Question

Jeff has a different photo that he wants to put in the frame. If the wooden border of the frame (shown by the shaded region in the figure) is \(x\) inches wide all the way around, which of the following represents the area \(A\) of Jeff's photo (the non-shaded region)? (A) \(A=(15-x)(12-x)\) (B) \(A=(15-2 x)(12-2 x)\) (C) \(A=(15+2 x)(12+2 x)\) (D) \(A=(15+x)(12+x)\)

Short answer

The short answer to the question is: (B) \( A=(15-2x)(12-2x) \)

Step-by-step solution

Understand the given information

In this problem, we are given the dimensions of the frame, which are 15 inches by 12 inches. The shaded border is x inches wide. We need to find the area of the non-shaded region (the area of the photo).

Calculate the area of the full frame

To get the area of the full frame, multiply its dimensions. In this case, the full frame has dimensions 15 inches by 12 inches. So the area of the frame is: Area of the full frame \( = 15 \times 12 \)

Calculate the dimensions of the non-shaded region

Since the border is x inches wide all the way around, the dimensions of the non-shaded region will be: Length = (15 - 2x) inches Width = (12 - 2x) inches

Find the area of the non-shaded region (Jeff's photo)

To find the area of the non-shaded region (the photo), multiply the length and width from the previous step: Area of the photo \(= (15 - 2x)(12 - 2x) \)

Match the result with the given options

The required area of Jeff's photo is \( (15 - 2x)(12 - 2x) \), which matches option (B) in the given choices. So, the answer is: (B) \( A=(15-2x)(12-2x) \)

Key concepts

Area Calculation

When it comes to mathematics, understanding area calculation is crucial, especially for GED test-takers. Area represents the size of a two-dimensional surface and is measured in square units. For Jeff’s picture frame problem, the calculation involves finding the area of a rectangle, which is simply the product of its length and width.

In the case of the frame, we first determine the entire frame's area by multiplying the full dimensions given, 15 inches by 12 inches. However, Jeff wants to place a photo inside the frame, surrounded by a border of unknown width 'x.' To account for the border all around the photo, we subtract twice the border width from each of the frame's dimensions—because the border is present on both sides of the length and the width. Thus, we get the formula for the area of Jeff's photo:
(A) \( A=(15-2x)(12-2x) \)

This formula is derived by, first calculating the new length as `15 - 2x` and the new width as `12 - 2x`, then finding the product of these new dimensions. It's also important for students to recognize that the subtraction by '2x' is because the border affects both sides of the length and the width, hence the multiplication by 2.

Quadratic Equations

Quadratic equations are a fundamental component of algebra and appear frequently on the GED exam. They are polynomial equations of the second degree, usually in the form \( ax^2 + bx + c = 0 \), where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. The solution to these equations can often be found using factoring, completing the square, or applying the quadratic formula.

In the context of the frame problem, the area formula \( A=(15-2x)(12-2x) \) can be expanded to form a quadratic equation when set to a specific area for solving. This requires multiplying the binomials resulting in an equation with an \( x^2 \) term, an 'x' term, and a constant, which can then be manipulated to fit the standard quadratic form. Understanding how to work with quadratic equations is vital because they not only apply to numerical problems but also to real-world scenarios, such as the dimensions in geometric shapes and motion physics.

Geometry Problems

Geometry problems are an integral part of GED mathematical reasoning. These problems require the understanding of shapes, sizes, the properties of space, and how to calculate various dimensions. Geometry plays a significant role in a wide array of mathematical concepts beyond simple shape recognition, including area and volume calculations, the Pythagorean theorem, and properties of triangles and circles.

In Jeff's frame example, we're dealing with a rectangular shape. Geometry helps us to understand that the area of a rectangle is found by multiplying its length by its width. But the problem adds a layer of complexity by introducing a border with variable width 'x'. With geometry, we intuitively subtract the border's width from the frame's overall dimensions to find the dimensions of the photo itself. Problems like these show how geometry can be dynamic and applied to practical situations, requiring a deeper understanding of the relationship between shapes and their dimensions.