Figure Efgh On The Grid Below Represents A Trapezoidal Plate

Figure EFGH on the Grid Below Represents a Trapezoidal Plate – A Step‑by‑Step Guide to Its Geometry and Physical Properties

When a shape is drawn on a coordinate grid, every vertex can be read off as an ordered pair ((x, y)). In many engineering and physics problems, the figure EFGH shown on the grid below is used as a model for a trapezoidal plate—a flat, uniform‑thickness object whose two parallel sides lie on horizontal grid lines. Understanding how to extract the plate’s dimensions, area, centroid, and moment of inertia from the grid is essential for solving problems related to load distribution, bending stress, and rotational dynamics. The following sections walk you through each calculation, explain the underlying theory, and answer common questions that arise when working with this type of plate.

1. Reading the Coordinates from the Grid

The first step is to identify the exact positions of the four vertices. Suppose the grid spacing is 1 unit (the same in the x‑ and y‑directions). By counting squares from the origin (0,0) we obtain:

Why these numbers?

• Points E and F lie on the lower horizontal line (y = 1).
• Points H and G lie on the upper horizontal line (y = 5).
• The left‑hand side EH runs from (x = 2) to (x = 3); the right‑hand side FG runs from (x = 8) to (x = 7).

Because the top and bottom edges are parallel to the x‑axis, the figure is indeed a trapezoid (a quadrilateral with one pair of parallel sides).

2. Determining the Plate’s Basic Dimensions

2.1 Length of the Parallel Sides (Bases)

• Bottom base (b_1 = |x_F - x_E| = |8 - 2| = 6) units.
• Top base (b_2 = |x_G - x_H| = |7 - 3| = 4) units. ### 2.2 Height (Distance Between the Bases)

Since the bases are horizontal, the height is simply the vertical difference:

[h = y_{\text{top}} - y_{\text{bottom}} = 5 - 1 = 4 \text{ units}. ]

2.3 Slanted Side Lengths (Optional)

If needed for stress analysis, the lengths of the non‑parallel sides are found with the distance formula:

[ \begin{aligned} |EH| &= \sqrt{(3-2)^2 + (5-1)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12 \text{ units},\[4pt] |FG| &= \sqrt{(7-8)^2 + (5-1)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12 \text{ units}. \end{aligned} ]

Notice the symmetry: the left and right slanted edges are equal, which simplifies later centroid calculations.

3. Computing the Area of the Trapezoidal Plate

The area (A) of a trapezoid with bases (b_1, b_2) and height (h) is:

[ A = \frac{(b_1 + b_2)h}{2}. ]

Substituting the values:

[A = \frac{(6 + 4) \times 4}{2} = \frac{10 \times 4}{2} = 20 \text{ square units}. ]

If the plate has a uniform thickness (t) (say, (t = 0.01) m), the volume would be (V = A \times t). For a pure‑geometry problem, we keep the area as the primary measure.

4. Locating the Centroid (Center of Mass) For a homogeneous plate, the centroid coincides with the center of mass. The centroid ((\bar{x}, \bar{y})) of a trapezoid can be found using the formulas:

[ \bar{x} = \frac{b_2 + 2b_1}{3(b_1 + b_2)} , h_{\text{shift}} + x_{\text{left}}, \qquad \bar{y} = \frac{h}{3} \frac{2b_1 + b_2}{b_1 + b_2} + y_{\text{bottom}}, ]

where (x_{\text{left}}) is the x‑coordinate of the left‑hand vertical line that would align with the left base if the trapezoid were “sheared” into a rectangle, and (h_{\text{shift}}) is the horizontal offset between the midpoints of the two bases. A more intuitive method is to decompose the trapezoid into a rectangle and two right triangles, compute each part’s centroid, and then take a weighted average.

4.1 Decomposition Approach

1. Rectangle: width = top base (b_2 = 4), height = (h = 4).

   • Area (A_R = 4 \times 4 = 16).
   • Centroid: at ((x_H + b_2/2,; y_{\text{bottom}} + h/2) = (3 + 2,; 1 + 2) = (5, 3)).

2. Left Triangle: base = (b_1 - b_2 = 6 - 4 = 2), height = (h = 4).

   • Area (A_{L} = \frac{1}{2} \times 2 \times 4 = 4).
   • Centroid of a right triangle (measured from the right angle) lies at ((\frac{b}{3}, \frac{h}{3})) from the right‑angle corner.
   • Placing the right angle at point H (3,5) and extending left‑downward, the centroid is at ((3 - \frac{2}{3},; 5 - \frac{4}{3}) = (2.33,; 3.67)).

3. Right Triangle: identical to the left triangle (symmetry).

   • Area (A_{Rtri} = 4).
   • Centroid: from point G (7,5) extending right‑downward: ((7 + \frac{2}{3},; 5 - \frac{4}{3}) = (7.67,; 3.67)).

4.2 Weighted Average [

\bar{x} = \frac{A_R x_R + A_{L} x_{L} + A_{Rtri} x_{Rtri}}{A