How to Find the Centroid of a Trapezoid

A trapezoid (also termed trapezium in some regions) is a quadrilateral with exactly one pair of parallel sides. The centroid of a trapezoid is the point at which the entire area of the figure is considered to be concentrated if the shape is of uniform density. The centroid location is fundamental in both pure geometry and applications such as mechanics.

Coordinate Geometry Definition of a Trapezoid and Its Centroid Position

Let a trapezoid have parallel sides of lengths $a$ and $b$ $(a \neq b)$, and let the perpendicular distance (height) between these sides be $h$. Denote the two parallel sides by $AB$ and $CD$, such that $AB = a$, $CD = b$, and $AB \parallel CD$. For precise geometric treatment, position the trapezoid in the Cartesian plane so that side $CD$ lies along the $x$-axis from $(0, 0)$ to $(b, 0)$ and side $AB$ is parallel to $CD$ at a height $h$ above the $x$-axis. Let side $AB$ extend from $(k, h)$ to $(k + a, h)$, where $k$ is a non-negative horizontal shift ensuring parallelism. The vertices of the trapezoid are thus: \[ C = (0, 0), \quad D = (b, 0), \quad A = (k, h), \quad B = (k + a, h). \]

Explicit Stepwise Derivation of the Centroid Coordinates

The centroid $(G_x, G_y)$ of a plane figure of uniform density can be determined by evaluating the coordinates as follows: \[ G_x = \frac{1}{A} \iint_{R} x\, dA, \qquad G_y = \frac{1}{A} \iint_{R} y\, dA, \] where $A$ is the area of the trapezoid and $R$ is its region in the $xy$-plane.

The area $A$ of the trapezoid is: \[ A = \frac{1}{2} (a+b)h. \]

For an arbitrary horizontal strip at height $y$ such that $0 \le y \le h$, the length of the strip, $L(y)$, is derived as follows. The bottom base ($CD$) lies between $x=0$ and $x=b$ at $y=0$. The top base ($AB$) lies between $x=k$ and $x=k+a$ at $y=h$. The $x$-coordinates of intersection at any height $y$ can be found using similar triangles: \[ \text{Left side:} \quad x_L(y) = k \cdot \frac{y}{h} \] \[ \text{Right side:} \quad x_R(y) = b + (a - b + k) \cdot \frac{y}{h} \] But with $k=0$ (no horizontal shift, i.e., $A$ directly above $C$ and $B$ directly above $D$), for the standard trapezoid: \[ x_L(y) = 0 \] \[ x_R(y) = b + \frac{(a-b)y}{h} \] Thus, the width of the horizontal strip at height $y$ is: \[ L(y) = x_R(y) - x_L(y) = b + \frac{(a-b)y}{h}. \]

The area element is $dA = L(y)dy$. The $y$-coordinate of the centroid is: \[ G_y = \frac{1}{A} \int_{0}^{h} y \cdot L(y) \, dy \] \[ = \frac{1}{A} \int_{0}^{h} y\left(b + \frac{(a-b)y}{h}\right) dy \] \[ = \frac{1}{A} \left[ b \int_{0}^{h} y\,dy + \frac{a-b}{h} \int_{0}^{h} y^2\,dy \right] \] \[ = \frac{1}{A} \left[ b \cdot \frac{y^2}{2} \Big|_{0}^{h} + \frac{a-b}{h} \cdot \frac{y^3}{3}\Big|_{0}^{h} \right] \] \[ = \frac{1}{A} \left[ b \cdot \frac{h^2}{2} + \frac{a-b}{h} \cdot \frac{h^3}{3} \right] \] \[ = \frac{1}{A} \left[ \frac{b h^2}{2} + \frac{(a-b) h^2}{3} \right] \] \[ = \frac{1}{A} \left[ \frac{3b h^2 + 2(a-b)h^2}{6} \right] \] \[ = \frac{1}{A} \cdot \frac{(3b + 2a - 2b) h^2}{6} \] \[ = \frac{1}{A} \cdot \frac{(2a + b) h^2}{6} \] Since $A = \frac{1}{2}(a+b)h$, \[ G_y = \frac{(2a + b) h^2}{6 \cdot \frac{1}{2}(a+b)h} \] \[ = \frac{(2a + b) h^2}{3 (a+b) h} \] \[ = \frac{(2a + b) h}{3(a+b)} \]

Result: The centroid, measured vertically from the base of length $b$, is located at a distance \[ G_y = \frac{b + 2a}{3(a+b)}h \] from the side of length $b$, and at a distance \[ h - G_y = h - \frac{b + 2a}{3(a + b)}h = \frac{2b + a}{3(a+b)}h \] from the side of length $a$.

For trapezoids with the parallel sides horizontal and no horizontal shift (as in competitive examples), the centroid's $x$-coordinate is at the midpoint of the parallel sides: \[ G_x = \frac{1}{2}b \] when measured from $x = 0$ to $x = b$ at $y = 0$ and $x = a$ at $y = h$.

Standard Formula for the Centroid of a Trapezoid

The most commonly used result, validated above, is: \[ \boxed{ G_y = \frac{b + 2a}{3(a+b)}h } \] where $G_y$ is the centroid's distance from the base of length $b$ (the longer or shorter parallel side, depending on orientation), $h$ is the perpendicular height, $a$ and $b$ are the lengths of the parallel sides. This corresponds with the standard for both the Area Of Trapezoid Formula and basic centroid calculations in JEE-level geometry.

Worked Example: Finding the Centroid of a Trapezoid with Given Dimensions

Given: A trapezoid with parallel sides of length $a = 6$ cm and $b = 8$ cm, and height $h = 5$ cm.

Substitution into the formula: \[ G_y = \frac{b + 2a}{3(a + b)}h = \frac{8 + 2 \times 6}{3(6 + 8)} \times 5 \] \[ = \frac{8 + 12}{3 \times 14} \times 5 \] \[ = \frac{20}{42} \times 5 \] \[ = \frac{100}{42} \] \[ = 2.38 \text{ cm (rounded to two decimal places)} \]

Result: The centroid is located $2.38$ cm above the base of length $8$ cm.

Worked Example: Centroid Location Relative to the Shorter Parallel Side

Given: Parallel sides are $a = 4$ cm, $b = 8$ cm, and height $h = 4.5$ cm.

Substitution into the formula: \[ G_y = \frac{8 + 2 \times 4}{3 (4 + 8)} \times 4.5 \] \[ = \frac{8 + 8}{3 \times 12} \times 4.5 \] \[ = \frac{16}{36} \times 4.5 \] \[ = \frac{72}{36} \] \[ = 2 \text{ cm} \]

Result: The centroid is located $2$ cm above the base of length $8$ cm.

Special Cases: Isosceles and Right Trapezoids

For an isosceles trapezoid, the centroid's vertical position remains $G_y = \frac{b + 2a}{3(a+b)}h$. For a right trapezoid (one non-parallel side perpendicular to the bases), the formula remains unchanged, provided the height is measured perpendicularly between the parallel sides. No formula modification is necessary, since only the parallel side lengths and height affect the vertical centroid position. For further formula-based queries, refer to Area Formula For Quadrilateral.

Centroid Position Using Vertex Coordinates

If all four vertices $(x_1, y_1),\, (x_2, y_2),\, (x_3, y_3),\, (x_4, y_4)$ are given, the centroid $G$ can also be found by decomposing the trapezoid into two triangles, calculating their centroids and areas, and taking the area-weighted mean of their centroids. The explicit step-by-step method is as follows:

1. Assign the vertices in order and divide the trapezoid into two triangles along a diagonal.
2. Compute the centroid $(G_{x_i}, G_{y_i})$ and area $A_i$ of each triangle using standard triangle centroid and area formulas.
3. The centroid of the whole figure is \[ G_x = \frac{A_1 G_{x_1} + A_2 G_{x_2}}{A_1 + A_2}, \quad G_y = \frac{A_1 G_{y_1} + A_2 G_{y_2}}{A_1 + A_2}. \]

Related Concepts in Area and Centroid Calculations

The trapezoid centroid formula is a special case in the broader class of centroid calculations for polygons. For additional formulas related to quadrilaterals and polygons, see Area And Perimeter Formula and Difference Between Area And Surface Area.

Summary of the Center of Mass of a Trapezoid

The centroid of a trapezoid depends solely on the lengths of the parallel sides and the perpendicular distance (height) separating them. The explicit formula remains valid regardless of side slant, as long as the bases and height are known. This centroid formula is extensively used in geometry, mechanics, and competitive exams to determine the axis of symmetry, balance points, and in the calculation of moments. For deeper insights into geometric centroids, consult Difference Between Area And Volume.