Geometry Problem Solving: Finding the Area of Rectangle ABCD

Introduction

This article presents a detailed exploration of a geometry problem, focusing on finding the area of a rectangle ABCD when certain dimensions are provided. We will use the Pythagorean Theorem to solve the problem, providing clear steps and explanations for each part of the solution.

Problem Statement

In rectangle ABCD, E is the midpoint of AB. Given DB 2√10 and DE √13, what is the area of ABCD?

Solution

Let's delve into the solution step-by-step.

Step 1: Identifying Relevant Triangles

Consider the triangles in our problem: △ADE and △ADB.

Step 2: Applying the Pythagorean Theorem to △ADE

For △ADE, we have :

[(x/2)2 y2 (√13)2]

[(x/2)2 y2 13]

[(x2/4) y2 13]

From this equation, we can find x2 in terms of y:

x2 4y2 52

Step 3: Applying the Pythagorean Theorem to △ADB

For △ADB:

x2 y2 (2√10)2

x2 y2 40

Step 4: Solving for y

To find y, we subtract the second equation from the first:

(x2 4y2) - (x2 y2) 52 - 40

3y2 12

y2 4

y ±2

Since the width of a rectangle cannot be negative, y 2.

Step 5: Solving for x

Now, substituting y 2 into x2 y2 40:

x2 22 40

x2 4 40

x2 36

x ±6

Again, since the length of a rectangle cannot be negative, x 6.

Step 6: Calculating the Area of Rectangle ABCD

With x 6 and y 2, we can now find the area of rectangle ABCD:

Area x * y 6 * 2 12 square units.

Alternative Solutions Using Trigonometry

Although the initial solution used the Pythagorean Theorem, it’s interesting to explore how trigonometric functions could have been used. Here are a few alternative approaches:

Alternative 1

Using sin(∠AED) y / √13 and rearranging, we could attempt to find the values of x and y. However, this method didn't lead to a significant simplification.

Alternative 2

[AB^2 AD^2 40]

[AE^2 AD^2 13]

[AE AB/2]

[[AB^2/4] AD^2 13]

Subtracting the equation from the first equation, we get:

[3/4 * AB^2 27]

[AB^2 4/3 * 27 108/3 36]

[AB 6]

From the first equation:

[40 - 36 AD^2]

[AD^2 4]

[AD 2]

The area of ABCD is AD * AB 2 * 6 12.

Alternative 3

Let AB 2y and DB x. Then, x2 y2 13 and (1/4)AB2 y2 13. Subtracting the second equation from the first, we get 3/4 * AB2 27, leading to AB 6 and AD 2. The area of ABCD is 2 * 6 12 square units.

Conclusion

The area of rectangle ABCD is 12 square units. The problem provides an excellent opportunity to understand the application of the Pythagorean Theorem and algebraic manipulation in solving geometric problems.