Problem 36 - Thomas Calculus

Question

Find the areas of the parallelograms whose vertices are given in Exercises \(35-38 .\) $$ A(0,0), \quad B(7,3), \quad C(9,8), \quad D(2,5) $$

Step-by-Step Solution

Answer

The area of the parallelogram is 22.5 square units.

Step 1: Understand the Formula for the Area of a Parallelogram

The area of a parallelogram defined by vertices \(A(x_1, y_1), B(x_2, y_2), C(x_3, y_3), \text{ and } D(x_4, y_4)\) can be calculated using the formula: \[ Area = \frac{1}{2} |x_1(y_2-y_4) + x_2(y_4-y_3) + x_3(y_3-y_1) + x_4(y_1-y_2)|. \] We will use this formula to find the area of the given parallelogram.

Step 2: Assign the Coordinates to the Formula

Assign the given coordinates to the variables in the formula: \((x_1, y_1) = A(0,0), (x_2, y_2) = B(7,3), (x_3, y_3) = C(9,8), (x_4, y_4) = D(2,5)\). Now we have \(x_1 = 0, y_1 = 0, x_2 = 7, y_2 = 3, x_3 = 9, y_3 = 8, x_4 = 2, \text{ and } y_4 = 5\).

Step 3: Plug the Values into the Formula

Substitute the values into the area formula: \[ Area = \frac{1}{2} |0(3-5) + 7(5-8) + 9(8-0) + 2(0-3)|. \] Calculate each component: - First term: \(0(3-5) = 0\),- Second term: \(7(5-8) = 7(-3) = -21\),- Third term: \(9(8-0) = 72\),- Fourth term: \(2(0-3) = -6\).

Step 4: Simplify the Expression

Now sum the results of the components: \(0 + (-21) + 72 + (-6)\). Simplify the expression to get: \(-21 + 72 - 6 = 45\).

Step 5: Calculate the Final Area

Use the formula to find the area: \[ Area = \frac{1}{2} |45| = \frac{1}{2} \times 45 = 22.5. \] The area of the parallelogram is 22.5 square units.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, plays a crucial role in finding the area of a parallelogram when its vertices are defined using coordinates. This branch of geometry allows us to calculate distances and areas using formulas derived from the coordinates of points. In the context of the parallelogram, we make use of a specific area formula that involves plugging in the coordinates of the vertices directly. The formula for the area is derived from determinants, and it provides a simple way to compute the area based on the position of the points on a plane. The formula is given by:\[ Area = \frac{1}{2} |x_1(y_2-y_4) + x_2(y_4-y_3) + x_3(y_3-y_1) + x_4(y_1-y_2)|. \] This technique is not just limited to parallelograms; it can be extended to other polygons as well. The power of coordinate geometry lies in its ability to transform geometrical problems into algebraic equations, making calculations more straightforward and accessible. For students, mastering coordinate geometry is essential for solving a wide range of geometry problems efficiently.

Vector Calculus

Vector calculus is a field that extends the understanding of calculus to multiple dimensions and involves differentiating and integrating vector fields. In the context of finding the area of a parallelogram, vectors provide a useful way to understand and compute the properties of shapes. One approach using vectors is to define two vectors originating from a common vertex and pointing towards two other vertices. For example, with vertices \(A(0,0)\), \(B(7,3)\), \(C(9,8)\), and \(D(2,5)\), the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\) can be created as:\[ \overrightarrow{AB} = B - A = (7 - 0, 3 - 0) = (7, 3), \]\[ \overrightarrow{AD} = D - A = (2 - 0, 5 - 0) = (2, 5). \]The area of the parallelogram can then be found using the cross product of these vectors, which involves calculating the determinant of their components:\[ Area = | \overrightarrow{AB} \times \overrightarrow{AD} | = |7 \cdot 5 - 3 \cdot 2| = |35 - 6| = 29. \] Though our earlier method used a different formula, vector cross products can also provide a robust technique to solve such problems, highlighting the versatility of vector calculus in geometry.

Plane Geometry

Plane geometry, often called Euclidean geometry, deals with shapes such as squares, triangles, and parallelograms on a flat surface. Understanding the intrinsic properties of these shapes, such as parallel sides and equal angles, forms the basis for finding areas and other attributes. For parallelograms specifically, the base and height method is traditionally used to find the area. However, when given coordinates, the formula involving points directly is more convenient. Plane geometry concepts ensure the understanding of why this formula works. When calculating the area using coordinates, we implicitly consider how each point relates to the others on the plane, ensuring that the shape remains a true parallelogram. Additionally, knowing the properties of parallelograms, such as opposite sides being equal and parallel, can help visualize and confirm that the quadrilateral in question is indeed a parallelogram. Thus, combining plane geometry concepts with coordinate geometry and vector calculus can provide a well-rounded approach to analyzing and solving geometrical problems in various contexts.