- Count the unit squares by rows or columns. - You can also split the figure into two rectangles.

1. The bottom row contains \(3\) unit squares. 2. Two more unit squares are above the leftmost square. 3. The area is \(3 + 2 = 5\) square units.

The area is \(5\) square units.

- Count the unit squares inside the figure. - You can separate the figure into a horizontal row and a vertical column without counting the corner square twice.

1. The bottom row contains \(3\) unit squares. 2. Three more unit squares are above the leftmost square. 3. The total is \(3 + 3 = 6\) square units.

\(6\) square units

- Use the small reference square to identify one square unit. - Count how many unit squares are in one row of the rectangle. - Count the rows, then combine the equal row counts.

1. Count \(4\) unit squares across each row. 2. There are \(3\) rows. 3. Count all the unit squares: \(4 + 4 + 4 = 12\).

The rectangle is covered by \(12\) unit squares, so its area is \(12\) square units.

- Look for the unit squares inside the boundary. - Count the squares in each row.

1. The shape is \(3\) unit squares across and \(2\) down. 2. Count all the squares: \(3+3=6\) square units.

\(6\) square units

- Find the area covered by one square. - Count how many of those same squares cover the card.

1. Each square covers \(1\) square inch. 2. Four squares cover \(4\) square inches altogether.

\(4\,\text{in}^2\)

- Imagine removing the ribbon. Would that change the card's front surface? - What would happen if Ana counted the tiles again instead?

1. The \(12\) square tiles cover the front, so its area is \(12\) square inches. 2. The \(14\) inches of ribbon measure the distance around the edge, not the area. Ana is not right.

No. The area is \(12\,\text{in}^2\).

- Does each tile cover its own separate part? - Check whether the entire board is covered.

1. Each area unit must cover a different part of the board. 2. Overlapping tiles do not cover six distinct squares, and a gap is left uncovered. 3. The count of tiles does not establish an area of six square units.

No. Area tiles must cover the board without gaps or overlaps.

- Count the unit squares inside each boundary. - Split a shape into rectangular parts to organize the count.

1. Split Figure A into small rectangles and count its \(5\) unit squares. 2. Split Figure B into narrow rectangular parts and count its \(5\) unit squares.

Figure A: \(5\) square units Figure B: \(5\) square units

- Count the unit squares inside each figure. - You may split a figure into smaller rectangles or subtract a missing rectangle from a larger one. - Compare the three areas.

1. Figure A can be split into a vertical rectangle with area \(3\) square units and a horizontal rectangle with area \(2\) square units. Its total area is \(3 + 2 = 5\) square units. 2. Figure B can be split into a vertical rectangle with area \(2\) square units and a horizontal rectangle with area \(3\) square units. Its total area is \(2 + 3 = 5\) square units. 3. Figure C is a \(3 \times 3\) square with a \(1 \times 2\) rectangle removed. Its area is \(9 - 2 = 7\) square units. 4. Figure C has the greatest area.

a) Figure A: \(5\) square units Figure B: \(5\) square units Figure C: \(7\) square units b) Figure C has the greatest area.

- Count the unit squares inside each figure. - You can organize the count by rows or columns. - Compare the three totals to find the matching pair.

1. Figure A is a \(3 \times 2\) rectangle, so its area is \(3 \times 2 = 6\) square units. 2. Figure B has \(3\) unit squares in the bottom row, \(2\) in the middle row, and \(1\) in the top row. Its area is \(3 + 2 + 1 = 6\) square units. 3. Figure C has \(3\) unit squares in the bottom row and \(2\) more above the leftmost square. Its area is \(3 + 2 = 5\) square units. 4. Figures A and B have the same area.

a) Figure A: \(6\) square units Figure B: \(6\) square units Figure C: \(5\) square units b) Figures A and B have the same area.

- Count the unit squares in each figure separately. - You can count by rows or columns. - Compare the two totals.

1. Figure a) is a rectangle with \(2\) rows of \(4\) unit squares. Its area is \(4 + 4 = 8\) square units. 2. Figure b) has \(5\) unit squares in the bottom row and \(3\) in the top row. Its area is \(5 + 3 = 8\) square units. 3. Therefore, the figures have the same area.

Yes. Figures a) and b) each have an area of \(8\) square units.

- Count the unit squares inside each figure. - You can split Figure b) into rectangles. - Compare the two area totals.

1. Figure a) is a square with side length \(3\) units, so its area is \(3 \times 3 = 9\) square units. 2. Figure b) can be split into two rectangles with areas \(2 \times 2 = 4\) square units and \(2 \times 3 = 6\) square units. Its total area is \(4 + 6 = 10\) square units. 3. Since \(10 > 9\), Figure b) has the greater area.

Figure b) has an area of \(10\) square units, which is greater than Figure a)'s area of \(9\) square units.

- Count how many unit squares each figure covers. - Break each figure into individual unit squares. - Compare the two counts.

1. Figure a) is a \(3 \times 2\) rectangle, so its area is \(3 \times 2 = 6\) square units. 2. Figure b) can be split into a \(3 \times 1\) rectangle and a \(1 \times 2\) rectangle, so its area is \(3 + 2 = 5\) square units. 3. Since \(6 > 5\), Figure a) has the greater area.

Figure a) has the greater area: \(6\) square units compared with \(5\) square units for Figure b).

- Separate the size of the board region from the number of magnets currently on it. - Each removed unit square exposes one square unit of the region.

1. The region's area remains \(16\) square units because removing magnets does not change the region. 2. Each removed unit-square magnet leaves \(1\) square unit uncovered, so \(4\) square units are uncovered.

a) \(16\) square units b) \(4\) square units

- Count only complete unit squares inside the boundary. - Break the figure into a bottom rectangle and an upper-left rectangle. - Add the square counts from the two non-overlapping parts.

1. Count \(8\) unit squares in the bottom \(2\) rows. 2. Count \(4\) more unit squares in the upper-left part. 3. Add the two counts: \(8 + 4 = 12\) square units.

The area of the figure is \(12\) square units.

- Think about what the word unit means in a measurement. - Check whether every counted piece represents the same amount of area.

1. A square unit must have one fixed size throughout a measurement. 2. Counting different-size squares as equal units gives an inconsistent total. 3. The region should be covered with squares of the same size, without gaps or overlaps, and those equal squares should be counted.

The measurement is invalid because the counted squares are not equal-size units. Use one consistent square-unit size and cover the region without gaps or overlaps.

- Compare the amount of space covered before and after each replacement. - Two half-square pieces can cover the same space as one unit square. - Area depends on the region covered, not only on the number of pieces.

1. Replacing one unit square with two half-square pieces does not change the space covered. 2. Each of the three replacements covers the same area as before. 3. The region itself has not changed, so its area remains \(12\) square units.

Maya is not correct. The number of pieces increased, but the area is still \(12\) square units.

- Think about whether the objects being counted cover the entire region. - Recall the shape and consistency required of an area unit.

1. Dots do not cover equal square regions, so they are not square units. 2. Area must measure the surface inside the boundary. 3. A valid method is to cover the region without gaps or overlaps using equal unit squares and count those squares.

Counting dots does not measure area because dots are not square units that cover the surface. Cover the region with equal unit squares without gaps or overlaps, then count them.

- Compare the size of each surface with the size of each square unit. - A useful unit should not make the count unnecessarily huge or too small to describe the surface well.

1. An index card is a small surface, so square inches give a useful count. 2. A classroom floor is a much larger surface, so square feet give a useful count.

Use square inches for the index card and square feet for the classroom floor.

- Count the complete unit squares inside each boundary. - For figure b), count the squares row by row. - Compare the two totals rather than the shapes’ appearances.

1. Figure a) has \(3\) unit squares in each of \(2\) rows, for \(6\) unit squares. 2. Figure b) has \(3\) unit squares in the bottom row, \(2\) in the next row, and \(1\) in the top row. 3. Figure b) has \(3 + 2 + 1 = 6\) unit squares. 4. Both figures have an area of \(6\) square units.

Yes. Both figures have an area of \(6\) square units.

- Decide which changes move a figure without changing its size. - Compare how many unit-square spaces each copy covers after the change.

1. Sliding Copy A changes only its location, so its area remains \(1\) square unit. 2. Turning Copy B changes only its orientation, so its area remains \(1\) square unit. 3. Stretching Copy C changes the amount of surface it covers to \(2\) square units.

Copies A and B still have an area of \(1\) square unit. Copy C does not.

- Use the grid to determine how many rows and columns of unit squares are shaded. - Area includes interior unit squares as well as boundary unit squares. - Count or multiply to include every shaded unit square exactly once.

1. Area counts every unit square inside the shaded region, not only the squares along its boundary. 2. The grid has \(5\) rows and \(5\) columns of unit squares. 3. Multiply: \(5 \times 5 = 25\) square units.

Eli counted only boundary squares and left out the interior squares. The area is \(25\) square units.

- Hold the region and unit size constant. - Ask whether a valid area measurement can change only because a different student counts it.

1. The region and the unit-square size are the same for both measurements. 2. A complete covering without gaps or overlaps must use the same number of those unit squares. 3. Therefore, at least one count is incorrect and the covering should be checked.

No. Both cannot be correct. The same region covered with the same unit squares without gaps or overlaps must have one fixed square-unit count.

- Find the total area represented by each partition separately. - Different partitions can measure the same whole region when their parts do not overlap.

1. The first partition gives a total area of \(6 + 9 = 15\) square units. 2. The second partition gives a total area of \(7 + 8 = 15\) square units. 3. Both partitions cover the same whole region without overlap, so the matching totals confirm the area.

Yes. Both partitions give an area of \(15\) square units.

- Identify how many quarter-square tiles make one whole square unit. - Use the same number of tiles for each of the \(4\) square units. - Multiply the number of square units by the tiles needed per square unit.

1. Each square unit needs \(4\) quarter-square tiles. 2. For \(4\) square units, multiply: \(4 \times 4 = 16\).

\(16\) quarter-square tiles are needed.

- Count only complete unit squares inside the shaded boundary. - Count the figure row by row. - Add the row counts to find the total area.

1. Count \(6\) unit squares in the bottom row. 2. Count \(4\) unit squares in each of the next \(3\) rows, for \(12\) more squares. 3. Add: \(6 + 12 = 18\) square units.

The area of the figure is \(18\) square units.

- Look at how each slanted edge cuts through a unit square. - How much of a whole grid square is covered by each triangular piece?

1. The diamond is made of four triangular pieces, each half a unit square. 2. Pair the four halves into two whole squares. 3. The area is \(2\) square units.

\(2\) square units; four half-squares make two whole square units.

- Count squares inside each outline separately. - An L-shaped region can be split into two smaller rectangles.

1. Shape a) covers \(2\) rows of \(3\) squares, or \(6\) square units. 2. Shape b) covers \(3\) squares in its bottom strip and \(3\) more above the left strip, also \(6\) square units. 3. The areas are equal although the shapes look different.

They have equal area: \(6\) square units each.

- Imagine putting one of Luis's tiles on top of one of Mia's. - Does using the same number of tiles guarantee the same amount of covered surface?

1. Tile count alone does not tell area unless the square-unit size is the same. 2. Eight larger tiles cover more area than eight smaller tiles if there are no gaps or overlaps.

Mia's card has more area because her \(8\) tiles are larger, and both cards are completely covered.

- The number of tiles alone does not determine area when tile sizes differ. - Find how many smaller tiles combine to cover one square foot. - Compare the total areas rather than the tile counts.

1. The first section has area \(12\) square feet. 2. Four \(\frac{1}{4}\)-square-foot tiles make one square foot. 3. The second section has area \(48\div4=12\) square feet.

Yes. Each floor section has an area of \(12\) square feet.

- Compare what perimeter measures with what area measures. - Think about whether two differently shaped regions could have the same distance around them.

1. Perimeter measures the distance around a shape, not the surface it covers. 2. Different shapes can have a perimeter of \(20\) units but cover different numbers of unit squares. 3. The area can be determined only with information about the region inside, such as a unit-square covering or enough dimensions.

No. A perimeter of \(20\) units does not determine the area. The number of square units covering the inside, or equivalent size information, is also needed.

- Check the area represented by one tile before counting the tiles. - Convert the covering into a count of unit squares.

1. The tiles are not unit squares; each tile covers \(2\) square units. 2. Multiply: \(6 \times 2 = 12\) square units.

Aaliyah counted tiles instead of unit squares. The area is \(12\) square units.

- Check whether one unit square in Region A is the same size as one unit square in Region B. - Numerical area counts are directly comparable only when their units match.

1. The two tile counts use different-size square units. 2. A square inch and a square centimeter do not cover the same amount of surface. 3. The areas must be expressed in the same square unit before their numerical values can be compared.

No. The counts cannot be compared directly because the square units are different sizes. Both areas must first be measured in the same square unit.

- A valid unit-square measurement must cover the entire region. - Decide whether uncovered spaces inside the boundary contribute to the region’s area.

1. The \(14\) squares show only the area currently covered. 2. Because gaps remain inside the region, the covering is incomplete. 3. The region’s full area cannot be determined from the given tile count alone.

No. The gaps mean the region is not completely covered, so \(14\) square units is not enough information to determine the region’s area.

- Find the area that remains after the full squares are counted. - Think about how many half-square pieces make one whole square unit.

1. The part not covered by full squares has area \(11 - 8 = 3\) square units. 2. Each square unit requires \(2\) half-square pieces. 3. The number of half-square pieces is \(3 \times 2 = 6\).

The design uses \(6\) half-square pieces.

- Combine the partial-square pieces into a fractional square unit. - Compare the distance from the exact area to each neighboring whole number.

1. The \(3\) quarter-square pieces have a combined area of \(\frac{3}{4}\) square unit. 2. The exact area is \(5\frac{3}{4}\) square units. 3. The area is \(\frac{1}{4}\) square unit less than \(6\) and \(\frac{3}{4}\) square unit greater than \(5\), so it is closer to \(6\).

The area is \(5\frac{3}{4}\) square units, and it is closer to \(6\) square units.

- Group pieces according to the fraction of a square unit each one covers. - Do not count a partial square as a full square unit.

1. The \(2\) half-square pieces combine to make \(1\) square unit. 2. The \(4\) quarter-square pieces combine to make \(1\) square unit. 3. The total area is \(7 + 1 + 1 = 9\) square units. Lucas counted pieces of different sizes as though each were one square unit.

The actual area is \(9\) square units.

- Combine the half-shaded parts before adding them to the fully shaded squares. - Use the total area to find the part that is not shaded.

1. The \(6\) half-shaded squares contribute \(6 \times \frac{1}{2} = 3\) square units of shaded area. 2. The total shaded area is \(5 + 3 = 8\) square units. 3. The unshaded area is \(12 - 8 = 4\) square units.

The shaded area is \(8\) square units, and the unshaded area is \(4\) square units.

- Track what happens to each board when the cards are moved. - Check whether moving cards changes the total amount of card surface.

1. Board A has \(14 - 4 = 10\) cards, so \(10\) square units are covered. 2. Board B has \(6 + 4 = 10\) cards, so \(10\) square units are covered. 3. Moving cards does not change their combined area: \(10 + 10 = 20\) square units.

a) Board A: \(10\) square units; Board B: \(10\) square units b) \(20\) square units

- Count the complete unit squares first. - Two half-square pieces make one square unit. - Add the full-square area and the area made by the paired halves.

1. There are \(15\) full unit squares inside the triangle. 2. The \(6\) half-square pieces make \(6 \div 2 = 3\) whole square units. 3. Add the areas: \(15 + 3 = 18\) square units.

The area of the triangle is \(18\) square units.

- Use the shaded reference square on the geoboard as the unit. - Find all the full squares first, then look at the slanted edges. - Two half-squares can form one full square unit.

1. Count \(8\) complete unit squares inside the shaded outline. 2. The slanted edges leave \(2\) triangular halves of a unit square. 3. Pair the halves into one whole square unit: \(8+\frac{1}{2}+\frac{1}{2}=9\).

\(9\) square units