- Count the equal-area parts in the whole badge. - A unit fraction has \(1\) as its numerator. - Use the number of equal parts as the denominator.

1. The badge is divided into \(2\) equal-area parts. 2. One of \(2\) equal parts is \(\frac{1}{2}\) of the whole.

Each part is \(\frac{1}{2}\) of the badge.

- Compare the widths of the two parts in each panel. - Equal-height rectangles have equal area when their widths are equal.

1. In a), the vertical segment is halfway across the rectangle, so it makes two rectangles with the same width and height. 2. In b), the segment is not halfway across, so one region is wider than the other. 3. Therefore, only a) makes two equal-area parts.

Partition a).

- Compare the sizes of the four sectors in the picture. - A unit-fraction name applies only after you know the parts are equal in area. - Use the number of equal parts in the whole to choose the denominator.

1. The four sectors in the circle are the same size, so the regions have equal area. 2. The whole is divided into \(4\) equal-area parts. 3. One of \(4\) equal parts is \(\frac{1}{4}\) of the whole.

Yes. The regions are equal in area, and one region is \(\frac{1}{4}\) of the whole circle.

- Count the columns in the whole rectangle. - Find a place that leaves the same number of columns on each side.

1. The rectangle is \(4\) grid squares wide. 2. A vertical line halfway across gives each part the same width and height. 3. Draw the line \(2\) grid squares from the left side, from top to bottom.

Draw a vertical line \(2\) grid squares from the left edge, all the way from top to bottom.

- Compare the number of grid cells in each strip. - Use the number of equal parts to name one part as a fraction.

1. Each strip is \(3\) grid squares across and \(1\) square tall, so each contains \(3\) unit squares. 2. All three areas are equal. 3. Each strip is one of \(3\) equal parts, so its fraction is \(\frac{1}{3}\).

Yes. Each strip is \(\frac{1}{3}\) of the square.

- Count the unit squares inside the shaded region. - Count how many equal-size regions make the whole. - Use the number of equal-size regions as the denominator.

1. The shaded strip contains \(2\) unit squares, so its area is \(2\,\text{square units}\). 2. The rectangle is divided into \(4\) equal vertical strips. 3. One of \(4\) equal-area strips is \(\frac{1}{4}\) of the rectangle.

The shaded strip has area \(2\,\text{square units}\) and is \(\frac{1}{4}\) of the rectangle.

- Compare the two triangles on the grid. - Look for a way one region could fit exactly onto the other. - Use the number of equal-area regions as the fraction denominator.

1. A diagonal of a square divides it into \(2\) matching triangles, so the triangles have equal area. 2. Share the total area equally: \(16 \div 2 = 8\). 3. Each triangle has area \(8\,\text{square units}\). 4. One of \(2\) equal-area parts is \(\frac{1}{2}\) of the square.

Each triangle has area \(8\,\text{square units}\) and is \(\frac{1}{2}\) of the square.

- Compare the sizes of the regions in the diagram. - Count how many equal regions make the whole. - Use the number of equal regions as the denominator of a unit fraction.

1. The three regions are the same size, so they have equal area. 2. The whole is divided into \(3\) equal-area regions. 3. One of \(3\) equal parts is \(\frac{1}{3}\) of the whole.

a) Yes. The three regions are the same size. b) Each region is \(\frac{1}{3}\) of the whole.

- Choose a simple way to split a rectangle into three matching shares. - Make sure the three regions have the same amount of area. - Count the equal regions to name one region as a unit fraction.

1. One valid description is to divide the rectangle into three equal-width strips that run from one side of the rectangle to the opposite side. 2. The strips have equal area because they have the same width and the same height. 3. One of three equal-area parts is \(\frac{1}{3}\) of the whole.

For example, divide the rectangle into three equal-width strips. Each part represents \(\frac{1}{3}\) of the whole.

- Count the regions in each circle. - Then compare their sizes. - “Thirds” means three equal-area parts, not merely three parts.

1. Circle a) has three equal-area sectors. 2. Circle b) has three sectors, but one is larger than the other two. 3. Thirds require three equal-area parts, so only a) is partitioned into thirds.

Circle a).

- Compare the widths of the left and right regions. - Equal-area rectangles need matching dimensions when their heights are the same. - Check all four regions, not just the top two.

1. In a), the vertical and horizontal segments both pass through the middle of the rectangle, making four equal rectangles. 2. In b), the vertical segment is off center, so the left regions have less area than the right regions. 3. Therefore, only a) makes four equal-area parts.

Partition a).

- Share the total area equally among the three regions. - Use the number of equal regions as the fraction denominator.

1. Divide the total area equally: \(18 \div 3 = 6\,\text{square units}\) per region. 2. One of three equal-area regions is \(\frac{1}{3}\) of the poster.

Each region has area \(6\,\text{square units}\) and is \(\frac{1}{3}\) of the poster.

- Having four regions is only one requirement for fourths. - Compare the sizes of the regions. - Use the meaning of equal-area parts.

1. Fourths require four equal-area parts. 2. The diagram has four regions, but one region is twice the size of each of the three smaller regions. 3. Therefore, the regions are not fourths.

No. There are \(4\) regions, but they do not have equal area.

- Look at where the dividing segment meets the bottom side. - Compare the two smaller triangles as mirror-image shapes. - If a whole is split into two equal-area parts, think about the unit fraction for one part.

1. The dividing segment runs through the middle of this left-right symmetric triangle. 2. The two regions are matching mirror images, so they have equal area. 3. The whole triangle is divided into two equal-area parts, so each part is \(\frac{1}{2}\) of the whole.

Yes. The two parts have equal area, and each part is \(\frac{1}{2}\) of the whole triangle.

- Check where the new line touches the bottom side. - Compare the widths of the two parts along that side. - Halves must have equal area.

1. The dividing line meets the bottom very close to the left corner. 2. The two triangles reach the same top point, but one has a much longer part of the bottom side than the other. 3. Their areas are not equal, so neither part is half of the whole.

No. One piece has a bottom side \(1\) grid square long and the other has a bottom side \(5\) squares long. They rise to the same top point, so the wider piece has more area. The pieces are not halves.

- Count the columns in the strip that is already separated. - Would a strip of a different width have the same area if its height stayed the same?

1. The whole rectangle is \(6\) grid squares wide, and the left strip is \(2\) squares wide. 2. For three equal strips, each must be \(2\) squares wide. 3. Put the second line \(2\) squares from the right edge, straight from top to bottom.

Draw a vertical line \(2\) grid squares from the right edge, from top to bottom.

- Picture folding the square to match the two large triangles. - What could you draw to make the smaller pieces match as well?

1. The drawn line connects \(A\) and \(C\). 2. Connect the other two opposite corners, \(B\) and \(D\). 3. The lines cross in the center. All four triangles have an outside side \(4\) squares long and reach \(2\) squares to the center. 4. Turning the square a quarter turn moves each triangle onto the next one. They have equal area and fill the whole square, so each is \(\frac{1}{4}\) of it.

Connect \(B\) to \(D\). The lines meet at the square's center. A quarter turn fits each small triangle exactly onto the next, so all four have equal area. Each is \(\frac{1}{4}\) of the square.

- Count the square units below the drawn line. - Compare the area above the line with the area below it. - What must be true of the two pieces you make from the larger part?

1. The bottom strip is \(4\) squares across and \(1\) square tall, so its area is \(4\) square units. 2. The top part covers \(8\) square units. Split it into two parts of \(4\) square units each. 3. One way is to draw a vertical line halfway across the upper part, \(2\) squares from the left edge, from the old line to the top. 4. This makes a \(4\)-square bottom rectangle and two \(2\)-by-\(2\) squares. Their areas are equal, but their shapes differ.

For example, draw a vertical line \(2\) squares from the left edge, starting at the existing line and ending at the top edge. Each piece has an area of \(4\) square units; the bottom part is a long rectangle, and the two top parts are squares.

- Count or group the unit squares in each region. - Look for a larger familiar shape that can help you compare two regions. - Equal-area regions do not have to have the same outline.

1. The left rectangle is \(2\) squares wide and \(2\) squares tall, so its area is \(2 \times 2 = 4\,\text{square units}\). 2. Each triangle on the right is half of a \(4 \times 2\) rectangle. That rectangle has area \(8\,\text{square units}\), so each triangle has area \(8 \div 2 = 4\,\text{square units}\). 3. All three parts have area \(4\,\text{square units}\), so they are equal in area even though they are not the same shape. 4. The whole is divided into \(3\) equal-area parts, so each part is \(\frac{1}{3}\) of the whole.

Yes. Each part has area \(4\,\text{square units}\), so each part is \(\frac{1}{3}\) of the whole.

- Count or group the unit squares in A. - Count or group the unit squares in B. - When two regions have equal area and make the whole, each is one half.

1. Part A is \(3\) squares wide and \(2\) squares tall, so its area is \(3 \times 2 = 6\,\text{square units}\). 2. Part B contains the other \(6\) unit squares, so its area is also \(6\,\text{square units}\). 3. The parts have equal area even though their shapes are different. 4. Each of the \(2\) equal-area parts is \(\frac{1}{2}\) of the whole.

a) A has area \(6\,\text{square units}\), and B has area \(6\,\text{square units}\). b) Yes. A and B have equal area. c) Each part is \(\frac{1}{2}\) of the whole.

- Check whether the three numbers in each row are equal. - Do not confuse having the correct total with having equal parts. - Use the number of equal regions to name one region as a unit fraction.

1. Each plan totals \(24\,\text{square units}\), but an equal-area partition needs all three region areas to match. 2. Only Plan A has three equal areas: \(8\,\text{square units}\) each. 3. One of \(3\) equal regions is \(\frac{1}{3}\) of the board.

Plan A. Each region has area \(8\,\text{square units}\) and is \(\frac{1}{3}\) of the board.

- Use the grid to find the area of each region. - Compare the area values before naming the fraction. - Equal area describes the amount of space, not the outline shape.

1. The left and right regions are each \(2\) squares wide and \(4\) squares tall, so each has area \(2 \times 4 = 8\,\text{square units}\). 2. The two middle regions together make a \(4 \times 4\) rectangle with area \(16\,\text{square units}\). The diagonal divides that rectangle into two equal-area triangles, so each has area \(16 \div 2 = 8\,\text{square units}\). 3. All four regions have area \(8\,\text{square units}\), so they are equal in area. 4. One of \(4\) equal-area regions is \(\frac{1}{4}\) of the whole. 5. The regions include rectangles and triangles, so equal-area regions do not have to have the same shape.

a) Yes. Each region has area \(8\,\text{square units}\). b) Each region is \(\frac{1}{4}\) of the whole. c) No. Equal-area regions can have different shapes.

- Count the final regions shown by the boundary lines. - Compare how many unit squares are in each region. - For part c), compare one smallest region with all the unit squares in the whole rectangle.

1. The boundary lines divide the rectangle into \(5\) regions. 2. Three regions contain \(2\) unit squares each, while the two smallest regions contain \(1\) unit square each, so the regions are not all equal in area. 3. The whole rectangle contains \(8\) equal unit squares. Each smallest region contains \(1\) of them, so each is \(\frac{1}{8}\) of the whole rectangle.

a) \(5\) regions b) No. c) Each smallest region is \(\frac{1}{8}\) of the whole rectangle.

- Use the grid to find the dimensions of the large region. - Find the area of one narrow region from its width and height. - Compare one narrow region with the area of the whole grid.

1. The large region is \(3\) squares wide and \(4\) squares tall, so its area is \(3 \times 4 = 12\,\text{square units}\). 2. The three smaller regions together also cover a \(3 \times 4\) area, so their total area is \(12\,\text{square units}\). 3. Each smaller region is \(1\) square wide and \(4\) squares tall, so each has area \(1 \times 4 = 4\,\text{square units}\). 4. The whole has area \(6 \times 4 = 24\,\text{square units}\). Since \(24 \div 4 = 6\), each smaller region is \(\frac{1}{6}\) of the whole.

a) The large region has area \(12\,\text{square units}\), and the three smaller regions have total area \(12\,\text{square units}\). b) Each smaller region has area \(4\,\text{square units}\). c) Each smaller region is \(\frac{1}{6}\) of the whole.

- Find the area of each rectangle size from its side lengths. - Compare the two area results before deciding whether the partition is equal in area. - For part b), decide which measurement matters in the phrase “equal-area.”

1. Each \(1\,\text{unit} \times 6\,\text{units}\) region has area \(1 \times 6 = 6\,\text{square units}\). 2. Each \(2\,\text{units} \times 3\,\text{units}\) region has area \(2 \times 3 = 6\,\text{square units}\). 3. All four regions therefore have the same area. 4. Equal-area partitions depend on area, not perimeter, so the different perimeters do not change the conclusion.

a) Yes. Each region has area \(6\,\text{square units}\). b) No. Different perimeters do not prevent the regions from having equal areas.

- Use the grid to find the area of each region. - Compare the wider regions with the narrower regions. - Compare each region's area with the area of the whole when finding its fraction.

1. Each of the two wider regions is \(2\) squares wide and \(2\) squares tall, so each has area \(2 \times 2 = 4\,\text{square units}\). 2. Each of the four narrower regions is \(1\) square wide and \(2\) squares tall, so each has area \(1 \times 2 = 2\,\text{square units}\). 3. The regions do not all have the same area because \(4 \ne 2\). 4. The whole has area \(8 \times 2 = 16\,\text{square units}\). Each wider region is \(\frac{4}{16} = \frac{1}{4}\) of the whole, while each narrower region is \(\frac{2}{16} = \frac{1}{8}\) of the whole.

a) No. Two regions have area \(4\,\text{square units}\), while four regions have area \(2\,\text{square units}\). b) Each wider region is \(\frac{1}{4}\) of the whole. Each narrower region is \(\frac{1}{8}\) of the whole.

- Find one piece's area before thinking about the rearrangement. - Ask what measurements stay unchanged when pieces are moved. - Count the same pieces in the new whole.

1. Each original piece has area \(24 \div 6 = 4\,\text{square units}\). 2. Rearranging the pieces does not change their areas or the total area. 3. The new whole still contains the same \(6\) equal-area pieces. 4. Each piece is \(\frac{1}{6}\) of the new whole.

a) \(4\,\text{square units}\) b) \(\frac{1}{6}\)

- Compare the region labels before adding them. - Check whether their total matches the whole area. - Divide the whole area by the required number of equal regions.

1. The labels use two different region areas, \(6\,\text{square units}\) and \(5\,\text{square units}\), so they do not describe equal areas. 2. Their total is \(5 \times 6 + 3 \times 5 = 30 + 15 = 45\,\text{square units}\), not \(48\,\text{square units}\). 3. Eight equal regions in an area of \(48\,\text{square units}\) must each have area \(48 \div 8 = 6\,\text{square units}\). 4. Every region should be labeled \(6\,\text{square units}\).

a) No. b) No; the labels total \(45\,\text{square units}\). c) \(6\,\text{square units}\) per region.

- Read the denominator as the number of equal parts in the whole. - Then share the total area equally among that many parts.

1. The denominator \(6\) means the whole is divided into \(6\) equal-area parts. 2. Divide the total area by \(6\): \(24 \div 6 = 4\,\text{square units}\).

a) \(6\) parts b) \(4\,\text{square units}\) per part

- Keep the existing equal left and right halves. - Think about how to cut both halves into equal top and bottom pieces at the same time. - The new segment should cross the entire rectangle.

1. The vertical segment already makes two equal-width halves. 2. To split both halves equally again, draw a horizontal segment halfway between the top and bottom edges. 3. The new line creates four rectangles with equal width-and-height pairs, so all four areas are equal.

Draw a horizontal segment through the middle of the rectangle from the left edge to the right edge.

- Count the unit squares in each region separately. - Equal-area parts can have different orientations or outlines. - After checking the areas, use the number of equal parts to name the unit fraction.

1. The left region covers \(2\) grid squares, so its area is \(2\,\text{square units}\). 2. The middle region also covers \(2\) grid squares, and the right region covers \(2\) grid squares. 3. The three regions therefore have equal area. 4. Since the whole is divided into \(3\) equal-area parts, each region is \(\frac{1}{3}\) of the whole.

Yes. Each region has area \(2\,\text{square units}\), so each region is \(\frac{1}{3}\) of the whole.

- Equal parts would need the same whole-number count of unit squares. - Test nearby equal group sizes. - Compare their totals with \(21\).

1. If each part had \(5\) unit squares, the total would be \(4 \times 5 = 20\). 2. If each part had \(6\) unit squares, the total would be \(4 \times 6 = 24\). 3. There is no whole-number square count between \(5\) and \(6\) for each part. 4. Therefore, \(21\) unit squares cannot be split into \(4\) equal-area parts without cutting a square.

No. Four equal groups of whole unit squares cannot total \(21\).

- Count the equal strips separately from the unit squares. - Find the area of the shaded strip from its width and height. - A fraction denominator names the number of equal parts in the partition.

1. The whole grid has area \(8 \times 8 = 64\,\text{square units}\). 2. The partition has \(8\) equal strips, not \(64\) equal strips. 3. One strip has area \(64 \div 8 = 8\,\text{square units}\). 4. One of \(8\) equal strips is \(\frac{1}{8}\) of the grid. Sam used the number of unit squares instead of the number of equal parts as the denominator.

One strip has area \(8\,\text{square units}\) and is \(\frac{1}{8}\) of the grid. Sam incorrectly used the total number of unit squares as the fraction denominator.

- First find the area each of the three regions must have. - Describe one simple region with that area. - Think about how the remaining area could be split into two equal parts with a different outline.

1. Each of three equal-area regions must have area \(18 \div 3 = 6\,\text{square units}\). 2. One valid description is to make a \(2 \times 3\) rectangle on the left, then divide the remaining \(4 \times 3\) rectangle with a diagonal. 3. The left rectangle has area \(6\,\text{square units}\). The remaining rectangle has area \(12\,\text{square units}\), and its diagonal makes two equal triangles of area \(6\,\text{square units}\) each. 4. The three regions therefore have equal area, and the rectangle region has a different shape from the two triangular regions.

For example, use a \(2 \times 3\) rectangle plus two triangles made by a diagonal across the remaining \(4 \times 3\) rectangle. Each region has area \(6\,\text{square units}\).