Question

the diagram shows a border that is 4 inches wide surrounding annie’s painting. write a simplified expression that represents the area of the border. 3x in. 12 in. 4 in. 4 in.

Explanation:

Step1: Find dimensions of outer rectangle

The border is 4 inches wide on all sides. So, the length of the outer rectangle (including border) is the painting's length plus \( 4 + 4 = 8 \) inches. The painting's length is \( 3x \) inches, so outer length \( L = 3x + 8 \). The width of the outer rectangle (including border) is the painting's width plus \( 4 + 4 = 8 \) inches. The painting's width is 12 inches, so outer width \( W = 12 + 8 = 20 \) inches. Wait, no—wait, actually, the border is 4 inches on each side, so for the length (horizontal) of the outer rectangle: the painting's length is \( 3x \), and we add 4 inches on the left and 4 inches on the right, so outer length is \( 3x + 4 + 4 = 3x + 8 \). For the width (vertical) of the outer rectangle: the painting's width is 12, and we add 4 inches on the top and 4 inches on the bottom, so outer width is \( 12 + 4 + 4 = 20 \). Wait, but maybe I misread the diagram. Wait, the painting is a rectangle with length \( 3x \) and width 12, and the border is 4 inches wide around it. So area of border is area of outer rectangle minus area of inner rectangle (painting).

So first, area of inner rectangle (painting) is \( \text{length} \times \text{width} = 3x \times 12 = 36x \) square inches.

Then, outer rectangle: length is \( 3x + 4 + 4 = 3x + 8 \), width is \( 12 + 4 + 4 = 20 \). So area of outer rectangle is \( (3x + 8) \times 20 \). Let's expand that: \( 20 \times 3x + 20 \times 8 = 60x + 160 \).

Then, area of border is outer area minus inner area: \( (60x + 160) - 36x \).

Step2: Simplify the expression

Combine like terms: \( 60x - 36x + 160 = 24x + 160 \). Wait, is that right? Wait, maybe I made a mistake in the outer width. Wait, maybe the painting's length is 12 and width is \( 3x \)? Wait, the diagram shows "3x in." and "12 in."—maybe the painting is length 12 and width \( 3x \)? Wait, the labels: the inner rectangle has 3x in. (maybe height) and 12 in. (width). Wait, maybe I mixed up length and width. Let's re-examine.

Alternative approach: Let's assume the inner rectangle (painting) has length \( l = 12 \) inches and width \( w = 3x \) inches. Then the outer rectangle (including border) has length \( l' = 12 + 4 + 4 = 20 \) inches, and width \( w' = 3x + 4 + 4 = 3x + 8 \) inches. Then area of inner rectangle: \( 12 \times 3x = 36x \). Area of outer rectangle: \( 20 \times (3x + 8) = 60x + 160 \). Then area of border: \( (60x + 160) - 36x = 24x + 160 \). Wait, but maybe the inner rectangle is length \( 3x \) and width 12, so outer length \( 3x + 8 \), outer width \( 12 + 8 = 20 \), same result.

Wait, another way: the border's area can be calculated as the sum of the areas of the top, bottom, left, and right rectangles, plus the four corner squares (but the corners are 4x4 squares, so four of them: 444=64? Wait, no, the top and bottom rectangles: top has length \( 3x + 8 \) (outer length) and width 4, so area \( 4(3x + 8) \). Bottom same as top: \( 4(3x + 8) \). Left and right rectangles: left has height 12 (inner height) and width 4, so area \( 4*12 \). Right same as left: \( 4*12 \). Wait, but that would be top + bottom + left + right: \( 24(3x + 8) + 24*12 \). Let's compute that: \( 8(3x + 8) + 96 = 24x + 64 + 96 = 24x + 160 \). Same result. So that's correct.

Answer:

\( 24x + 160 \)