2. Find the centroids of these two rectangles by cartoon the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line AB.
2. Split the shape into 2 other rectangles, as shown in fig 3. Find the centroids of these two rectangles past drawing the diagonals. Draw a line joining the centroids. The centroid of the L-shape must lie on this line CD.
3. As the centroid of the shape must lie along AB and also along CD, it must be at the intersection of these ii lines, at O. The point O might lie inside or outside the L-shaped object.

Of a triangle [edit]

The centroid of a triangle is the signal of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side).^[4] The centroid divides each of the medians in the ratio two:1, which is to say information technology is located ⅓ of the altitude from each side to the opposite vertex (run into figures at right).^{[9] [10]} Its Cartesian coordinates are the means of the coordinates of the three vertices. That is, if the three vertices are ${\displaystyle L=(x_{50},y_{L}),}$ ${\displaystyle Chiliad=(x_{M},y_{M}),}$ and ${\displaystyle N=(x_{N},y_{N}),}$ then the centroid (denoted C hither but most commonly denoted G in triangle geometry) is

$${\displaystyle C={\frac {1}{iii}}(L+Grand+Due north)=\left({\frac {i}{3}}(x_{L}+x_{M}+x_{N}),\;\;{\frac {one}{iii}}(y_{L}+y_{Grand}+y_{N})\right).}$$

The centroid is therefore at ${\displaystyle {\tfrac {i}{3}}:{\tfrac {1}{3}}:{\tfrac {i}{3}}}$ in barycentric coordinates.

In trilinear coordinates the centroid can exist expressed in any of these equivalent ways in terms of the side lengths a, b, c and vertex angles L, M, N:^[11]

$${\displaystyle {\brainstorm{aligned}C&={\frac {1}{a}}:{\frac {1}{b}}:{\frac {ane}{c}}=bc:ca:ab=\csc Fifty:\csc M:\csc N\\[6pt]&=\cos L+\cos M\cdot \cos N:\cos M+\cos North\cdot \cos L:\cos N+\cos 50\cdot \cos M\\[6pt]&=\sec L+\sec One thousand\cdot \sec Due north:\sec Chiliad+\sec N\cdot \sec L:\sec N+\sec L\cdot \sec M.\end{aligned}}}$$

The centroid is also the physical center of mass if the triangle is made from a uniform sheet of cloth; or if all the mass is concentrated at the three vertices, and evenly divided among them. On the other paw, if the mass is distributed along the triangle's perimeter, with uniform linear density, then the eye of mass lies at the Spieker center (the incenter of the medial triangle), which does non (in general) coincide with the geometric centroid of the full triangle.

The surface area of the triangle is ane.v times the length of any side times the perpendicular distance from the side to the centroid.^[12]

A triangle's centroid lies on its Euler line between its orthocenter H and its circumcenter O, exactly twice as shut to the latter every bit to the erstwhile:^{[13] [fourteen]}

$${\displaystyle {\overline {CH}}=2{\overline {CO}}.}$$

In addition, for the incenter I and nine-point middle N, nosotros have

$${\displaystyle {\begin{aligned}{\overline {CH}}&=4{\overline {CN}}\\[5pt]{\overline {CO}}&=2{\overline {CN}}\\[5pt]{\overline {IC}}&<{\overline {HC}}\\[5pt]{\overline {IH}}&<{\overline {HC}}\\[5pt]{\overline {IC}}&<{\overline {IO}}\end{aligned}}}$$

If G is the centroid of the triangle ABC, so:

$${\displaystyle ({\text{Area of }}\triangle \mathrm {ABG} )=({\text{Area of }}\triangle \mathrm {ACG} )=({\text{Area of }}\triangle \mathrm {BCG} )={\frac {1}{3}}({\text{Area of }}\triangle \mathrm {ABC} )}$$

The isogonal cohabit of a triangle's centroid is its symmedian point.

Whatever of the three medians through the centroid divides the triangle'southward area in half. This is non truthful for other lines through the centroid; the greatest divergence from the equal-area division occurs when a line through the centroid is parallel to a side of the triangle, creating a smaller triangle and a trapezoid; in this case the trapezoid's area is 5/nine that of the original triangle.^[15]

Let P be any indicate in the plane of a triangle with vertices A, B, and C and centroid M. Then the sum of the squared distances of P from the 3 vertices exceeds the sum of the squared distances of the centroid G from the vertices by 3 times the squared altitude between P and Yard:^[16]

$${\displaystyle PA^{2}+Atomic number 82^{2}+PC^{2}=GA^{2}+GB^{two}+GC^{ii}+3PG^{2}.}$$

The sum of the squares of the triangle's sides equals three times the sum of the squared distances of the centroid from the vertices:^[xvi]

$${\displaystyle AB^{2}+BC^{2}+CA^{2}=3(GA^{two}+GB^{2}+GC^{2}).}$$

A triangle'southward centroid is the point that maximizes the production of the directed distances of a point from the triangle'south sidelines.^[17]

Let ABC be a triangle, let Thou be its centroid, and let D, E, and F be the midpoints of BC, CA, and AB, respectively. For whatever point P in the plane of ABC then^[18]

$${\displaystyle PA+Atomic number 82+PC\leq two(PD+PE+PF)+3PG.}$$

Of a polygon [edit]

The centroid of a not-self-intersecting airtight polygon defined by due north vertices (x ₀,y ₀), (ten ₁,y ₁), ..., (x _n−one,y _n−1) is the point (C _x, C _y),^[19] where

$${\displaystyle C_{\mathrm {ten} }={\frac {i}{6A}}\sum _{i=0}^{northward-i}(x_{i}+x_{i+i})(x_{i}\ y_{i+1}-x_{i+1}\ y_{i}),}$$

and

$${\displaystyle C_{\mathrm {y} }={\frac {ane}{6A}}\sum _{i=0}^{n-i}(y_{i}+y_{i+one})(x_{i}\ y_{i+i}-x_{i+1}\ y_{i}),}$$

and where A is the polygon's signed area,^[19] equally described by the shoelace formula:

$${\displaystyle A={\frac {1}{2}}\sum _{i=0}^{northward-1}(x_{i}\ y_{i+i}-x_{i+1}\ y_{i}).}$$

In these formulae, the vertices are assumed to be numbered in order of their occurrence forth the polygon's perimeter; furthermore, the vertex ( 10 _n , y _northward ) is assumed to be the same every bit (x ₀, y ₀), meaning ${\displaystyle i+1}$ on the final case must loop around to ${\displaystyle i=0}$ . (If the points are numbered in clockwise gild, the area A, computed every bit above, will exist negative; notwithstanding, the centroid coordinates will be right even in this case.)

Of a cone or pyramid [edit]

The centroid of a cone or pyramid is located on the line segment that connects the noon to the centroid of the base. For a solid cone or pyramid, the centroid is 1/4 the distance from the base of operations to the apex. For a cone or pyramid that is simply a crush (hollow) with no base, the centroid is 1/3 the distance from the base plane to the apex.

Of a tetrahedron and n-dimensional simplex [edit]

A tetrahedron is an object in three-dimensional space having four triangles equally its faces. A line segment joining a vertex of a tetrahedron with the centroid of the contrary confront is called a median, and a line segment joining the midpoints of two opposite edges is called a bimedian. Hence in that location are four medians and three bimedians. These vii line segments all meet at the centroid of the tetrahedron.^[twenty] The medians are divided by the centroid in the ratio 3:ane. The centroid of a tetrahedron is the midpoint between its Monge bespeak and circumcenter (middle of the circumscribed sphere). These three points ascertain the Euler line of the tetrahedron that is coordinating to the Euler line of a triangle.

These results generalize to any northward-dimensional simplex in the post-obit way. If the set of vertices of a simplex is ${\displaystyle {v_{0},\ldots ,v_{n}}}$ , then because the vertices as vectors, the centroid is

$${\displaystyle C={\frac {i}{n+1}}\sum _{i=0}^{n}v_{i}.}$$

The geometric centroid coincides with the center of mass if the mass is uniformly distributed over the whole simplex, or concentrated at the vertices as n+one equal masses.

Of a hemisphere [edit]

The centroid of a solid hemisphere (i.due east. half of a solid ball) divides the line segment connecting the sphere's center to the hemisphere's pole in the ratio 3:5 (i.e. it lies iii/8 of the manner from the eye to the pole). The centroid of a hollow hemisphere (i.e. half of a hollow sphere) divides the line segment connecting the sphere's centre to the hemisphere'southward pole in half.

Run across also [edit]

• Chebyshev center
• Fréchet mean
• chiliad-means algorithm
• List of centroids
• Locating the center of mass
• Medoid
• Pappus'due south centroid theorem
• Spectral centroid
• Triangle center

Notes [edit]

1. ^ Protter & Morrey, Jr. (1970, p. 521)
2. ^ ^{a b} Protter & Morrey, Jr. (1970, p. 520)
3. ^ Court, Nathan Altshiller (1960). "Notes on the centroid". The Mathematics Teacher. 53 (i): 33–35. doi:10.5951/MT.53.1.0033. JSTOR 27956057.
4. ^ ^{a b} Altshiller-Courtroom (1925, p. 66)
5. ^ ^{a b} Protter & Morrey, Jr. (1970, p. 526)
6. ^ Protter & Morrey, Jr. (1970, p. 527)
7. ^ Protter & Morrey, Jr. (1970, p. 528)
8. ^ Larson (1998, pp. 458–460)
9. ^ Altshiller-Court (1925, p. 65)
10. ^ Kay (1969, p. 184)
11. ^ Clark Kimberling's Encyclopedia of Triangles "Encyclopedia of Triangle Centers". Archived from the original on 2012-04-19. Retrieved 2012-06-02 .
12. ^ Johnson (2007, p. 173)
13. ^ Altshiller-Court (1925, p. 101)
14. ^ Kay (1969, pp. 18, 189, 225–226)
15. ^ Bottomley, Henry. "Medians and Expanse Bisectors of a Triangle". Retrieved 27 September 2013.
16. ^ ^{a b} Altshiller-Courtroom (1925, pp. 70–71)
17. ^ Kimberling, Clark (201). "Trilinear distance inequalities for the symmedian bespeak, the centroid, and other triangle centers". Forum Geometricorum. 10: 135–139.
18. ^ Gerald A. Edgar, Daniel H. Ullman & Douglas B. W (2018) Issues and Solutions, The American Mathematical Monthly, 125:one, 81-89, DOI: 10.1080/00029890.2018.1397465
19. ^ ^{a b} Bourke (1997)
20. ^ Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong Academy Press, 1994, pp. 53–54

References [edit]

• Altshiller-Courtroom, Nathan (1925), College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circumvolve (2nd ed.), New York: Barnes & Noble, LCCN 52013504
• Bourke, Paul (July 1997). "Computing the area and centroid of a polygon".
• Johnson, Roger A. (2007), Advanced Euclidean Geometry, Dover
• Kay, David C. (1969), Higher Geometry, New York: Holt, Rinehart and Winston, LCCN 69012075
• Larson, Roland E.; Hostetler, Robert P.; Edwards, Bruce H. (1998), Calculus of a Unmarried Variable (6th ed.), Houghton Mifflin Company
• Protter, Murray H.; Morrey, Jr., Charles B. (1970), College Calculus with Analytic Geometry (2nd ed.), Reading: Addison-Wesley, LCCN 76087042

External links [edit]

• Weisstein, Eric Westward. "Geometric Centroid". MathWorld.
• Encyclopedia of Triangle Centers by Clark Kimberling. The centroid is indexed as 10(2).
• Feature Property of Centroid at cut-the-knot
• Interactive animations showing Centroid of a triangle and Centroid construction with compass and straightedge
• Experimentally finding the medians and centroid of a triangle at Dynamic Geometry Sketches, an interactive dynamic geometry sketch using the gravity simulator of Cinderella.

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Source: https://en.wikipedia.org/wiki/Centroid

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