- Read the side length from the diagram. - A square has the same length and width. - Multiply the side length by itself and use square meters.

1. The diagram shows a side length of \(15\,\text{m}\). 2. The area of a square is side length times side length: \(15 \times 15=225\). 3. The garden's area is \(225\,\text{m}^2\).

The area of the garden is \(225\,\text{m}^2\).

- Think about where Sam will place the edging. - Which measurement tells the distance around a shape? - How is that different from the space inside?

1. Edging goes along the outside boundary of the bed. 2. Perimeter is the distance around that boundary. Area measures the surface inside instead.

Use the perimeter because it measures the distance around the garden bed.

- Think about what carpet covers: a boundary or a surface. - Recall what area measures and what perimeter measures. - No side lengths are needed to make this decision.

1. Carpet covers the surface inside the room. 2. Area measures the amount of surface inside a shape, while perimeter measures distance around its boundary.

Area, because carpet covers the surface inside the room.

- Area can be found by counting the unit squares that cover the region. - Look for equal rows and equal numbers of cells in each row. - A multiplication expression can represent the rows of unit squares.

1. The rectangle has \(3\) rows with \(5\) square units in each row. 2. Its area is \(3 \times 5 = 15\) square units.

\(15\) square units

- Area tells how many square units cover the inside. - Use the two lengths marked on the rectangle.

1. The rectangle is four centimeters wide and two centimeters tall. 2. Its area is \(4\times2=8\,\text{cm}^2\).

\(8\,\text{cm}^2\)

- Follow the outside edge, not the squares inside. - Count the spaces along all four edges.

1. The shape has two horizontal sides of three units each and two vertical sides of two units each. 2. Add its boundary: \(3+2+3+2=10\) units.

\(10\) units

- Count how many squares fit in one row. - Use the number of rows to find the total without counting twice.

1. There are two rows with three squares in each row. 2. The area is \(2\times3=6\) square units.

\(6\) square units

- Is the question about one edge or the surface inside it? - Which unit names a small square rather than a line segment?

1. The amount of a surface covered is its area. 2. Area is measured in square units, so use \(\text{cm}^2\).

\(\text{cm}^2\), because the question is about area.

- Picture tracing the frame with a piece of string. - How many equal side lengths would the string follow?

1. The distance around is the perimeter. 2. Four sides of five centimeters make \(4\times5=20\,\text{cm}\).

\(20\,\text{cm}\)

- The question asks about the floor inside the rug's edges. - Use square feet in the answer, not feet.

1. The area of a rectangle is length times width. 2. The rug covers \(5\times3=15\,\text{ft}^2\).

\(15\,\text{ft}^2\)

- What makes a square different from a rectangle with unequal sides? - Check your side length by going all the way around the square.

1. Perimeter is the total of four equal sides. 2. One side is \(16\div4=4\,\text{cm}\).

\(4\,\text{cm}\)

- Find the distance around the entire rectangle. - No fencing is needed across the gate opening. - Subtract the opening from the full perimeter.

1. The garden perimeter is \(50+30+50+30=160\,\text{ft}\). 2. Subtract the gate opening: \(160-6=154\,\text{ft}\).

\(154\,\text{ft}\)

- Think of the playground as a rectangular array of square units. - Multiply the length by the width. - Break \(125\) into easier parts if needed.

1. Use the rectangle area formula: \(A=\ell \times w\). 2. Compute \(125 \times 80=(100 \times 80)+(20 \times 80)+(5 \times 80)=8000+1600+400=10{,}000\). 3. The playground area is \(10{,}000\,\text{ft}^2\).

\(10{,}000\,\text{ft}^2\)

- All four sides of a square are equal. - Perimeter counts the distance around the shape. - Area counts the square units inside.

1. Its perimeter is \(4\times14=56\,\text{in}\). 2. Its area is \(14\times14=196\,\text{in}^2\).

Perimeter: \(56\,\text{in}\). Area: \(196\,\text{in}^2\).

- Find the area of each separate rectangle. - Their areas do not overlap, so you can add them.

1. The first rectangle covers \(5\times2=10\,\text{ft}^2\). 2. The second covers \(3\times1=3\,\text{ft}^2\). 3. The areas add to \(10+3=13\,\text{ft}^2\).

\(13\,\text{ft}^2\)

- Find the first rectangle's area. - Use the same area for the second rectangle. - Treat the unknown width as the missing factor in \(10 \times ? = 60\).

1. The first rectangle's area is \(15 \times 4=60\,\text{cm}^2\). 2. The second rectangle also has area \(60\,\text{cm}^2\). Since \(10 \times 6=60\), its width is \(6\,\text{cm}\).

The width is \(6\,\text{cm}\).

- A rectangle has two of each side length. - Subtract the known side pair from the whole perimeter. - Share the rest equally between the other two sides.

1. The two \(8\)-foot sides total \(2\times8=16\,\text{ft}\). 2. The remaining sides total \(26-16=10\,\text{ft}\). 3. Each remaining side is \(10\div2=5\,\text{ft}\).

\(5\,\text{ft}\)

- Split each shape into small rectangles or single squares. - Count the rows and columns in each part. - Add all the parts without counting any square twice.

1. a) The lower \(4\times1\) rectangle covers \(4\) square units. Each upper arm covers \(1\times2=2\) more square units. Total: \(4+2+2=8\). 2. b) The figure consists of \(5\) unit squares, so its area is \(5\) square units.

a) \(8\) square units b) \(5\) square units

- Count the unit squares in each figure. - Split a complicated figure into rectangles. - Use the shaded square as one square unit.

1. Figure 1 is made of \(5\) unit squares, so its area is \(5\) square units. 2. Figure 2 is a \(2 \times 3\) rectangle, so its area is \(6\) square units. 3. Figure 3 can be split into a \(3 \times 1\) rectangle and a \(1 \times 2\) rectangle, so its area is \(3+2=5\) square units. 4. Figures 1 and 3 have the same area.

Figures 1 and 3; \(5\) square units

- Divide the figure horizontally into two rectangles. - Use the labeled width and height of each rectangle. - Add the two rectangle areas.

1. The lower rectangle measures \(24\,\text{ft}\times 9\,\text{ft}\), so its area is \(216\,\text{ft}^2\). 2. The upper rectangle measures \(6\,\text{ft}\times 12\,\text{ft}\), so its area is \(72\,\text{ft}^2\). 3. The total area is \(216+72=288\,\text{ft}^2\).

\(288\,\text{ft}^2\)

- Read both side lengths from the diagram. - Express both side lengths in centimeters before multiplying. - Use square centimeters for the area unit.

1. The diagram shows side lengths \(14\,\text{cm}\) and \(50\,\text{mm}\). Convert \(50\,\text{mm}=5\,\text{cm}\). 2. The area is \(14 \times 5=70\,\text{cm}^2\).

The area is \(70\,\text{cm}^2\).

- Read the width value and the length expression from the diagram. - Substitute the width value into the length expression. - Use both side lengths to find the distance around the rectangle.

1. The diagram shows \(w=8\,\text{ft}\) and the length as \(w+4\,\text{ft}\). 2. The length is \(8+4=12\,\text{ft}\). 3. The perimeter is \(2 \times (12+8)=40\,\text{ft}\).

The perimeter is \(40\,\text{ft}\).

- Read both side lengths from the diagram. - Multiply the two side lengths. - Use square centimeters for the area unit.

1. The diagram shows side lengths \(25\,\text{cm}\) and \(14\,\text{cm}\). 2. Multiply the length by the width: \(25 \times 14=350\). 3. The poster's area is \(350\,\text{cm}^2\).

The poster's area is \(350\,\text{cm}^2\).

- Follow the entire outside boundary once. - Combine horizontal segments and vertical segments separately if that helps. - Check that every boundary segment is counted exactly once.

1. The three horizontal step segments total \(2 + 2 + 2 = 6\,\text{cm}\), and the three vertical step segments also total \(6\,\text{cm}\). 2. The bottom side and left side are each \(6\,\text{cm}\). 3. Add the boundary lengths: \(6 + 6 + 6 + 6 = 24\,\text{cm}\).

The perimeter is \(24\,\text{cm}\).

- Find the area of each field using length times width. - Compare the two areas. - Subtract to find how much greater one area is.

1. Field a) has area \(60\,\text{m} \times 40\,\text{m} = 2400\,\text{m}^2\). 2. Field b) has area \(50\,\text{m} \times 50\,\text{m} = 2500\,\text{m}^2\). 3. Field b) is greater, and the difference is \(2500 - 2400 = 100\,\text{m}^2\).

Field b) is \(100\,\text{m}^2\) greater than Field a).

- Draw a line where the square room meets the larger rectangle. - Find each room's area separately. - Add the areas without counting shared edges.

1. The left rectangle measures \(12\,\text{ft}\) by \(15\,\text{ft}\), with area \(12\times15=180\,\text{ft}^2\). 2. The attached square is \(6\,\text{ft}\) by \(6\,\text{ft}\), with area \(36\,\text{ft}^2\). 3. Add the two areas: \(180+36=216\,\text{ft}^2\).

\(216\,\text{ft}^2\)

- Use the diagram to distinguish outside edges from shared interior edges. - Count how many small-triangle sides make up the outside boundary. - Equilateral means every small-triangle side has the same length.

1. Count only the small-triangle sides that lie on the outside boundary. There are \(6\) such sides. 2. Each outside side is \(8\,\text{cm}\), so the perimeter is \(6 \times 8=48\,\text{cm}\).

The perimeter of the figure is \(48\,\text{cm}\).

- Use the known area and one side to recover the missing side. - Check the missing side by multiplying the two side lengths. - Perimeter uses all four sides of the rectangle.

1. Since \(48 \div 6 = 8\), the length is \(8\,\text{ft}\). 2. The perimeter is \(2 \times (8 + 6) = 28\,\text{ft}\).

Length: \(8\,\text{ft}\) Perimeter: \(28\,\text{ft}\)

- Think of the display as six equal strips. - What width gives the stated number of square feet?

1. The area is length times width. 2. The missing width is \(24\div6=4\,\text{ft}\).

\(4\,\text{ft}\)

- Notice where the L-shape gets narrower. - Can you find the area without counting each square one at a time?

1. A complete rectangle four units wide and three units tall would contain \(12\) unit squares. 2. One unit square is missing from the top right, so the shape has area \(12-1=11\) square units.

\(11\) square units

- Decide which boundary is already covered by the wall. - Count only the sides that actually need a fence.

1. The side along the wall needs no fence. 2. Add the opposite length and both widths: \(8+5+5=18\,\text{ft}\).

\(18\,\text{ft}\)

- Adding the four side lengths measures what part of a rectangle? - Area counts square units covering the inside.

1. Adding \(6+2+6+2=16\) gives the perimeter, measured in feet. 2. The area is \(6\times2=12\,\text{ft}^2\).

Avery found the perimeter, \(16\,\text{ft}\). The area is \(12\,\text{ft}^2\).

- Use side length times side length for the area of a square. - Think of the number of square feet as a missing factor in a multiplication equation with \(144\,\text{in.}^2\). - Check that the numerical value becomes smaller when the same area is expressed in the larger unit.

1. The area is \(24 \times 24=576\,\text{in.}^2\). 2. Since \(4 \times 144\,\text{in.}^2=576\,\text{in.}^2\), the area is \(4\,\text{ft}^2\).

The area is \(576\,\text{in.}^2\), or \(4\,\text{ft}^2\).

- Find the distance around the schoolyard once. - Multiply by the number of laps in one recess. - Multiply the daily distance by the number of school days.

1. One lap is the perimeter of the rectangle: \(100+60+100+60=320\,\text{ft}\). 2. Two laps cover \(2 \times 320=640\,\text{ft}\). 3. In \(5\) days, Felix runs \(5 \times 640=3200\,\text{ft}\).

a) \(640\,\text{ft}\) b) \(3200\,\text{ft}\)

- A full lap follows the entire boundary of the rectangle. - Use the corrected lap length for the four-lap distance. - Express the kilometer comparison in the same unit.

1. A lap follows all four sides: \(20\,\text{m} + 10\,\text{m} + 20\,\text{m} + 10\,\text{m} = 60\,\text{m}\). Keisha added only one length and one width. 2. Four laps are \(4 \times 60\,\text{m} = 240\,\text{m}\). Since \(1\,\text{km} = 1000\,\text{m}\), \(240\,\text{m}\) is less than \(1\,\text{km}\).

1. One lap is \(60\,\text{m}\). Keisha forgot to include the opposite length and width. 2. Four laps are \(240\,\text{m}\), which is less than \(1\,\text{km}\).

- Think about the size of one kilometer compared with a classroom. - Decide which metric unit is reasonable for a room. - A walk around a rectangle follows how many sides?

1. The measurements are not reasonable. A classroom is not several kilometers long or wide; meters would be an appropriate unit. 2. The perimeter calculation is also incorrect. Using the stated measurements, the perimeter would be \(8 + 10 + 8 + 10 = 36\,\text{km}\), not \(18\,\text{km}\). The value \(18\) comes from adding only two sides.

1. Kilometers are not a reasonable unit for classroom dimensions; meters would be appropriate. 2. A perimeter includes all four sides. With the stated measurements, the perimeter would be \(36\,\text{km}\), not \(18\,\text{km}\).

- First find the width from the relationship in the problem. - Then use all four sides of the rectangle. - Check that opposite sides have equal lengths.

1. The width is \(10 \div 2=5\,\text{ft}\). 2. The perimeter is \(10+5+10+5=30\,\text{ft}\).

\(30\,\text{ft}\)

- A frame goes around all four sides. - Find the perimeter in inches first. - Use \(12\,\text{in}=1\,\text{ft}\) and a related multiplication fact to convert.

1. Find the perimeter: \(36+24+36+24=120\,\text{in}\). 2. Since \(10\times12\,\text{in}=120\,\text{in}\), \(120\,\text{in}=10\,\text{ft}\).

\(120\,\text{in}\), or \(10\,\text{ft}\)

- Find the area of each rectangle separately. - Multiply length by width. - Compare the two areas.

1. City Hall Plaza has area \(45 \times 60=2700\,\text{ft}^2\). 2. Theater Plaza has area \(55 \times 50=2750\,\text{ft}^2\). 3. Since \(2750>2700\), Theater Plaza has the greater area.

City Hall Plaza has area \(2700\,\text{ft}^2\). Theater Plaza has area \(2750\,\text{ft}^2\). Theater Plaza is larger.

- Find the area of the whole garden. - Find the area covered by the shed. - Subtract the shed area from the garden area. - Check that your answer is smaller than the total garden area.

1. The garden area is \(65 \times 40=2600\,\text{ft}^2\). 2. The shed area is \(12 \times 25=300\,\text{ft}^2\). 3. The uncovered area is \(2600-300=2300\,\text{ft}^2\).

\(2300\,\text{ft}^2\)

- Find the perimeter of the rectangle first. - A square has four equal sides. - Divide the square perimeter equally among its sides.

1. The rectangle perimeter is \(24+16+24+16=80\,\text{ft}\). 2. A square has four equal sides, so each side is \(80 \div 4=20\,\text{ft}\).

\(20\,\text{ft}\)

- Find the width first. - Calculate the full perimeter. - Compare the perimeter with the coverage from whole cans. - You must round up because a partial can cannot be purchased.

1. The width is \(120 \div 2=60\,\text{ft}\). 2. The perimeter is \(2 \times (120+60)=360\,\text{ft}\). 3. Two cans cover \(300\,\text{ft}\), which is not enough. Three cans cover \(450\,\text{ft}\), so \(3\) cans are needed.

\(3\) cans

- Picture the two crease lines after the paper is unfolded. - Find the side length of one small square. - A square has four equal sides.

1. Each crease halves a side of the large square, so a small square has side length \(16 \div 2=8\,\text{cm}\). 2. Its perimeter is \(4 \times 8=32\,\text{cm}\).

\(32\,\text{cm}\)

- Picture the square divided into three equal strips. - Find the length and width of one strip. - Calculate the two perimeters before comparing them.

1. The square perimeter is \(4\times18=72\,\text{ft}\). 2. One strip is \(18\,\text{ft}\) by \(18\div3=6\,\text{ft}\). 3. The strip perimeter is \(2\times(18+6)=48\,\text{ft}\). 4. The difference is \(72-48=24\,\text{ft}\).

a) \(48\,\text{ft}\) b) \(24\,\text{ft}\)

- Find how many \(1\)-foot-square pavers cover the walkway. - Compare the total number of pavers with whole packages of \(10\). - Choose the least whole number of packages that provides enough pavers.

1. The walkway needs \(12\) pavers along its length and \(4\) pavers across its width, so it needs \(12 \times 4=48\) pavers. 2. Four packages contain \(4 \times 10=40\) pavers, which is not enough. 3. Five packages contain \(5 \times 10=50\) pavers, which is enough. Therefore, \(5\) packages are needed.

\(5\) packages are needed.

- Each dimension gets a \(6\)-inch extension at both ends. - Combine the two \(6\)-inch extensions before converting them to feet. - Find the perimeter of the enlarged rectangle and compare it with \(31\,\text{ft}\).

1. Along each dimension, the border adds \(6\) inches at both ends. The two extensions total \(12\,\text{in.}=1\,\text{ft}\). 2. The outside length is \(8+1=9\,\text{ft}\), and the outside width is \(6+1=7\,\text{ft}\). 3. The outside perimeter is \(2 \times (9+7)=32\,\text{ft}\). 4. Since \(32\,\text{ft}>31\,\text{ft}\), the boards are not long enough.

No. The outside perimeter is \(32\,\text{ft}\), so the set is \(1\,\text{ft}\) too short.

- Find the dimensions after placing the squares side by side. - Convert all measurements to millimeters first. - Use the rectangle perimeter formula. - Convert the final result back to mixed units.

1. Convert the side length: \(9\,\text{cm}\,4\,\text{mm}=94\,\text{mm}\). 2. The rectangle is \(188\,\text{mm}\) long and \(94\,\text{mm}\) wide. 3. Its perimeter is \(2 \times (188+94)=564\,\text{mm}\). 4. Convert back: \(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\).

\(564\,\text{mm}=56\,\text{cm}\,4\,\text{mm}\)

- Use the square perimeter to find one side. - Cutting through the middle halves one dimension. - Find the perimeter of the new rectangle. - Check that you used two lengths and two widths.

1. The square side length is \(32 \div 4=8\,\text{cm}\). 2. Each small rectangle is \(8\,\text{cm}\) by \(4\,\text{cm}\). 3. Its perimeter is \(2 \times (8+4)=24\,\text{cm}\).

\(24\,\text{cm}\)

- Try rectangles with the same area but different sides. - Find more than one factor pair of \(25\). - Think about whether rearranging pieces changes their total covering.

1. a) False. A \(2\,\text{cm}\times8\,\text{cm}\) rectangle and a \(4\,\text{cm}\times4\,\text{cm}\) square each have area \(16\,\text{cm}^2\), but perimeters \(20\,\text{cm}\) and \(16\,\text{cm}\). 2. b) False. A \(1\,\text{cm}\times25\,\text{cm}\) rectangle has area \(25\,\text{cm}^2\) but is not a square. 3. c) True. Rearranging all the same pieces without gaps or overlaps keeps their combined area unchanged.

a) False b) False c) True

- Each little square has sides one unit long. - Count only outside edges in each arrangement. - Compare the areas and perimeters of both arrangements.

1. a) A row of four unit squares is a \(1\times4\) rectangle. Its perimeter is \(2\times(1+4)=10\) units. 2. b) The \(2\times2\) arrangement has perimeter \(2\times(2+2)=8\) units. 3. c) Both have area \(4\) square units, but their perimeters differ.

a) \(10\) units b) \(8\) units c) Shapes can have the same area but different perimeters.

- Think of a body measurement or familiar object with a length you know approximately. - Estimate the wall's width and height separately. - What gives the area of a rectangle from its dimensions?

1. Choose a familiar length whose approximate size you know, such as your arm span or the width of a door. 2. Estimate the wall's width and height by comparing them with that length. 3. Express both estimates in the same unit. 4. Multiply the estimated width by the estimated height. The result is an approximate area in square units.

One possible method is to use your arm span as a length estimate, find the wall's approximate width and height in the same unit, and multiply the estimates. Other reasonable methods are possible.

- Use the two dimensions of one rectangular space before working with the number of spaces. - Find how many spaces the lot contains. - Keep track of square feet when you scale the area of one space to all spaces.

1. The area of one space is \(9\times18=162\,\text{ft}^2\). 2. The number of spaces is \(8\times15=120\). 3. The total area is \(120\times162=19{,}440\,\text{ft}^2\).

a) \(162\,\text{ft}^2\) b) \(19{,}440\,\text{ft}^2\)

- Find the area of the entire playground from its dimensions. - Find the area of each rectangular activity section separately. - Compare the combined activity area with the whole playground area.

1. The playground area is \(80\times75=6000\,\text{ft}^2\). 2. The activity areas are \(25\times40=1000\,\text{ft}^2\), \(20\times30=600\,\text{ft}^2\), and \(50\times40=2000\,\text{ft}^2\). 3. Their total area is \(1000+600+2000=3600\,\text{ft}^2\). 4. The open play space is \(6000-3600=2400\,\text{ft}^2\).

\(2400\,\text{ft}^2\)

- First express both side lengths in the same unit. - Perimeter is the total distance around the rectangle. - Area is the number of square units it covers.

1. Convert \(50\,\text{mm}=5\,\text{cm}\). 2. Perimeter: \(2\times(12+5)=34\,\text{cm}\). 3. Area: \(12\times5=60\,\text{cm}^2\).

The perimeter is \(34\,\text{cm}\), and the area is \(60\,\text{cm}^2\).

- What number multiplied by itself makes the area? - Remember that all four sides of a square are equal. - Once you know one side, find the boundary length.

1. The square has side length \(8\,\text{ft}\), since \(8\times8=64\). 2. Its perimeter is \(4\times8=32\,\text{ft}\).

The perimeter is \(32\,\text{ft}\).

- A rectangle has two sides of each length. - Half the perimeter tells you the total of two adjacent sides. - Find the other side before finding the area.

1. Half the perimeter is one length plus one width: \(28\div2=14\,\text{ft}\). 2. The other side is \(14-6=8\,\text{ft}\). 3. The area is \(6\times8=48\,\text{ft}^2\).

The area is \(48\,\text{ft}^2\).

- Divide the perimeter equally among four sides. - Multiply the side length by itself to find area. - Use feet for length and square feet for area.

1. A square has four equal sides, so one side is \(48\div4=12\,\text{ft}\). 2. The area is \(12\times12=144\,\text{ft}^2\).

a) \(12\,\text{ft}\) b) \(144\,\text{ft}^2\)

- Convert the side length to inches before calculating. - A square has four equal sides. - Area counts square inches, not inches around the edge.

1. Convert the side length: \(2\,\text{ft}\ 1\,\text{in}=25\,\text{in}\). 2. The perimeter is \(4\times25=100\,\text{in}\). 3. The area is \(25\times25=625\,\text{in}^2\).

a) \(100\,\text{in}\) b) \(625\,\text{in}^2\)

- Use the stated rectangular model to estimate the area of one sheet first. - Break \(32\) into tens and ones when multiplying by the one-sheet area. - Use \(144\,\text{in.}^2=1\,\text{ft}^2\) to convert the estimated total area.

1. Using the stated estimation model, one sheet has area \(9\times12=108\,\text{in.}^2\). 2. For \(32\) sheets, \(108\times32=(108\times30)+(108\times2)=3240+216=3456\,\text{in.}^2\). 3. Since \(24\times144\,\text{in.}^2=3456\,\text{in.}^2\), the estimated total area is \(24\,\text{ft}^2\). 4. \(24\,\text{ft}^2\) is closer to \(21\,\text{ft}^2\) than to \(8\,\text{ft}^2\), so the sheets cover about the area of an interior door.

The sheets would cover an area closer to an interior door. Their estimated total area is \(24\,\text{ft}^2\).

- Find the area of one square note first. - Multiply by the number of notes. - Build the square-foot equivalent with convenient multiples of \(144\,\text{in.}^2\) instead of using a large divisor.

1. One note has area \(3 \times 3=9\,\text{in.}^2\). 2. The \(400\) notes have total area \(400 \times 9=3600\,\text{in.}^2\). 3. Twenty square feet is \(20 \times 144=2880\,\text{in.}^2\), and another \(5\) square feet is \(5 \times 144=720\,\text{in.}^2\). Since \(2880+720=3600\), the notes cover \(25\,\text{ft}^2\). 4. \(25\,\text{ft}^2\) is much closer to \(24\,\text{ft}^2\) than to \(2\,\text{ft}^2\).

The total area is closer to the area of the large wall mirror. The notes have a combined area of \(25\,\text{ft}^2\).

- First find the side length of the second square. - Use four equal sides for perimeter and side times side for area. - Compare the two areas with a multiplication equation rather than a division equation.

1. For the smaller square, the perimeter is \(4 \times 5=20\,\text{in.}\), and the area is \(5 \times 5=25\,\text{in.}^2\). 2. The larger square has side length \(2 \times 5=10\,\text{in.}\). Its perimeter is \(4 \times 10=40\,\text{in.}\), and its area is \(10 \times 10=100\,\text{in.}^2\). 3. Since \(4 \times 25\,\text{in.}^2=100\,\text{in.}^2\), the larger area is \(4\) times as great.

1) Smaller square: perimeter \(20\,\text{in.}\), area \(25\,\text{in.}^2\); larger square: perimeter \(40\,\text{in.}\), area \(100\,\text{in.}^2\) 2) The larger area is \(4\) times the smaller area.

- Each square has four sides of equal length. - Find a number that multiplied by itself gives Square B's area. - Compare the two side lengths after finding them.

1. Square A has side length \(20\div4=5\,\text{ft}\) and area \(5\times5=25\,\text{ft}^2\). 2. Square B has side length \(10\,\text{ft}\), since \(10\times10=100\), and perimeter \(4\times10=40\,\text{ft}\). 3. Square B has the longer side because \(10>5\).

Square A has area \(25\,\text{ft}^2\). Square B has perimeter \(40\,\text{ft}\) and has the longer sides.

- Half the perimeter equals the sum of two adjacent sides. - Think about which units can describe a side length. - Change millimeters to centimeters before comparing.

1. Half the perimeter is \(15\,\text{cm}\). One length and one width must total \(15\,\text{cm}\), so the length is \(15-8=7\,\text{cm}\). Choice 1 is too long. 2. Square centimeters measure area, not side length. 3. \(40\,\text{mm}=4\,\text{cm}\), which is too short. 4. A \(7\,\text{cm}\) length works because \(7+8=15\,\text{cm}\).

1) No; too long. 2) No; square centimeters are area units. 3) No; \(40\,\text{mm}=4\,\text{cm}\), too short. 4) Yes; the length is \(7\,\text{cm}\).

- Divide the perimeter among four equal sides. - Distinguish length units from square units. - Convert each proposed length to centimeters.

1. The correct side length is \(200\div4=50\,\text{cm}\), so choice 1) is correct. 2. Choice 2) measures area, not a side length. 3. Choice 3) is \(200\,\text{cm}\), equal to the entire perimeter rather than one side. 4. Choice 4) is \(500\,\text{mm}=50\,\text{cm}\), so it is correct.

Choices 2) and 3) are incorrect. Choice 2) gives an area unit; choice 3) gives the full perimeter.

- Half the perimeter is one length plus one width. - Subtract the known side from that total. - For the impossible width, consider just its two equal sides.

1. Half the perimeter is \(22\div2=11\,\text{ft}\). 2. a) The width is \(11-6=5\,\text{ft}\). 3. b) If the width were \(16\,\text{ft}\), its two sides would total \(32\,\text{ft}\), already more than the entire perimeter.

a) \(5\,\text{ft}\) b) Two \(16\)-foot sides total \(32\,\text{ft}\), exceeding the given perimeter.

- Use the area and known side to find the other side. - Add all four lengths once both sides are known. - Check that the answer for perimeter uses units of length.

1. The missing side is \(72\div9=8\,\text{cm}\). 2. The perimeter is \(2\times(9+8)=34\,\text{cm}\).

The missing side is \(8\,\text{cm}\), and the perimeter is \(34\,\text{cm}\).

- Divide the area by the given side to find each width. - Calculate both perimeters, not just the areas. - Compare the two perimeter lengths.

1. Rug A has width \(12\div4=3\,\text{ft}\), so its perimeter is \(2\times(4+3)=14\,\text{ft}\). 2. Rug B has width \(12\div6=2\,\text{ft}\), so its perimeter is \(2\times(6+2)=16\,\text{ft}\). 3. Rug B has the greater perimeter by \(16-14=2\,\text{ft}\).

Rug B has the greater perimeter by \(2\,\text{ft}\).

- Change the perimeter to centimeters first. - Half the perimeter equals the sum of two adjacent sides. - Find the width before calculating the area.

1. Convert \(400\,\text{mm}=40\,\text{cm}\). 2. Half the perimeter is \(40\div2=20\,\text{cm}\). 3. The width is \(20-12=8\,\text{cm}\). 4. The area is \(12\times8=96\,\text{cm}^2\).

Width: \(8\,\text{cm}\). Area: \(96\,\text{cm}^2\).

- First turn the entire perimeter into inches. - A square has four equal side lengths. - Once you know the side length, find its area.

1. The perimeter is \(10\times12+8=128\,\text{in}\). 2. One side is \(128\div4=32\,\text{in}\). 3. The area is \(32\times32=1024\,\text{in}^2\).

\(1024\,\text{in}^2\)

- Change the perimeter to centimeters first. - Half the perimeter is one length plus one width. - Find the missing side before the area.

1. Convert the perimeter: \(1\,\text{m}=100\,\text{cm}\). 2. Half the perimeter is \(100\div2=50\,\text{cm}\). 3. The other side is \(50-35=15\,\text{cm}\). 4. The area is \(35\times15=525\,\text{cm}^2\).

The other side is \(15\,\text{cm}\); the area is \(525\,\text{cm}^2\).

- Treat the shorter side as the missing factor in the area equation. - Check which number multiplied by \(30\) gives \(600\). - Use both side lengths to find the distance around the lot.

1. Since \(20 \times 30=600\), the shorter side is \(20\,\text{ft}\). 2. The perimeter is \(2 \times (30+20)=100\,\text{ft}\).

The shorter side is \(20\,\text{ft}\), and the perimeter is \(100\,\text{ft}\).

- Use the place value in \(200\) to simplify the area multiplication. - For part b), think of the number of plots as a missing factor. - Check your plot count by multiplying it by \(500\,\text{ft}^2\).

1. For the total area, \(150 \times 2=300\). Since \(200=2 \times 100\), \(150 \times 200=30{,}000\,\text{ft}^2\). 2. Since \(60 \times 500\,\text{ft}^2=30{,}000\,\text{ft}^2\), the area can be divided into \(60\) study plots.

a) \(30{,}000\,\text{ft}^2\) b) \(60\) study plots

- Treat the width as a missing factor in the area equation. - Use the place values in \(400\) and \(80{,}000\) to test a likely factor. - Use both dimensions to find the distance around the field.

1. Since \(200 \times 400=80{,}000\), the width is \(200\,\text{ft}\). 2. The perimeter is \(2 \times (400+200)=1200\,\text{ft}\).

a) \(200\,\text{ft}\) b) \(1200\,\text{ft}\) of fencing

- Break \(30\) into \(3 \times 10\) when finding the cleared area. - Compare the two areas in the same unit. - Subtract to find how much larger the replanted area is.

1. First, \(4000 \times 3=12{,}000\). Since \(30=3 \times 10\), the cleared area is \(120{,}000\,\text{ft}^2\). 2. Since \(150{,}000>120{,}000\), the replanted area is large enough. 3. The difference is \(150{,}000-120{,}000=30{,}000\,\text{ft}^2\).

a) \(120{,}000\,\text{ft}^2\) b) Yes. The replanted area is \(30{,}000\,\text{ft}^2\) larger.

- Each square has four sides of the same length. - Multiply a side by itself to find the square's area. - To work backward from perimeter, share the boundary equally among the four sides.

1. Square A's perimeter is \(4\times12=48\,\text{cm}\), and its area is \(12\times12=144\,\text{cm}^2\). 2. Square B's side length is \(36\div4=9\,\text{m}\), and its area is \(9\times9=81\,\text{m}^2\).

<table> <tr><th>Square</th><th>Side length</th><th>Perimeter</th><th>Area</th></tr> <tr><td>A</td><td>\(12\,\text{cm}\)</td><td>\(48\,\text{cm}\)</td><td>\(144\,\text{cm}^2\)</td></tr> <tr><td>B</td><td>\(9\,\text{m}\)</td><td>\(36\,\text{m}\)</td><td>\(81\,\text{m}^2\)</td></tr> </table>

- The wire's length becomes the distance around the first square. - What number times itself gives the second square's area? - Compare the two side lengths, not their different given measurements.

1. The wire forms the first square's perimeter, so one side is \(28\div4=7\,\text{in}\). 2. The second square has side length \(8\,\text{in}\), since \(8\times8=64\). 3. Because \(8>7\), the second square has the longer side.

The second square: \(8\,\text{in}\) compared with \(7\,\text{in}\) for the wire square.

- Find the first poster's side from its area. - Double that length for the second poster. - Find the new perimeter and area separately.

1. The first side is \(7\,\text{in}\), because \(7\times7=49\). 2. The second side is \(2\times7=14\,\text{in}\). 3. Its perimeter is \(4\times14=56\,\text{in}\). 4. Its area is \(14\times14=196\,\text{in}^2\).

Perimeter: \(56\,\text{in}\); area: \(196\,\text{in}^2\).

- Find the area of one square photo first. - Multiply by the number of photos. - Treat the number of square feet as the missing factor in a multiplication equation with \(144\,\text{in.}^2\).

1. One photo has area \(6 \times 6=36\,\text{in.}^2\). 2. All \(40\) photos have area \(40 \times 36=1440\,\text{in.}^2\). 3. Since \(10 \times 144\,\text{in.}^2=1440\,\text{in.}^2\), the total area is \(10\,\text{ft}^2\).

a) \(36\,\text{in.}^2\) b) \(1440\,\text{in.}^2\), or \(10\,\text{ft}^2\)

- Split each \(2.5\)-foot width into two feet and half a foot. - Use the width of each banner to find the entire row's length. - Find the area of a banner before multiplying by twelve.

1. a) Each banner is \(2\) feet plus half a foot wide. Twelve whole \(2\)-foot parts total \(24\,\text{ft}\); twelve half-foot parts total \(6\,\text{ft}\). Together that is \(30\,\text{ft}\). 2. b) One banner has area \(8\times2=16\,\text{ft}^2\) plus half of \(8\,\text{ft}^2\), another \(4\,\text{ft}^2\). That is \(20\,\text{ft}^2\) per banner. 3. All twelve banners cover \(12\times20=240\,\text{ft}^2\).

a) \(30\,\text{ft}\) b) \(240\,\text{ft}^2\)

- Find the area of the entire square. - Subtract the strip from both ends to find the inner square's side. - Compare inner and outer areas before counting flowers.

1. The full lawn has area \(10\times10=100\,\text{ft}^2\). 2. The inner square has side \(10-1-1=8\,\text{ft}\) and area \(8\times8=64\,\text{ft}^2\). 3. The flower beds cover \(100-64=36\,\text{ft}^2\). 4. The number of flowers is \(36\times8=288\).

\(288\) flowers

- Find the area of one sticker for part a). - For part b), first determine how many \(4\)-inch sticker sides fit along one \(24\)-inch poster side. - Use the same number of stickers in each row and column.

1. One sticker has area \(4 \times 4=16\,\text{in.}^2\). 2. Along one side of the poster, \(6\) stickers fit because \(6 \times 4\,\text{in.}=24\,\text{in.}\). 3. The square arrangement has \(6\) rows of \(6\) stickers, so \(6 \times 6=36\) stickers are needed.

a) \(16\,\text{in.}^2\) b) \(36\) stickers

- Put both dimensions in feet first. - Area is length multiplied by width in matching units.

1. Convert \(96\,\text{in}=8\,\text{ft}\). 2. The area is \(15\times8=120\,\text{ft}^2\).

\(120\,\text{ft}^2\)

- What number multiplied by itself makes the square's area? - A square has four equal sides.

1. Each side is \(100\,\text{ft}\), because \(100\times100=10{,}000\). 2. The perimeter is \(4\times100=400\,\text{ft}\).

\(400\,\text{ft}\)

- Find the square's perimeter. - The rectangle has the same distance around it. - Half a rectangle's perimeter is the total of two adjacent sides.

1. The square's perimeter is \(4\times8=32\,\text{ft}\). 2. Half that perimeter is \(32\div2=16\,\text{ft}\), one length plus one width of the rectangle. 3. The length is \(16-6=10\,\text{ft}\).

\(10\,\text{ft}\)

- First find the rectangle's area. - What side length makes a square with that area? - The square's perimeter is four times its side length.

1. a) The rectangle's area is \(16\times9=144\,\text{in}^2\). 2. The square has side length \(12\,\text{in}\), since \(12\times12=144\). 3. b) Its perimeter is \(4\times12=48\,\text{in}\).

a) \(144\,\text{in}^2\) b) \(48\,\text{in}\)

- Find the original area and then the new total area. - Use the unchanged length to find the new width. - Compare the original and new perimeters.

1. The original area is \(5 \times 3 = 15\,\text{ft}^2\). 2. The new area is \(15 + 10 = 25\,\text{ft}^2\). 3. The new width is \(25 \div 5 = 5\,\text{ft}\). 4. The original perimeter is \(2 \times (5 + 3) = 16\,\text{ft}\), and the new perimeter is \(2 \times (5 + 5) = 20\,\text{ft}\). 5. The perimeter increases by \(20 - 16 = 4\,\text{ft}\).

a) \(5\,\text{ft}\) b) The perimeter increases by \(4\,\text{ft}\).

- Find the area of each part separately. - Add the two areas. - Think about how many tile-sized square-foot groups fit that total.

1. The first rectangle covers \(12\times5=60\,\text{ft}^2\). 2. The square covers \(6\times6=36\,\text{ft}^2\). 3. Together they cover \(60+36=96\,\text{ft}^2\). 4. The number of tiles is \(96\div4=24\).

\(24\) tiles

- Split \(8.5\) into eight and one-half. - Find the area of the whole-number rectangle and the half-foot strip. - Compare the two areas using square feet.

1. a) The rectangle's area is \(8\times6=48\,\text{ft}^2\) plus half of \(6\,\text{ft}^2\), another \(3\,\text{ft}^2\), for \(51\,\text{ft}^2\). 2. The square has area \(7\times7=49\,\text{ft}^2\). 3. b) The rectangular rug is larger by \(51-49=2\,\text{ft}^2\).

a) Rectangle: \(51\,\text{ft}^2\); square: \(49\,\text{ft}^2\) b) The rectangle has \(2\,\text{ft}^2\) more area.

- Find the area of one poster first. - Use \(50=5 \times 10\) to simplify the multiplication by \(50\). - Build the square-foot equivalent with convenient multiples of \(144\,\text{in.}^2\).

1. One poster has area \(12 \times 18=216\,\text{in.}^2\). 2. First, \(216 \times 5=1080\). Since \(50=5 \times 10\), all \(50\) posters have area \(10{,}800\,\text{in.}^2\). 3. To convert, \(144 \times 70=10{,}080\) and \(144 \times 5=720\). Since \(10{,}080+720=10{,}800\), the total area is \(75\,\text{ft}^2\).

The posters cover \(75\,\text{ft}^2\).

- Use the array to determine the patio's side length. - Put the patio dimensions in feet before comparing areas. - Calculate both areas; matching one side length is not enough to decide the comparison.

1. Four pavers lie along each patio side, so the side length is \(4\times18=72\,\text{in}\). 2. Convert the patio side: \(72\,\text{in}=6\,\text{ft}\). 3. The patio area is \(6\times6=36\,\text{ft}^2\). 4. The shed area is \(5\times7=35\,\text{ft}^2\). 5. The patio is larger by \(36-35=1\,\text{ft}^2\).

The patio is larger by \(1\,\text{ft}^2\).

- Find the original perimeter first. - Write the doubled side lengths before finding the new perimeter. - Compare the two perimeters with a multiplication relationship.

1. The original perimeter is \(2 \times (6+4)=20\,\text{cm}\). 2. The new side lengths are \(12\,\text{cm}\) and \(8\,\text{cm}\). The new perimeter is \(2 \times (12+8)=40\,\text{cm}\). 3. Since \(2 \times 20=40\), doubling both side lengths doubles the perimeter.

a) \(20\,\text{cm}\) b) \(40\,\text{cm}\) c) The perimeter doubles when both side lengths are doubled.

- Think of the longer bed as the shorter one with an extra strip at one end. - What is the length of the extra strip? - Its area is given, so how could you find its width?

1. The longer bed has \(60-45=15\,\text{ft}\) of extra length. 2. Both beds have the same width, so the extra area is a rectangle \(15\,\text{ft}\) long with that common width. 3. Find a width that makes an area of \(375\,\text{ft}^2\). Since \(15\times20=300\) and \(15\times5=75\), the width is \(20+5=25\,\text{ft}\).

The common width is \(25\,\text{ft}\).

- Picture the full rectangle before the corner was cut out. - Find the width and height of the missing corner from the labels. - Subtract the missing area from the full rectangle's area.

1. A surrounding rectangle would measure \(10\,\text{ft}\) by \(8\,\text{ft}\). Its area is \(10\times8=80\,\text{ft}^2\). 2. The missing corner is \((10-4)=6\,\text{ft}\) by \((8-4)=4\,\text{ft}\), so its area is \(6\times4=24\,\text{ft}^2\). 3. The patio's area is \(80-24=56\,\text{ft}^2\).

\(56\,\text{ft}^2\)

- Use the labeled segments in the diagram to find the full rectangle dimensions. - Find the full perimeter, then account for the open gate segment shown in the diagram. - Whole rolls must provide at least as much fencing as the required length.

1. From the diagram, the garden length is \(12 + 4 + 14 = 30\,\text{ft}\), and the width is \(18\,\text{ft}\). 2. The full perimeter is \(2 \times (30 + 18) = 96\,\text{ft}\). 3. Leave the marked \(4\)-foot gate segment unfenced: \(96 - 4 = 92\,\text{ft}\) of fencing is needed. 4. Nine rolls provide \(90\,\text{ft}\), which is not enough. Ten rolls provide \(100\,\text{ft}\), so \(10\) rolls are required. 5. The total cost is \(10 \times \$45.00 = \$450.00\).

a) \(92\,\text{ft}\) of fencing b) \(10\) rolls, costing \(\$450.00\)

- Which rectangles could cover the playground without overlapping? - The middle bar reaches from the far left to the far right. - For the perimeter, follow the entire outside edge, including the corners where the arms meet.

1. The middle bar is \(12\,\text{ft}\) long and \(4\,\text{ft}\) high, so its area is \(12\times4=48\,\text{ft}^2\). 2. The upper and lower parts each measure \(4\,\text{ft}\times3\,\text{ft}\), adding \(12+12=24\,\text{ft}^2\). Total area: \(48+24=72\,\text{ft}^2\). 3. Going around the outside gives four \(4\)-foot edges, four \(3\)-foot edges, and four more \(4\)-foot edges: \(8\times4+4\times3=44\,\text{ft}\).

a) \(72\,\text{ft}^2\) b) \(44\,\text{ft}\)

- Trace and add every outer and inner boundary segment. - Use subtraction to find the unlabeled arm width and bottom-strip height. - Split the U-shape into three nonoverlapping rectangles and add their areas.

1. Add all eight boundary lengths: \(7+5+2+3+3+3+2+5=30\,\text{ft}\). 2. The left side width is \(7-2-3=2\,\text{ft}\), and the bottom strip height is \(5-3=2\,\text{ft}\). Split the shape into two \(2\,\text{ft}\times 5\,\text{ft}\) side rectangles and one \(3\,\text{ft}\times 2\,\text{ft}\) bottom rectangle. 3. The area is \(2\times(2\times 5)+3\times 2=20+6=26\,\text{ft}^2\).

a) \(30\,\text{ft}\) b) \(26\,\text{ft}^2\)

- Read the outer patio dimensions and inner pool dimensions from the diagram. - Subtract the pool area from the full patio area. - Use the tiled area, not the full patio area, when finding the cost.

1. From the diagram, the full patio measures \(12\,\text{ft}\times8\,\text{ft}\), so its area is \(12 \times 8=96\,\text{ft}^2\). 2. The pool measures \(6\,\text{ft}\times4\,\text{ft}\), so its area is \(6 \times 4=24\,\text{ft}^2\). 3. The tiled area is \(96-24=72\,\text{ft}^2\). 4. The total tile cost is \(72 \times \$35.00=\$2520.00\).

a) \(72\,\text{ft}^2\) b) \(\$2520.00\)

- Split the T-shape into two nonoverlapping rectangles. - Find each rectangle's dimensions. - Add the two areas.

1. Split the T-shape into a top rectangle measuring \(12\,\text{ft}\times 4\,\text{ft}\) and a lower rectangle measuring \(4\,\text{ft}\times 6\,\text{ft}\). 2. Their areas are \(12\times 4=48\,\text{ft}^2\) and \(4\times 6=24\,\text{ft}^2\). 3. The total area is \(48+24=72\,\text{ft}^2\).

The patio area is \(72\,\text{ft}^2\).

- Draw imaginary lines to split each shape into rectangles that do not overlap. - Use the grid marks to count each rectangle's rows and columns. - Add the areas of the parts.

1. a) Figure A consists of a \(5\times2\) lower rectangle and two \(2\times2\) upper rectangles. Its area is \(10+4+4=18\) square units. 2. b) Figure B consists of a \(5\times2\) middle rectangle and two \(3\times1\) rectangles. Its area is \(10+3+3=16\) square units.

a) \(18\) square units b) \(16\) square units

- Perimeter is the total distance around the outside boundary. - Count peg-to-peg segments once around each figure. - Do not skip or double-count an edge. - Each adjacent-peg segment is \(1\,\text{cm}\).

1. Figure A has side lengths \(4,3,1,2,2,2,1,3\), so its perimeter is \(4+3+1+2+2+2+1+3=18\,\text{cm}\). 2. Figure B has side lengths \(4,1,1,1,1,1,1,1,1,4\), so its perimeter is \(4+1+1+1+1+1+1+1+1+4=16\,\text{cm}\).

Figure A: \(18\,\text{cm}\) Figure B: \(16\,\text{cm}\)

- Trace the outside edge to find each perimeter. - Pay special attention to inward and outward corners. - Count unit squares or split each figure into rectangles. - Compare both perimeter and area.

1. Figure A is a \(4 \times 2\) rectangle. Its perimeter is \(4+2+4+2=12\) units, and its area is \(4 \times 2=8\) square units. 2. Figure B has perimeter \(3+2+2+2+1+4=14\) units. It can be split into a \(1 \times 4\) rectangle and a \(2 \times 2\) rectangle, so its area is \(4+4=8\) square units. 3. Figure B has the greater perimeter, but the two figures have equal areas.

a) Figure A: \(12\) units; Figure B: \(14\) units b) Figure B c) Each figure has area \(8\) square units. Equal-area figures can have different perimeters.

- For area, count unit squares or split the figure into rectangles. - For perimeter, trace the outside edge once. - Each grid segment is \(1\,\text{cm}\) long.

1. Split the figure into vertical strips containing \(3\), \(2\), and \(1\) unit squares. The area is \(3+2+1=6\,\text{cm}^2\). 2. Add the lengths around the outside boundary: \(3+1+1+1+1+1+1+3=12\,\text{cm}\).

a) \(6\,\text{cm}^2\) b) \(12\,\text{cm}\)

- Draw a line where the crossbar meets the stem. - Find the area of each separate rectangle. - Look for different factor pairs of the total area.

1. a) The stem is \(2\times4=8\) square units, and the crossbar is \(8\times2=16\) square units. 2. The total is \(8+16=24\) square units. 3. b) Two possible side-length pairs are \(3\) and \(8\), and \(4\) and \(6\), since both products equal \(24\).

a) \(24\) square units b) For example, \(3\) by \(8\) and \(4\) by \(6\)

- To find perimeter, count the grid-unit lengths along the boundary. - To find area, count the unit squares inside each figure. - Organize the perimeter and area values so you can compare them. - Think about how changing a rectangle from compact to long and narrow affects its area.

1. Figure 1 is a square. Its perimeter is \(3 + 3 + 3 + 3 = 12\) units, and its area is \(3 \times 3 = 9\) square units. 2. Figure 2 is a rectangle. Its perimeter is \(4 + 2 + 4 + 2 = 12\) units, and its area is \(4 \times 2 = 8\) square units. 3. Figure 3 is a rectangle. Its perimeter is \(5 + 1 + 5 + 1 = 12\) units, and its area is \(5 \times 1 = 5\) square units. 4. All three figures have the same perimeter, but their areas are different. 5. Figure 1 has the greatest area for the same perimeter.

a) Figure 1: perimeter \(12\) units, area \(9\) square units. Figure 2: perimeter \(12\) units, area \(8\) square units. Figure 3: perimeter \(12\) units, area \(5\) square units. b) The perimeters are equal, but the areas are different. c) Figure 1 encloses the greatest area.

- For area, count the squares inside each shape. - For perimeter, follow the outside edges only. - Compare what the two measurements tell you.

1. Figure 1 is a \(4\,\text{cm}\) by \(2\,\text{cm}\) rectangle. Its area is \(4\times2=8\,\text{cm}^2\) and its perimeter is \(2\times(4+2)=12\,\text{cm}\). 2. Figure 2 is made of a \(5\times1\) strip and a \(3\times1\) strip. Its area is \(5+3=8\,\text{cm}^2\). 3. Figure 2's boundary is \(5+1+2+1+3+2=14\,\text{cm}\). 4. Jordan is incorrect: the areas match, but the perimeters are different.

Figure 1: area \(8\,\text{cm}^2\), perimeter \(12\,\text{cm}\). Figure 2: area \(8\,\text{cm}^2\), perimeter \(14\,\text{cm}\). Jordan is not correct.

- Find the area and perimeter of each figure separately. - Split each L-shape into rectangles. - For perimeter, count only the outside edges.

1. Figure a) can be split into a \(4 \times 1\) rectangle and a \(1 \times 3\) rectangle. Its area is \(4+3=7\) square units. Its perimeter is \(4+1+3+3+1+4=16\) units. 2. Figure b) can be split into a \(3 \times 2\) rectangle and a \(2 \times 1\) rectangle. Its area is \(6+2=8\) square units. Its perimeter is \(3+2+1+1+2+3=12\) units. 3. Since \(8>7\), Figure b) has the greater area.

Figure b) has the greater area: \(8\) square units compared with \(7\) square units for Figure a). The perimeters are \(16\) units for Figure a) and \(12\) units for Figure b).

- Split the U-shape into three rectangles that do not overlap. - For perimeter, trace every edge around the indentation. - Check that the area is smaller than the surrounding full rectangle.

1. The two upright arms each cover \(1\times3=3\) square units. The base between them covers \(4\times1=4\) square units. 2. The total area is \(3+3+4=10\) square units. 3. Add the boundary, including the indentation: \(6+3+1+2+4+2+1+3=22\) length units.

Area: \(10\) square units Perimeter: \(22\) length units

- First count the boundary in grid segments and the area in grid squares. - Use the given peg spacing for boundary lengths. - Find the area of one grid square before scaling to the whole figure.

1. The boundary contains \(12\) grid segments, so the perimeter is \(12 \times 3\,\text{in} = 36\,\text{in}\). 2. The figure contains \(5\) grid squares. Each grid square has area \(3\,\text{in} \times 3\,\text{in} = 9\,\text{in}^2\). 3. The total area is \(5 \times 9\,\text{in}^2 = 45\,\text{in}^2\).

Perimeter: \(36\,\text{in}\) Area: \(45\,\text{in}^2\)

- Look for the three horizontal levels of the floor plan. - Each step changes the width of the rectangle above it. - Check that your rectangles cover the floor once, without gaps.

1. Split the floor into three horizontal rectangles, one at each level of the steps. 2. From bottom to top, their dimensions are \(12\,\text{ft}\times4\,\text{ft}\), \(8\,\text{ft}\times5\,\text{ft}\), and \(4\,\text{ft}\times3\,\text{ft}\). 3. Add the areas: \(48+40+12=100\,\text{ft}^2\).

\(100\,\text{ft}^2\)

- What rectangle would fill the whole space from one outer edge to the other? - Which part of that rectangle is not planted? - Check whether your answer is smaller than the area of the complete rectangle.

1. Complete the outline to a \(12\,\text{ft}\times10\,\text{ft}\) rectangle. Its area is \(120\,\text{ft}^2\). 2. The open section is \(8\,\text{ft}\times4\,\text{ft}\), with area \(32\,\text{ft}^2\). 3. Subtract: \(120-32=88\,\text{ft}^2\).

\(88\,\text{ft}^2\)

- Divide the figure into three equal-width rectangles. - Use the labeled step lengths to determine each rectangle's height. - Add the three rectangle areas.

1. Each step is \(8\,\text{in}\) wide and \(8\,\text{in}\) high. Decompose the panel into three vertical rectangles. 2. Their widths are each \(8\,\text{in}\), and their heights are \(24\,\text{in}\), \(16\,\text{in}\), and \(8\,\text{in}\). 3. The total area is \(8 \times 24 + 8 \times 16 + 8 \times 8 = 384\,\text{in}^2\).

\(384\,\text{in}^2\)

- Convert both side lengths to feet before finding the areas. - Find the area of the outer square and the square opening. - Subtract the opening's area from the outer square's area.

1. Convert the side lengths: \(36\,\text{in}=3\,\text{ft}\) and \(12\,\text{in}=1\,\text{ft}\). 2. The area of the outer square is \(3\,\text{ft}\times3\,\text{ft}=9\,\text{ft}^2\). 3. The area of the opening is \(1\,\text{ft}\times1\,\text{ft}=1\,\text{ft}^2\). 4. The remaining area is \(9-1=8\,\text{ft}^2\).

\(8\,\text{ft}^2\)

- Perimeter is the total distance around the figure. - Find the two unlabeled side lengths from the full dimensions. - Include all six boundary sides.

1. The missing side lengths are \(5-3=2\,\text{ft}\) and \(6-2=4\,\text{ft}\). 2. Add all six boundary sides: \(6+2+4+3+2+5=22\,\text{ft}\).

\(22\,\text{ft}\)

- Separate the tall strip on the left from the parts sticking out to the right. - How wide is each part beyond the left strip? - Count each square centimeter only once.

1. The upright strip is \(3\,\text{cm}\) wide and \(10\,\text{cm}\) tall. Its area is \(30\,\text{cm}^2\). 2. Each of the three bars sticks out \(9-3=6\,\text{cm}\) beyond the upright strip and is \(2\,\text{cm}\) high. 3. Add the three extensions: \(30+3\times(6\times2)=66\,\text{cm}^2\).

\(66\,\text{cm}^2\)

- Find the area of rectangle A first. - Split Area B into two rectangles. - Use the total height to find the missing height of the upper rectangle. - Compare the two total areas.

1. Area A is \(8\,\text{m} \times 6\,\text{m} = 48\,\text{m}^2\). 2. Split Area B into two rectangles. The lower rectangle has area \(10\,\text{m} \times 4\,\text{m} = 40\,\text{m}^2\). The upper rectangle has height \(7 - 4 = 3\,\text{m}\), so its area is \(4\,\text{m} \times 3\,\text{m} = 12\,\text{m}^2\). 3. Area B is \(40 + 12 = 52\,\text{m}^2\). 4. Since \(52 > 48\), Area B is greater and will need more grass seed.

Area B is greater. It has an area of \(52\,\text{m}^2\), compared with \(48\,\text{m}^2\) for Area A.

- Can you see two long rectangles with a short one between them? - The space between the top and bottom gaps gives the width of the middle walkway. - Make sure that no paved space is counted twice.

1. Each long path is \(4\,\text{ft}\) wide and \(10\,\text{ft}\) long, so the two paths cover \(2\times(4\times10)=80\,\text{ft}^2\). 2. The connecting walkway is \(6\,\text{ft}\) long. Its width is \(10-3-3=4\,\text{ft}\), so it covers \(6\times4=24\,\text{ft}^2\). 3. The three rectangles do not overlap. Add their areas: \(80+24=104\,\text{ft}^2\).

\(104\,\text{ft}^2\)

- Find the area of the main room before including the nook. - Which two measurements describe the part that sticks out from the wall? - Check that the two parts meet but do not overlap.

1. The main room measures \(14\,\text{ft}\times14\,\text{ft}\), so its area is \(196\,\text{ft}^2\). 2. The nook extends \(6\,\text{ft}\) from the wall and is \(5\,\text{ft}\) wide, so its area is \(30\,\text{ft}^2\). 3. Add: \(196+30=226\,\text{ft}^2\).

\(226\,\text{ft}^2\)

- Use subtraction to find the unlabeled left width. - Add the labeled vertical steps to find each section's height. - Divide the stepped shape into three rectangles and add their areas.

1. The left section is \(36-12-9=15\,\text{ft}\) wide. The three section heights are \(6+6+12=24\,\text{ft}\), \(6+6=12\,\text{ft}\), and \(6\,\text{ft}\). 2. Divide the floor plan into three vertical rectangles measuring \(15\,\text{ft}\times 24\,\text{ft}\), \(12\,\text{ft}\times 12\,\text{ft}\), and \(9\,\text{ft}\times 6\,\text{ft}\). 3. Their areas are \(360\,\text{ft}^2\), \(144\,\text{ft}^2\), and \(54\,\text{ft}^2\). 4. The total area is \(360+144+54=558\,\text{ft}^2\).

\(558\,\text{ft}^2\)

- Look for two rectangles that meet along one edge. - Find each part's area before combining them. - The final unit is square feet, not feet.

1. The wide end is a rectangle measuring \(9\,\text{ft}\times3\,\text{ft}\). Its area is \(27\,\text{ft}^2\). 2. The narrow section measures \(4\,\text{ft}\times5\,\text{ft}\), with area \(20\,\text{ft}^2\). 3. Add the two parts: \(27+20=47\,\text{ft}^2\).

\(47\,\text{ft}^2\)

- Complete the shape to make one large rectangle. - Find the dimensions of the missing rectangle. - Subtract the missing area.

1. View the floor plan as a \(90\,\text{ft}\times 45\,\text{ft}\) rectangle with a \(30\,\text{ft}\times 15\,\text{ft}\) rectangle removed. 2. The large rectangle has area \(90\times 45=4050\,\text{ft}^2\). The missing rectangle has area \(30\times 15=450\,\text{ft}^2\). 3. The hall's area is \(4050-450=3600\,\text{ft}^2\).

\(3600\,\text{ft}^2\)

- Imagine a horizontal segment that makes two rectangles. - Use the labeled width and height of each rectangle. - Add the two areas.

1. The lower rectangle measures \(36\,\text{ft}\times 15\,\text{ft}\), so its area is \(540\,\text{ft}^2\). 2. The upper rectangle measures \(12\,\text{ft}\times 21\,\text{ft}\), so its area is \(252\,\text{ft}^2\). 3. The total area is \(540+252=792\,\text{ft}^2\).

a) The lower rectangle has area \(540\,\text{ft}^2\), and the upper rectangle has area \(252\,\text{ft}^2\). b) The total area is \(792\,\text{ft}^2\).

- Read the square's side length from the diagram. - Convert meters to centimeters before finding the area. - Use \(120=12 \times 10\) to organize the multiplication with place value.

1. The diagram shows a side length of \(1.2\,\text{m}\). Convert: \(1.2\,\text{m}=120\,\text{cm}\). 2. Write \(120=12 \times 10\). Then the area is \((12 \times 12) \times (10 \times 10)=144 \times 100=14{,}400\,\text{cm}^2\).

The area is \(14{,}400\,\text{cm}^2\).

- The rectangles have different units written on their sides. - Express all side lengths in one unit before multiplying. - Compare the resulting areas in square units.

1. Use millimeters. Rectangle A measures \(80\,\text{mm}\) by \(25\,\text{mm}\), so its area is \(80\times25=2000\,\text{mm}^2\). 2. Rectangle B measures \(40\,\text{mm}\) by \(60\,\text{mm}\), so its area is \(40\times60=2400\,\text{mm}^2\). 3. Since \(2400>2000\), Rectangle B has the greater area.

Rectangle B has the greater area: \(2400\,\text{mm}^2\) compared with \(2000\,\text{mm}^2\) for A.

- Trace only the outside edges, not the line where the beds meet. - Some short sides are missing labels. Use the matching horizontal and vertical distances to find them. - The cost per foot includes dollars and cents; you can work with those parts separately.

1. The missing left step is \(8-4=4\,\text{ft}\), so the outside edges are \(8\), \(2\), \(4\), \(2\), \(8\), \(2\), \(4\), and \(2\) feet long. 2. The perimeter is \(8+2+4+2+8+2+4+2=32\,\text{ft}\). 3. At \(\$12\) per foot, the cost is \(32\times12=384\) dollars. The extra \(50\) cents per foot adds \(32\times50=1600\) cents, or \(\$16\). Total: \(\$384+\$16=\$400\).

\(\$400.00\)

- Divide the floor plan into two nonoverlapping rectangles. - Add their areas. - Interpret \(\$0.50\) as one-half dollar and find half of the total area.

1. Decompose the floor plan into a \(15\,\text{ft}\times 12\,\text{ft}\) rectangle and a \(9\,\text{ft}\times 6\,\text{ft}\) rectangle. 2. The areas are \(15\times 12=180\,\text{ft}^2\) and \(9\times 6=54\,\text{ft}^2\), for a total of \(234\,\text{ft}^2\). 3. A rate of \(\$0.50\) per square foot is one-half dollar per square foot. Half of \(234\) is \(117\), so the cost is \(\$117.00\).

\(\$117.00\)

- Start with the area of the full panel. Which rectangle is missing? - Cutting inward creates two new horizontal boundary pieces. - Follow the cut edges as carefully as the four outside edges.

1. The full rectangle measures \(10\,\text{cm}\times8\,\text{cm}\), so its area is \(80\,\text{cm}^2\). 2. The cutout is \(3\,\text{cm}\times2\,\text{cm}\). The remaining area is \(80-6=74\,\text{cm}^2\). 3. The cut replaces a \(2\)-cm outer edge with a path that goes \(3\) cm in, \(2\) cm up, and \(3\) cm out. The perimeter is \(2\times(10+8)+2\times3=42\,\text{cm}\).

Perimeter: \(42\,\text{cm}\) Area: \(74\,\text{cm}^2\)

- Follow each edge around the outside, including the edges between the bars. - The wide top bar and the narrow upright share part of their width. - Check that you have counted all the horizontal and vertical edges.

1. The unlabeled upper inside edge is \(9-3=6\,\text{cm}\). The lower edge of the middle bar is \(4\,\text{cm}\). 2. The perimeter has horizontal lengths \(9\), \(6\), \(4\), \(4\), and \(3\) cm, for a total of \(26\,\text{cm}\). 3. Its vertical lengths are \(2\), \(2\), \(2\), \(4\), and \(10\) cm, totaling \(20\,\text{cm}\). 4. Add: \(26+20=46\,\text{cm}\).

\(46\,\text{cm}\)

- Which two rectangles make up the floor plan? - The connection between the rooms is inside the floor plan, not part of its outside boundary. - Check that the perimeter uses feet and the area uses square feet.

1. The large rectangle is \(18\,\text{ft}\times12\,\text{ft}\), with area \(216\,\text{ft}^2\). 2. The part sticking out is \(6\,\text{ft}\times9\,\text{ft}\), with area \(54\,\text{ft}^2\). Total area: \(216+54=270\,\text{ft}^2\). 3. Add the outside edges: \(6+9+6+12+18+12+6+9=78\,\text{ft}\).

Area: \(270\,\text{ft}^2\) Perimeter: \(78\,\text{ft}\)

- The shape can be covered by a top bar, a bottom bar, and a narrow middle strip. - The middle strip is narrower than the bars. Which widths tell you how much narrower? - For the perimeter, include the edges that turn inward toward the middle.

1. The top and bottom bars each measure \(11\,\text{cm}\times2\,\text{cm}\), so together they cover \(44\,\text{cm}^2\). 2. The middle strip is \(11-4-4=3\,\text{cm}\) wide and \(6\,\text{cm}\) high. Its area is \(18\,\text{cm}^2\). The total area is \(44+18=62\,\text{cm}^2\). 3. Around the outside, the horizontal edges total \(2\times11+4\times4=38\,\text{cm}\), and the vertical edges total \(4\times2+2\times6=20\,\text{cm}\). 4. The perimeter is \(38+20=58\,\text{cm}\).

Area: \(62\,\text{cm}^2\) Perimeter: \(58\,\text{cm}\)

- Start by drawing an imaginary rectangle around the whole U. - Subtract the empty part when finding area. - Include both inside arms of the U when finding perimeter.

1. The surrounding rectangle covers \(8\times6=48\,\text{cm}^2\). 2. The opening measures \(4\,\text{cm}\) by \(4\,\text{cm}\), so the remaining area is \(48-16=32\,\text{cm}^2\). 3. Count the full boundary, including the opening: \(8+6+2+4+4+4+2+6=36\,\text{cm}\).

Area: \(32\,\text{cm}^2\) Perimeter: \(36\,\text{cm}\)

- The floor would be one large rectangle without the closets. - The two missing corner spaces have different dimensions. - Check that you subtract each closet only once.

1. The full room is \((20+4)=24\,\text{ft}\) wide and \((11+4)=15\,\text{ft}\) long, with area \(24\times15=360\,\text{ft}^2\). 2. One closet takes \(4\times3=12\,\text{ft}^2\); the other takes \(6\times4=24\,\text{ft}^2\). 3. Subtract both closet areas: \(360-12-24=324\,\text{ft}^2\).

\(324\,\text{ft}^2\)

- Imagine a line that cuts the shape into two rectangles. - Use the marked lengths to find each rectangle's area. - Add the two areas without overlap.

1. One way is a \(10\,\text{cm}\) by \(4\,\text{cm}\) lower rectangle and a \(4\,\text{cm}\) by \(8\,\text{cm}\) upper rectangle. 2. The total area is \(10\times4+4\times8=40+32=72\,\text{cm}^2\). 3. Another way gives \(4\times12+6\times4=48+24=72\,\text{cm}^2\), confirming the result.

\(72\,\text{cm}^2\)

- Use subtraction to find any missing side length. - Find area by multiplying side lengths or subtracting a missing square. - For perimeter, add every outside edge. - Equal areas do not necessarily mean equal perimeters.

1. Figure 1 is a rectangle, so its area is \(4\,\text{cm} \times 3\,\text{cm} = 12\,\text{cm}^2\). 2. In Figure 2, the missing right-side length is \(4 - 2 = 2\,\text{cm}\), and the missing inward horizontal length is also \(4 - 2 = 2\,\text{cm}\). 3. Figure 2 is a \(4\,\text{cm} \times 4\,\text{cm}\) square with a \(2\,\text{cm} \times 2\,\text{cm}\) square removed. Its area is \(16 - 4 = 12\,\text{cm}^2\). 4. The figures have the same area. 5. The perimeter of Figure 1 is \(2 \times (4 + 3) = 14\,\text{cm}\). 6. The perimeter of Figure 2 is \(4 + 2 + 2 + 2 + 2 + 4 = 16\,\text{cm}\). 7. Figure 2 has the greater perimeter.

a) Both figures have an area of \(12\,\text{cm}^2\). b) Figure 1 has a perimeter of \(14\,\text{cm}\), and Figure 2 has a perimeter of \(16\,\text{cm}\). Figure 2 has the greater perimeter.

- Use the diagram to identify which side becomes an interior shared edge. - Interior segments are not part of the perimeter. - Add only the labeled side lengths that remain on the outside boundary.

1. The diagram shows that the two sides labeled \(b\) are joined and lie inside the combined figure, so they are not part of the perimeter. 2. Each tile contributes sides \(a\), \(c\), and \(d\) to the outside boundary. 3. The perimeter is \(2 \times (5+7+3)=30\,\text{cm}\).

The perimeter of the combined figure is \(30\,\text{cm}\).

- Read the full garden dimensions and the three-way partitions from the diagram. - Use the equal partitions to find one small bed's dimensions. - Use those dimensions for both the area and the center-bed perimeter.

1. From the diagram, the garden is \(12\,\text{ft}\) by \(9\,\text{ft}\), with each side split into \(3\) equal parts. The small-bed dimensions are \(12 \div 3=4\,\text{ft}\) and \(9 \div 3=3\,\text{ft}\). 2. One bed has area \(4 \times 3=12\,\text{ft}^2\). 3. The center bed's perimeter is \(2 \times (4+3)=14\,\text{ft}\).

a) \(4\,\text{ft}\) by \(3\,\text{ft}\) b) \(12\,\text{ft}^2\) c) \(14\,\text{ft}\)

- Read all five side lengths from the diagram. - Add the side lengths before converting units. - Use \(10\,\text{mm}=1\,\text{cm}\) to express the total in centimeters.

1. Add the five side lengths shown: \(50+40+35+50+35=210\,\text{mm}\). 2. Since \(21 \times 10\,\text{mm}=210\,\text{mm}\), the perimeter is \(21\,\text{cm}\).

The perimeter is \(21\,\text{cm}\).

- Find area and perimeter separately for each rectangle. - Compare like quantities: area with area and perimeter with perimeter. - Use the two results to decide whether equal perimeter determines area.

1. Rectangle A has perimeter \(2 \times (6 + 4) = 20\,\text{ft}\) and area \(6 \times 4 = 24\,\text{ft}^2\). 2. Rectangle B has perimeter \(2 \times (8 + 2) = 20\,\text{ft}\) and area \(8 \times 2 = 16\,\text{ft}^2\). 3. Rectangle A has \(24 - 16 = 8\,\text{ft}^2\) more area. 4. Equal perimeters do not force rectangles to have equal areas.

Rectangle A has the greater area by \(8\,\text{ft}^2\). Rectangles can have the same perimeter but different areas.

- Find the covered space for both gardens. - Then separately add the lengths around their boundaries.

1. Their areas are \(2\times6=12\,\text{m}^2\) and \(3\times4=12\,\text{m}^2\), so the areas match. 2. Their perimeters are \(2+6+2+6=16\,\text{m}\) and \(3+4+3+4=14\,\text{m}\), so the perimeters differ.

Same area: yes, \(12\,\text{m}^2\) each. Same perimeter: no, \(16\,\text{m}\) versus \(14\,\text{m}\).

- First find the area the outside rectangle would cover. - The hole does not count as part of the tile.

1. The outer rectangle is \(6\times5=30\,\text{cm}^2\). 2. The opening is \(2\times3=6\,\text{cm}^2\), so \(30-6=24\,\text{cm}^2\) remains.

\(24\,\text{cm}^2\)

- Which sides have known lengths, and which ones are still unknown? - How many copies of the unknown length go around the rectangle?

1. The two widths total \(2\times4=8\,\text{cm}\). 2. The two lengths total \(26-8=18\,\text{cm}\). Each length is \(18\div2=9\,\text{cm}\).

\(9\,\text{cm}\)

- The wire forms the entire perimeter. - A square has four equal sides. - For a rectangle, the length plus width is half the perimeter. - In part c), first consider the two opposite long sides.

1. For the square, divide the perimeter among four equal sides: \(60 \div 4=15\,\text{cm}\). 2. For the rectangle, the length and width together equal half the perimeter: \(60 \div 2=30\,\text{cm}\). The width is \(30-20=10\,\text{cm}\). 3. Two sides of length \(35\,\text{cm}\) would require \(2 \times 35=70\,\text{cm}\), which is already longer than the wire.

a) \(15\,\text{cm}\) b) \(10\,\text{cm}\) c) Two \(35\,\text{cm}\) sides require \(70\,\text{cm}\), but the wire is only \(60\,\text{cm}\) long.

- Use \(P=4s\) and \(A=s \times s\) to complete the table. - Compare the first and third columns with multiplication factors. - Check whether the perimeter factor and area factor are the same or different.

1. For \(s=3\,\text{in.}\), \(P=4 \times 3=12\,\text{in.}\) and \(A=3 \times 3=9\,\text{in.}^2\). 2. For \(s=6\,\text{in.}\), \(P=4 \times 6=24\,\text{in.}\) and \(A=6 \times 6=36\,\text{in.}^2\). 3. For \(s=9\,\text{in.}\), \(P=4 \times 9=36\,\text{in.}\) and \(A=9 \times 9=81\,\text{in.}^2\). 4. The side length is multiplied by \(3\). The perimeter is also multiplied by \(3\), because \(3 \times 12=36\). The area is multiplied by \(9\), because \(9 \times 9=81\).

<table> <tr><td>Side length \(s\)</td><td>\(3\,\text{in.}\)</td><td>\(6\,\text{in.}\)</td><td>\(9\,\text{in.}\)</td></tr> <tr><td>Perimeter \(P\)</td><td>\(12\,\text{in.}\)</td><td>\(24\,\text{in.}\)</td><td>\(36\,\text{in.}\)</td></tr> <tr><td>Area \(A\)</td><td>\(9\,\text{in.}^2\)</td><td>\(36\,\text{in.}^2\)</td><td>\(81\,\text{in.}^2\)</td></tr> </table> The perimeter is multiplied by \(3\), and the area is multiplied by \(9\).

- Area is the product of length and width. - Think about how each side-length change affects one factor in that product. - In each case, decide whether the two side changes reinforce each other, offset each other, or act only once before calculating.

1. Tripling one side while keeping the other side unchanged triples the area: \(48\times3=144\,\text{cm}^2\). 2. Doubling the length first doubles the area to \(96\,\text{cm}^2\). Halving the width then halves the area back to \(48\,\text{cm}^2\). 3. Halving one side halves the area from \(48\,\text{cm}^2\) to \(24\,\text{cm}^2\). Halving the other side halves it again: \(24\div2=12\,\text{cm}^2\).

1) \(144\,\text{cm}^2\) 2) \(48\,\text{cm}^2\) 3) \(12\,\text{cm}^2\)

- Find the original area and then the target area. - Look for different factor pairs of the target area. - Check each proposed pair by multiplying its dimensions.

1. The original area is \(4 \times 3 = 12\,\text{m}^2\), so the new area must be \(4 \times 12 = 48\,\text{m}^2\). 2. One option is \(16\,\text{m} \times 3\,\text{m}\), which has area \(48\,\text{m}^2\). 3. Another option is \(8\,\text{m} \times 6\,\text{m}\), which also has area \(48\,\text{m}^2\).

Possible answers include \(16\,\text{m}\) by \(3\,\text{m}\), \(4\,\text{m}\) by \(12\,\text{m}\), and \(8\,\text{m}\) by \(6\,\text{m}\). Any two valid pairs are acceptable.

- Express all sides in inches first. - Find the area of each rectangle. - Compare their perimeters separately from their areas.

1. Rectangle B's longer side is \(2\,\text{ft}\ 6\,\text{in}=30\,\text{in}\). 2. The areas are \(15\times20=300\,\text{in}^2\) and \(30\times10=300\,\text{in}^2\), so the areas are equal. 3. The perimeters are \(2\times(15+20)=70\,\text{in}\) and \(2\times(30+10)=80\,\text{in}\). 4. Rectangle B has the greater perimeter.

Their areas are equal at \(300\,\text{in}^2\) each. Rectangle B has the greater perimeter, \(80\,\text{in}\) compared with \(70\,\text{in}\).

- Find the square's area first. - Use that area to find the rectangle's unknown length. - Compare the lengths around each shape.

1. The square's area is \(20\times20=400\,\text{in}^2\). 2. The rectangle's length is \(400\div4=100\,\text{in}\). 3. The square's perimeter is \(4\times20=80\,\text{in}\). 4. The rectangle's perimeter is \(2\times(100+4)=208\,\text{in}\). 5. The rectangle has the greater perimeter.

The rectangle is \(100\,\text{in}\) long and has perimeter \(208\,\text{in}\), compared with \(80\,\text{in}\) for the square. The rectangle has the greater perimeter.

- For a missing side from an area, think of the side as a missing factor. - Half the perimeter equals one length plus one width. - Convert kilometers to meters before using the perimeter.

1. Rectangle 1: Since \(8 \times 12=96\), \(w=8\,\text{cm}\). Then \(P=2 \times (12+8)=40\,\text{cm}\). 2. Rectangle 2: Half of \(140\,\text{m}\) is \(70\,\text{m}\), so \(w=70-25=45\,\text{m}\). The area is \(25 \times 45=1125\,\text{m}^2\). 3. Rectangle 3: Since \(30 \times 50=1500\), \(w=30\,\text{m}\). Then \(P=2 \times (50+30)=160\,\text{m}\). 4. Rectangle 4: \(1\,\text{km}=1000\,\text{m}\), so half the perimeter is \(500\,\text{m}\). Thus \(w=500-200=300\,\text{m}\). For the area, \(200 \times 3=600\), then multiplying by \(100\) gives \(200 \times 300=60{,}000\,\text{m}^2\).

<table> <tr><th>Rectangle</th><th>Length \(l\)</th><th>Width \(w\)</th><th>Perimeter \(P\)</th><th>Area \(A\)</th></tr> <tr><td>1</td><td>\(12\,\text{cm}\)</td><td>\(8\,\text{cm}\)</td><td>\(40\,\text{cm}\)</td><td>\(96\,\text{cm}^2\)</td></tr> <tr><td>2</td><td>\(25\,\text{m}\)</td><td>\(45\,\text{m}\)</td><td>\(140\,\text{m}\)</td><td>\(1125\,\text{m}^2\)</td></tr> <tr><td>3</td><td>\(50\,\text{m}\)</td><td>\(30\,\text{m}\)</td><td>\(160\,\text{m}\)</td><td>\(1500\,\text{m}^2\)</td></tr> <tr><td>4</td><td>\(200\,\text{m}\)</td><td>\(300\,\text{m}\)</td><td>\(1\,\text{km}\)</td><td>\(60{,}000\,\text{m}^2\)</td></tr> </table>

- Half a rectangle's perimeter equals its length plus its width. - Express the sides of Rectangle B in centimeters. - Compare the three areas after finding their values.

1. A: Half the perimeter is \(11\,\text{cm}\), so width \(11-6=5\,\text{cm}\) and area \(6\times5=30\,\text{cm}^2\). 2. B: Convert \(80\,\text{mm}=8\,\text{cm}\); its area is \(8\times2=16\,\text{cm}^2\). 3. C: Half the perimeter is \(9\,\text{cm}\), so width \(9-5=4\,\text{cm}\) and area \(5\times4=20\,\text{cm}^2\). 4. A has the greatest area because \(30>20>16\).

Rectangle A: width \(5\,\text{cm}\), area \(30\,\text{cm}^2\) Rectangle B: area \(16\,\text{cm}^2\) Rectangle C: width \(4\,\text{cm}\), area \(20\,\text{cm}^2\) Rectangle A has the greatest area.

- Treat Plot 1's missing width as a factor that must make the given area. - Use half of Plot 2's perimeter to find its missing length. - Compare both perimeters and both areas after the missing dimensions are known.

1. For Plot 1, \(20 \times 24=480\), so the width is \(20\,\text{ft}\). Its perimeter is \(2 \times (24+20)=88\,\text{ft}\). 2. For Plot 2, half the perimeter is \(46\,\text{ft}\). Its length is \(46-20=26\,\text{ft}\), so its area is \(26 \times 20=520\,\text{ft}^2\). 3. Plot 2 needs more fencing because \(92\,\text{ft}>88\,\text{ft}\), and it has the greater area because \(520\,\text{ft}^2>480\,\text{ft}^2\).

Plot 2 needs more fencing and has the greater area. Its perimeter is \(92\,\text{ft}\), and its area is \(520\,\text{ft}^2\).

- What must the length and width add to for this perimeter? - Try a very short width and a longer length. - Check both the perimeter and area of the example you choose.

1. If the perimeter is \(20\,\text{ft}\), then one length plus one width must be \(10\,\text{ft}\). 2. Choose \(1\,\text{ft}\) and \(9\,\text{ft}\). The perimeter is \(2\times(1+9)=20\,\text{ft}\). 3. The area is \(1\times9=9\,\text{ft}^2\), which is less than \(10\,\text{ft}^2\).

Yes. A \(1\,\text{ft}\) by \(9\,\text{ft}\) rectangle has perimeter \(20\,\text{ft}\) and area \(9\,\text{ft}^2\).

- Find the square's side length from its area. - Find Rectangle B's length from its width and area. - Compare the distances around the two shapes.

1. a) A has sides \(10\,\text{in}\) long, since \(10\times10=100\). Its perimeter is \(4\times10=40\,\text{in}\). 2. b) B must be \(100\,\text{in}\) long. Its perimeter is \(2\times(100+1)=202\,\text{in}\). 3. c) The rectangles have equal areas but very different perimeters.

a) \(40\,\text{in}\) b) \(202\,\text{in}\) c) Equal-area rectangles can have different perimeters.

- Half the perimeter equals one length plus one width. - Model the length as five copies of the width. - Find the dimensions before the area.

1. Half the perimeter is \(96\div2=48\,\text{ft}\). That is one width plus five widths, or six equal width parts. 2. The width is \(48\div6=8\,\text{ft}\). 3. The length is \(5\times8=40\,\text{ft}\). 4. The area is \(40\times8=320\,\text{ft}^2\).

Length: \(40\,\text{ft}\); width: \(8\,\text{ft}\); area: \(320\,\text{ft}^2\).

- Think of the rectangle as three equal squares in a row. - Divide the area equally among the squares. - Use one square's side to work out both dimensions.

1. Picture three equal squares in a row. Each has area \(108\div3=36\,\text{cm}^2\). 2. Each small square has side \(6\,\text{cm}\), because \(6\times6=36\). This is the rectangle's width. 3. The length is \(3\times6=18\,\text{cm}\). 4. The perimeter is \(2\times(18+6)=48\,\text{cm}\).

Width: \(6\,\text{cm}\); length: \(18\,\text{cm}\); perimeter: \(48\,\text{cm}\).

- Begin by finding the square's area. - Four equal smaller squares can make the new rectangle. - Compare the perimeters after finding the rectangle's sides.

1. The square's area is \(6\times6=36\,\text{cm}^2\). 2. Picture the rectangle as four equal squares in a row. Each has area \(36\div4=9\,\text{cm}^2\) and side length \(3\,\text{cm}\). 3. The rectangle is \(3\,\text{cm}\) wide and \(4\times3=12\,\text{cm}\) long. 4. The square has perimeter \(4\times6=24\,\text{cm}\); the rectangle has perimeter \(2\times(12+3)=30\,\text{cm}\). 5. The perimeter difference is \(30-24=6\,\text{cm}\).

The rectangle is \(3\,\text{cm}\) by \(12\,\text{cm}\); its perimeter is \(6\,\text{cm}\) longer.

- Use the pen's area and length to find its width. - The walking path extends beyond both sides of each pen dimension. - Find the perimeter of the larger rectangle.

1. The pen's width is \(24 \div 6 = 4\,\text{ft}\). 2. The path is \(2\,\text{ft}\) beyond each side, so its dimensions are \(6 + 2 + 2 = 10\,\text{ft}\) by \(4 + 2 + 2 = 8\,\text{ft}\). 3. The path's perimeter is \(2 \times (10 + 8) = 36\,\text{ft}\).

a) \(4\,\text{ft}\) b) \(36\,\text{ft}\)

- Find the square's side length from its perimeter. - Use the square's area as the rectangle's area. - Find the rectangle's missing length before comparing perimeters.

1. The square's side length is \(48 \div 4 = 12\,\text{ft}\). 2. Its area is \(12 \times 12 = 144\,\text{ft}^2\). 3. The rectangle's length is \(144 \div 9 = 16\,\text{ft}\). 4. The rectangle's perimeter is \(2 \times (16 + 9) = 50\,\text{ft}\). 5. The difference is \(50\,\text{ft} - 48\,\text{ft} = 2\,\text{ft}\).

The rectangle's perimeter is \(2\,\text{ft}\) greater.

- Compare the area of the patio with the area of one tile. - Work out the number of complete packages. - Think of the price as whole dollars plus cents.

1. The patio covers \(12\times8=96\,\text{ft}^2\). 2. Each tile covers \(2\times2=4\,\text{ft}^2\), so \(96\div4=24\) tiles are needed. 3. There are \(24\div6=4\) packages. 4. Four packages cost four lots of \(\$24\) plus four lots of \(50\) cents: \(\$96+\$2=\$98\).

\(\$98.00\)

- Use half the rectangle's perimeter to find the missing side. - Use place value to break the rectangle's area multiplication into easier steps. - A square divides its perimeter equally among four sides; then compare the two areas.

1. Half the rectangle's perimeter is \(200\,\text{ft}\), so the other side is \(200-120=80\,\text{ft}\). 2. For the rectangle, \(120 \times 8=960\), then multiplying by \(10\) gives \(120 \times 80=9600\,\text{ft}^2\). 3. If the path is a square, each side is \(100\,\text{ft}\). One hundred groups of \(100\) make \(10{,}000\), so the square's area is \(10{,}000\,\text{ft}^2\). 4. The difference is \(10{,}000-9600=400\,\text{ft}^2\).

a) \(9600\,\text{ft}^2\) b) \(10{,}000\,\text{ft}^2\) c) \(400\,\text{ft}^2\)

- Make positive whole-number pairs that add to half the total string length. - Compare areas from unequal and nearly equal side lengths. - Remember that a square also counts as a rectangle here.

1. Half the perimeter is \(36\div2=18\,\text{in}\). The two adjacent sides must total \(18\,\text{in}\). 2. List all positive whole-number side pairs: \((1,17)\), \((2,16)\), \((3,15)\), \((4,14)\), \((5,13)\), \((6,12)\), \((7,11)\), \((8,10)\), and \((9,9)\). Each pair adds to \(18\). 3. Their areas, in square inches, are \(17,32,45,56,65,72,77,80,81\). 4. a) The smallest is \(17\,\text{in}^2\), from sides \(1\,\text{in}\) and \(17\,\text{in}\). 5. b) The largest is \(81\,\text{in}^2\), from sides \(9\,\text{in}\) and \(9\,\text{in}\).

a) \(1\,\text{in}\) by \(17\,\text{in}\), area \(17\,\text{in}^2\) b) \(9\,\text{in}\) by \(9\,\text{in}\), area \(81\,\text{in}^2\)

- What square side makes the given total area? - Remember the inside strip comes off both ends of each side. - Subtract inner area from outer area to get the planted area. - Working in cents avoids multiplying decimal amounts.

1. a) The original side is \(50\,\text{ft}\), since \(50\times50=2500\). 2. The open square has side length \(50-2-2=46\,\text{ft}\). 3. b) Its area is \(46\times46=2116\,\text{ft}^2\). 4. The planted strip covers \(2500-2116=384\,\text{ft}^2\). 5. c) At \(20\) cents per square foot, the cost is \(384\times20=7680\) cents, or \(\$76.80\).

a) \(50\,\text{ft}\) b) \(2116\,\text{ft}^2\) c) \(\$76.80\)

- Find the total rectangular area. Use the trailing zeros to simplify the multiplication. - Subtract the border width from both sides of each dimension. - Find the inner area, then subtract it from the total area.

1. Use place value to find the total area: \(220\times 90=(22\times 9)\times 100=19{,}800\,\text{ft}^2\). 2. The inner dimensions are \(220-10=210\,\text{ft}\) and \(90-10=80\,\text{ft}\). 3. Use place value again: \(210\times 80=(21\times 8)\times 100=16{,}800\,\text{ft}^2\). 4. Subtract the inner area from the total area. The difference between \(19{,}800\,\text{ft}^2\) and \(16{,}800\,\text{ft}^2\) is \(3000\,\text{ft}^2\).

a) \(19{,}800\,\text{ft}^2\) b) \(3000\,\text{ft}^2\)

- Use multiplication to find the missing width of Field B. - Subtract the border width twice from each dimension. - Find each border area by subtracting the inner area from the total area. - Break apart a large product using a nearby multiple of \(10\) when useful.

1. Since \(250\times 40=10{,}000\), Field B's width is \(40\,\text{ft}\). 2. Field A is \(100\,\text{ft}\) by \(100\,\text{ft}\). Its inner dimensions are \(98\,\text{ft}\) by \(98\,\text{ft}\), and \(98\times 98=9604\,\text{ft}^2\). The difference between the total area of \(10{,}000\,\text{ft}^2\) and the inner area of \(9604\,\text{ft}^2\) is \(396\,\text{ft}^2\). 3. Field B's inner dimensions are \(248\,\text{ft}\) by \(38\,\text{ft}\). Use the distributive property: \(248\times 38=248\times 40-248\times 2=9920-496=9424\,\text{ft}^2\). The difference between the total area of \(10{,}000\,\text{ft}^2\) and the inner area of \(9424\,\text{ft}^2\) is \(576\,\text{ft}^2\). 4. Field B loses more usable area.

a) \(40\,\text{ft}\) b) Field A: \(396\,\text{ft}^2\); Field B: \(576\,\text{ft}^2\). Field B loses more usable area.

- Add the path's width on both ends of each dimension. - Subtract the original patio area from the entire outer area. - Use the cost per square foot on the path area only.

1. The outside dimensions are \(8+2+2=12\,\text{ft}\) and \(5+2+2=9\,\text{ft}\). 2. The outside area is \(12\times9=108\,\text{ft}^2\). 3. The patio area is \(8\times5=40\,\text{ft}^2\). 4. The path covers \(108-40=68\,\text{ft}^2\). 5. The cost is \(68\times\$15=\$1020\).

\(\$1020\)

- The border adds width at both ends of each dimension. - Subtract the pool's area from the outer area. - Multiply the deck area by the mass per square foot, then check the truck limit.

1. Two \(1.5\)-foot strips add \(3\) feet to each dimension, giving \(13\,\text{ft}\) by \(8\,\text{ft}\) outside. 2. The outer area is \(13\times8=104\,\text{ft}^2\). 3. The pool covers \(10\times5=50\,\text{ft}^2\), leaving \(104-50=54\,\text{ft}^2\) of deck. 4. The tiles weigh \(54\times25=1350\,\text{lb}\). 5. Since \(1350>1000\), they exceed the truck's limit.

The tiles weigh \(1350\,\text{lb}\). No, the truck cannot carry them all in one trip.

- Find every positive whole-number factor pair of the area. - The perimeter includes two copies of each side. - Compare the perimeters after listing all cases.

1. a) The whole-number factor pairs of \(24\) are \(1\times24\), \(2\times12\), \(3\times8\), and \(4\times6\), in feet. 2. b) Their perimeters are \(2\times(1+24)=50\,\text{ft}\), \(2\times(2+12)=28\,\text{ft}\), \(2\times(3+8)=22\,\text{ft}\), and \(2\times(4+6)=20\,\text{ft}\). 3. c) The \(4\)-foot by \(6\)-foot rectangle has the least perimeter.

a) \(1\times24\), \(2\times12\), \(3\times8\), \(4\times6\), each in feet b) \(50\,\text{ft}\), \(28\,\text{ft}\), \(22\,\text{ft}\), \(20\,\text{ft}\), in that order c) \(4\,\text{ft}\) by \(6\,\text{ft}\)

- Begin with the area of the full rectangle. - For each kind of space, use area for both the total and one space. - Divide the available area by the area of a single space.

1. The total field area is \(120\times200=24{,}000\,\text{ft}^2\). 2. a) \(24{,}000\div300=80\) large zones. 3. One small zone covers \(14{,}400\div144=100\,\text{ft}^2\). 4. b) \(24{,}000\div100=240\) small zones.

a) \(80\) large zones b) \(240\) small zones

- Treat each missing side as a factor that must produce the given area. - After shortening the \(60\)-centimeter side, use the same area to determine the new other side. - Find both perimeters before comparing them.

1. Since \(40 \times 60=2400\), the original missing side is \(40\,\text{cm}\). 2. The shortened side is \(60-10=50\,\text{cm}\). 3. Since \(48 \times 50=2400\), the new other side is \(48\,\text{cm}\). 4. The original perimeter is \(2 \times (60+40)=200\,\text{cm}\). The new perimeter is \(2 \times (50+48)=196\,\text{cm}\). 5. The new perimeter is \(200-196=4\,\text{cm}\) less.

a) \(40\,\text{cm}\) b) \(48\,\text{cm}\) c) The perimeter decreases by \(4\,\text{cm}\), from \(200\,\text{cm}\) to \(196\,\text{cm}\).

- Find whole-number factor pairs of twenty-four. - Test the change with a pair whose width is even. - Compare the old and new areas.

1. a) Some whole-number factor pairs of \(24\) are \(8\) and \(3\), \(6\) and \(4\), \(12\) and \(2\), and \(24\) and \(1\). Any three different pairs work. 2. b) If the garden starts at \(6\,\text{ft}\) by \(4\,\text{ft}\), changing it to \(12\,\text{ft}\) by \(2\,\text{ft}\) keeps the area \(24\,\text{ft}^2\). 3. In general, doubling one side and halving the other gives the same number of square units.

a) For example, \(8\,\text{ft}\) by \(3\,\text{ft}\), \(6\,\text{ft}\) by \(4\,\text{ft}\), and \(12\,\text{ft}\) by \(2\,\text{ft}\) b) The area stays at \(24\,\text{ft}^2\); doubling one side and halving the other leave the product unchanged.

- For subtraction, use the enclosing rectangle and the square cutout. - For addition, find the height below the cutout and split the U into three nonoverlapping rectangles. - Include the inner edges of the cutout when finding perimeter.

1. Subtraction method: The enclosing rectangle has area \(10\times 6=60\,\text{cm}^2\). The cutout has area \(4\times 4=16\,\text{cm}^2\). The template area is \(60-16=44\,\text{cm}^2\). 2. Addition method: The two side rectangles each have area \(3\times 6=18\,\text{cm}^2\). The bottom middle height is \(6-4=2\,\text{cm}\), so the bottom middle rectangle has area \(4\times 2=8\,\text{cm}^2\). The total is \(18+18+8=44\,\text{cm}^2\). 3. Add all eight boundary lengths: \(10+6+3+4+4+4+3+6=40\,\text{cm}\).

a) \(44\,\text{cm}^2\) by either method b) \(40\,\text{cm}\)

- Read Rectangle A's dimensions from the diagram and find its area. - Find a square side length whose product with itself gives that same area. - Calculate and compare the two perimeters.

1. From the diagram, Rectangle A measures \(18\,\text{ft}\times8\,\text{ft}\), so its area is \(18 \times 8=144\,\text{ft}^2\). 2. Square B has the same area. Since \(12 \times 12=144\), \(s=12\,\text{ft}\). 3. Rectangle A has perimeter \(2 \times (18+8)=52\,\text{ft}\). 4. Square B has perimeter \(4 \times 12=48\,\text{ft}\). 5. Square B needs \(52-48=4\,\text{ft}\) less fencing.

a) \(144\,\text{ft}^2\) b) \(s=12\,\text{ft}\) c) Rectangle A: \(52\,\text{ft}\); Square B: \(48\,\text{ft}\). Square B saves \(4\,\text{ft}\) of fencing.

- For area, break the U-shape into rectangles that do not overlap. - Trace every boundary edge, including those around the gap. - Compare the square counts with the distance you traced around each boundary.

1. a) Figure 1 has two \(1\times4\) arms and a \(2\times1\) base between them. Its area is \(4+4+2=10\) square units. 2. Figure 2 is a \(5\times2\) rectangle, also with area \(10\) square units. 3. b) Figure 1's boundary totals \(4+4+1+3+2+3+1+4=22\) length units. 4. Figure 2's perimeter is \(2\times(5+2)=14\) length units. 5. c) Both areas are \(10\) square units, but Figure 1 has a longer perimeter. Its inward edges add to the distance around it.

a) \(10\) square units b) Figure 1: \(22\) length units; Figure 2: \(14\) length units c) Figure 2's area is \(10\) square units. The areas are equal, but Figure 1's perimeter is longer because of the inward edges.

- Subtract the notch's area from the area of the original rectangle. - Don't skip the edges inside the notch when counting perimeter. - Find any unlabeled sides by comparing aligned lengths.

1. The full plate is \(20\,\text{cm}\) by \(12\,\text{cm}\), with area \(240\,\text{cm}^2\). 2. The notch is \(6\,\text{cm}\) by \(4\,\text{cm}\), with area \(24\,\text{cm}^2\). The remaining area is \(240-24=216\,\text{cm}^2\). 3. Count every outside edge, including the notch: \(20+4+6+4+6+4+20+12=76\,\text{cm}\).

Area: \(216\,\text{cm}^2\) Perimeter: \(76\,\text{cm}\)

- Try one horizontal dividing line and one vertical dividing line. - Make sure the rectangles in each decomposition cover the shaded cells without overlap.

1. One decomposition gives a \(6 \times 3\) rectangle and a \(2 \times 7\) rectangle: \(18 + 14 = 32\) square units. 2. Another decomposition gives an \(8 \times 3\) rectangle and a \(2 \times 4\) rectangle: \(24 + 8 = 32\) square units.

The area is \(32\) square units. Two decompositions are \(6 \times 3 + 2 \times 7\) and \(8 \times 3 + 2 \times 4\).