Problem 47
Question
The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame. (IMAGE CANT COPY)
Step-by-Step Solution
Verified Answer
The width of the frame is 2 inches.
1Step 1: Understand the Problem
In this problem the rectangle formed by the painting and its frame has a perimeter of 72 inches. The painting within the frame is 12 inches by 16 inches. The width of the frame is unknown and is the same on all sides. The goal is to find out the width of the frame.
2Step 2: Setup equations Using Perimeter Formula
The perimeter formula for a rectangle is \(2*(length + width)\). Substitute the given dimensions into this formula. Given the painting's length and width as 16 and 12 inches respectively and width of the frame being 'x'. The total length and width of the framed painting would be \(16+2x\) and \(12+2x\). Substituting these into the perimeter formula gives \(2*((16+2x) + (12+2x)) = 72\)
3Step 3: Solve for 'x'
After simplification, the equation becomes \(2*(28+4x) = 72\). Solving this equation for 'x' by first dividing the equation by 2 gives \(28+4x = 36\). Subtracting 28 from both sides of the equation gives \(4x = 8\). Finally, divide by 4, to get \(x = 2\) which gives the width of the frame.
Key Concepts
Solving EquationsGeometryAlgebra
Solving Equations
Solving equations involves finding the unknown value that makes the equation true. In this exercise, the unknown value is the width of the frame, denoted as 'x'. You start by using information given in the problem to set up an equation. Here:
- The perimeter of the rectangle formed by the painting and its frame is provided, which is 72 inches.
- The dimensions of the painting are 12 inches by 16 inches.
- The width of the frame is the unknown variable 'x'.
Geometry
Geometry helps us understand spatial relationships and properties of shapes. In this problem, it helps us calculate perimeter, a fundamental property. The original painting is a rectangle, a shape with four sides at right angles. You need to apply knowledge of geometry to find out how the frame fits around the painting.
- The painting's dimensions are 12 by 16 inches, forming the inner rectangle part.
- With the frame, the entire object's perimeter changes.
- Every side of the rectangle is increased by the frame's width, on both ends, hence adding '2x' to each dimension.
Algebra
Algebra is crucial for mastering how to manipulate and solve equations involving unknowns. It involves using symbols to represent numbers and a variety of mathematical operations. In this exercise, algebraic techniques help to express the relationship between the painting, the frame, and the equation for perimeter.
- The unknown width of the frame, 'x', is the variable you solve for in the equation.
- Algebraic manipulation includes expanding expressions, simplifying terms, and ultimately isolating the variable on one side of the equation.
- In the equation \(2 \times (28 + 4x) = 72\), you first divide both sides by 2, simplifying to \(28 + 4x = 36\).
- By further isolating 'x', subtracting 28 from both sides gives \(4x = 8\), and dividing by 4 provides the solution \(x = 2\).
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