- The two regions cover the whole figure without overlapping. - What operation combines the areas of nonoverlapping parts?

1. The regions do not overlap, so their areas add. 2. \(14+9=23\), so the whole figure has area \(23\,\text{cm}^2\).

\(23\,\text{cm}^2\)

- How many corners in this outline turn inward instead of outward? - What shapes could cover the entire region without overlapping? - How could you check your pieces against the original outline?

1. Draw a horizontal split from the inside corner to the left edge. 2. The bottom rectangle measures \(8\,\text{cm}\) by \(2\,\text{cm}\). 3. The upper rectangle measures \(5\,\text{cm}\) by \(3\,\text{cm}\). Other correct splits are possible.

One choice: an \(8\,\text{cm}\times2\,\text{cm}\) rectangle and a \(5\,\text{cm}\times3\,\text{cm}\) rectangle.

- Does a straight edge have area? - Would adding the two interiors count any region twice?

1. Segment \(BE\) is a shared edge, not a shared region with area. 2. The rectangles do not overlap, so their areas can be added without subtracting anything.

They share only edge \(BE\), which has no area. Their areas can be added directly.

- Which horizontal pieces together span the full bottom edge? - Use the full width to find the missing part.

1. The total horizontal width is \(7\,\text{cm}\). 2. The upper horizontal part is \(4\,\text{cm}\), so the short step is \(7-4=3\,\text{cm}\).

\(3\,\text{cm}\).

- Does the inner rectangle contain material or empty space? - What must be removed from the area of the full sheet?

1. The smaller rectangle is missing from the sheet. 2. Subtract the hole's area from the outer rectangle's area.

Subtract the hole's area from the outer rectangle's area.

- Which part of the paper belongs to both rectangles? - How many times does straight addition count that part?

1. Adding both full areas counts the shared region twice. 2. The overlap must be counted once, so the combined area is \(15+12-3=24\,\text{cm}^2\).

The \(3\,\text{cm}^2\) overlap is counted twice; the combined area is \(24\,\text{cm}^2\).

- Look for a simple outer rectangle that contains the whole L-shape. - Identify the rectangular part that is missing from that outer rectangle. - Subtract the missing area from the outer area.

1. View the figure as an \(8\,\text{m}\) by \(6\,\text{m}\) rectangle with a \(3\,\text{m}\) by \(2\,\text{m}\) rectangle removed. 2. The outer rectangle has area \(8\cdot6=48\,\text{m}^2\). 3. The removed rectangle has area \(3\cdot2=6\,\text{m}^2\). 4. The L-shaped area is \(48-6=42\,\text{m}^2\).

\(42\,\text{m}^2\)

- Find the area of each rectangle separately. - Because the rectangles do not overlap, their areas can be added.

1. The larger rectangle has area \(6\cdot4=24\,\text{m}^2\). 2. The smaller rectangle has area \(2\cdot3=6\,\text{m}^2\). 3. Add the nonoverlapping areas: \(24+6=30\,\text{m}^2\).

\(30\,\text{m}^2\)

- Read the dimensions of the outer rectangle and the opening from the diagram. - Treat the opening as area that must be removed from the whole rectangle.

1. The outer rectangle has area \(8\cdot6=48\,\text{cm}^2\). 2. The rectangular opening has area \(3\cdot2=6\,\text{cm}^2\). 3. Subtract the opening: \(48-6=42\,\text{cm}^2\).

\(42\,\text{cm}^2\)

- Where are the two inward corners of this outline? - Try sketching one possible cut. What region remains? - Check that your suggested rectangles cover exactly the shaded region.

1. Two rectangles joined along an edge cannot make both arms and the gap of this U shape. At least three are needed. 2. The bottom rectangle is \(8\,\text{cm}\times2\,\text{cm}\). Each upper arm is \(2\,\text{cm}\times3\,\text{cm}\). 3. Their total area is \(8\cdot2+2(2\cdot3)=28\,\text{cm}^2\).

No. One split uses a bottom \(8\,\text{cm}\times2\,\text{cm}\) rectangle and two \(2\,\text{cm}\times3\,\text{cm}\) rectangles. Total area: \(28\,\text{cm}^2\).

- What simple shapes could cover the T-shaped region? - Which lengths belong to each part rather than the whole outline? - How could you check that your pieces leave no gaps?

1. Separate the top bar from the stem below it. 2. The bar is \(7\,\text{cm}\times2\,\text{cm}\), with area \(14\,\text{cm}^2\). The stem is \(3\,\text{cm}\times3\,\text{cm}\), with area \(9\,\text{cm}^2\). 3. The two rectangles meet only along an edge, so the total is \(14+9=23\,\text{cm}^2\).

A \(7\,\text{cm}\times2\,\text{cm}\) top rectangle and a \(3\,\text{cm}\times3\,\text{cm}\) stem cover the region. They meet along an edge but share no area. Total: \(23\,\text{cm}^2\).

- Does any part of one rectangle lie inside the other? - Which region belongs to both rectangles at once? - How can you count the combined region once instead of twice?

1. Each rectangle covers \(3\cdot5=15\,\text{ft}^2\). 2. Their common strip covers \(1\cdot5=5\,\text{ft}^2\), and was counted twice. 3. The combined region covers \(15+15-5=25\,\text{ft}^2\).

Quinn counted the \(5\,\text{ft}^2\) overlap twice. The total area is \(25\,\text{ft}^2\).

- Would a single cut make all the parts rectangles? - Can you use the heights of the steps to find convenient strips? - Check that every part of the shaded region is counted once.

1. Split the region into horizontal strips of height \(2\,\text{cm}\). 2. The bottom strip has area \(9\cdot2=18\,\text{cm}^2\), the middle \(6\cdot2=12\,\text{cm}^2\), and the top \(3\cdot2=6\,\text{cm}^2\). 3. Together, the strips have area \(18+12+6=36\,\text{cm}^2\).

One split uses three horizontal rectangles: \(9\times2\), \(6\times2\), and \(3\times2\), all in centimeters. They do not overlap, and their areas total \(36\,\text{cm}^2\).

- Find the area before the hole was removed. - Which smaller region is missing? - Why is subtraction appropriate here?

1. The outer rectangle has area \(9\cdot7=63\,\text{cm}^2\). 2. The hole has area \(3\cdot4=12\,\text{cm}^2\). 3. The shaded area is \(63-12=51\,\text{cm}^2\).

\(51\,\text{cm}^2\).

- Extend the inside corner's edge to split the shape into rectangles. - Alternatively, think of a complete outside rectangle. - What smaller corner would that outside rectangle include by mistake?

1. Adding parts gives \(7\cdot4+4\cdot2=28+8=36\,\text{cm}^2\). 2. Completing a rectangle gives \(7\cdot6=42\,\text{cm}^2\). 3. The missing corner has area \(3\cdot2=6\,\text{cm}^2\), so \(42-6=36\,\text{cm}^2\).

Both methods give \(36\,\text{cm}^2\).

- How do the two top pieces and the notch add to the full width? - Compare the total height with the height of the notch floor. - What rectangle is missing from the full sign?

1. The notch width is \(10-3-2=5\,\text{cm}\). 2. Its depth is \(8-5=3\,\text{cm}\). 3. The outer rectangle's area is \(10\cdot8=80\,\text{cm}^2\); the notch is \(5\cdot3=15\,\text{cm}^2\). 4. Remaining area: \(80-15=65\,\text{cm}^2\).

Width \(5\,\text{cm}\); depth \(3\,\text{cm}\); remaining area \(65\,\text{cm}^2\).

- Which familiar shapes could cover this outline without gaps? - What does the \(2\,\text{m}\) roof rise measure? - Do your parts share any area or just an edge?

1. The rectangle is \(8\,\text{m}\) wide and \(3\,\text{m}\) tall, so its area is \(24\,\text{m}^2\). 2. The roof triangle has base \(8\,\text{m}\) and height \(2\,\text{m}\), so its area is \(\frac{1}{2}\cdot8\cdot2=8\,\text{m}^2\). 3. The total is \(24+8=32\,\text{m}^2\).

\(32\,\text{m}^2\).

- How do the two card areas compare? - Which part would appear in both areas if you added them? - Count the area covered by either card exactly once.

1. The two cards have areas \(6\cdot4=24\,\text{cm}^2\) and \(5\cdot3=15\,\text{cm}^2\). 2. The corner overlap covers \(2\cdot2=4\,\text{cm}^2\). 3. Adding the full card areas counts the corner square twice. The covered area is \(24+15-4=35\,\text{cm}^2\).

\(35\,\text{cm}^2\).

- For each shape, complete the outer rectangle in your mind. - Which corner would have to be removed? - Does knowing the perimeter alone determine the area?

1. In a), the outside rectangle has area \(8\cdot4=32\,\text{cm}^2\), and the missing corner has area \(2\cdot2=4\,\text{cm}^2\). The L-shaped area is \(28\,\text{cm}^2\). 2. In b), the outside rectangle has area \(6\cdot6=36\,\text{cm}^2\), and the missing corner has area \(3\cdot2=6\,\text{cm}^2\). The L-shaped area is \(30\,\text{cm}^2\). 3. Shape b) has more area even though both perimeters are \(24\,\text{cm}\).

a) \(28\,\text{cm}^2\) b) \(30\,\text{cm}^2\); b) has the larger area.

- What kind of quantity is \(8\cdot6\)? - What shape is the piece that was removed? - Use consistent square units when subtracting.

1. The missing upper-right corner is a rectangle, not two line lengths to subtract. 2. The outside rectangle has area \(8\cdot6=48\,\text{cm}^2\). 3. The missing corner has area \(3\cdot2=6\,\text{cm}^2\), so the L shape has area \(48-6=42\,\text{cm}^2\).

The student subtracted lengths instead of the missing corner's area. Correct area: \(42\,\text{cm}^2\).

- Find the two familiar shapes inside the bed. - Which legs can serve as the triangle's base and height? - Are you counting any region twice?

1. The rectangle's area is \(9\cdot4=36\,\text{ft}^2\). 2. The right triangle's area is \(\frac{1}{2}\cdot3\cdot4=6\,\text{ft}^2\). 3. Add the areas: \(36+6=42\,\text{ft}^2\).

\(42\,\text{ft}^2\).

- Compare the units on the two diagrams. - How many centimeters are in one meter? - Are the two area measurements expressed in matching units?

1. The first section is \(2\,\text{m}\times1\,\text{m}\), so its area is \(2\,\text{m}^2\). 2. The second is \(100\,\text{cm}\times100\,\text{cm}\), or \(1\,\text{m}\times1\,\text{m}\), so its area is \(1\,\text{m}^2\). 3. Combined area is \(2+1=3\,\text{m}^2\).

\(3\,\text{m}^2\).

- Which parts of the outer rectangle belong to the panel? - Compare the panel's area with that of its outer rectangle.

1. The side notch is missing from the full rectangle, so it is not part of the panel's area. 2. Use \(8\cdot5-3\cdot2=40-6=34\,\text{cm}^2\).

\(8\cdot5-3\cdot2\), because the notch is missing; the area is \(34\,\text{cm}^2\).

- How could a few rectangles cover the entire outlined region without overlap? - Which corners might be useful when you draw a dividing line? - Check that every grid square inside the outline is counted once.

1. Count the squares to see a lower \(6\times2\) rectangle and an upper-left \(3\times2\) rectangle. 2. The total is \(6\cdot2+3\cdot2=18\) square feet.

One split makes a \(6\times2\) rectangle and a \(3\times2\) rectangle. The total is \(18\,\text{ft}^2\).

- Which two familiar shapes make up the wall? - Find the rectangle's area. - Find the triangle's area. - Add the two areas.

1. The rectangle's area is \(4\cdot2.5=10\,\text{m}^2\). 2. The triangle's area is \(\frac{1}{2}\cdot4\cdot1.8=3.6\,\text{m}^2\). 3. The total area is \(10+3.6=13.6\,\text{m}^2\).

The total area is \(13.6\,\text{m}^2\).

- Subtract the lawn's area from the rectangle's area. - The triangular bed's height is the garden's width. - Use the triangle area formula to solve for the unknown base.

1. The entire garden has area \(80\cdot30=2400\,\text{ft}^2\). 2. The flower bed has area \(2400-1800=600\,\text{ft}^2\). 3. Let \(b\) be the triangle's base. Its height is \(30\,\text{ft}\), so \(600=\frac{1}{2}b(30)\). 4. Solving gives \(600=15b\), so \(b=40\,\text{ft}\).

The flower bed has area \(600\,\text{ft}^2\), and its base is \(40\,\text{ft}\) long.

- Convert all lengths to the same unit before combining areas. - A removed piece must be subtracted from the original area. - Check that every area is expressed in square centimeters.

1. The full sheet has area \(80\cdot40=3200\,\text{cm}^2\). 2. Convert the cutout dimensions: \(200\,\text{mm}=20\,\text{cm}\) and \(150\,\text{mm}=15\,\text{cm}\). 3. The cutout area is \(20\cdot15=300\,\text{cm}^2\). 4. The remaining area is \(3200-300=2900\,\text{cm}^2\).

The remaining glass area is \(2900\,\text{cm}^2\).

- Visualize the shape and count only exposed side segments. - Add the areas of the five nonoverlapping squares. - Substitute the side length only after writing the expressions.

1. The four outer squares each contribute three exposed sides, for a total of \(12\) exposed segments of length \(a\). Thus \(P=12a\). 2. The figure contains five squares of area \(a^2\), so \(A=5a^2\). 3. When \(a=4\), \(P=12\cdot4=48\,\text{cm}\), and \(A=5\cdot4^2=80\,\text{cm}^2\).

a) \(P=12a\) b) \(A=5a^2\) c) \(P=48\,\text{cm}\); \(A=80\,\text{cm}^2\)

- View the annulus as a large circle with a smaller circle removed. - Subtract the two circle areas. - How do areas change when every length is doubled?

1. Subtract the inner circle's area from the outer circle's area: \(A = \pi r_o^2 - \pi r_i^2 = \pi(r_o^2-r_i^2)\). 2. Substitute the radii: \(A = \pi(8^2-5^2)\,\text{cm}^2 = 39\pi\,\text{cm}^2 \approx 122.52\,\text{cm}^2\). 3. After both radii are doubled, \(A_{\text{new}}=\pi(16^2-10^2)\,\text{cm}^2=156\pi\,\text{cm}^2\). 4. Since \(\frac{156\pi}{39\pi}=4\), the area is multiplied by \(4\).

a) \(39\pi\,\text{cm}^2 \approx 122.52\,\text{cm}^2\) b) The area is multiplied by \(4\).

- Model the path as an annulus. - Use the inner circumference to find the inner radius. - How is the outer radius related to the path width?

1. Find the inner radius: \(r_i = \frac{300\,\text{m}}{2\pi} \approx 47.75\,\text{m}\). 2. The outer radius is \(r_o = r_i + 4\,\text{m}\). 3. The path area is \(A = \pi r_o^2 - \pi r_i^2\). 4. Without rounding the inner radius, \(A = \pi\left[\left(\frac{300}{2\pi}+4\right)^2-\left(\frac{300}{2\pi}\right)^2\right] \approx 1250.27\,\text{m}^2\). 5. Rounded to the nearest square meter, the area is \(1250\,\text{m}^2\).

Approximately \(1250\,\text{m}^2\)

- View the annulus as a large circle with a smaller circle removed. - Check whether each given measurement is a radius or a diameter. - Subtract the inner circle's area from the outer circle's area.

1. The area of an annulus is \(A = \pi r_o^2-\pi r_i^2=\pi(r_o^2-r_i^2)\). 2. For a), \(A=\pi(10^2-6^2)\,\text{cm}^2=64\pi\,\text{cm}^2\approx201.06\,\text{cm}^2\). 3. For b), the inner radius is \(r_i=\frac{14\,\text{m}}{2}=7\,\text{m}\). 4. Then \(A=\pi(9^2-7^2)\,\text{m}^2=32\pi\,\text{m}^2\approx100.53\,\text{m}^2\).

a) \(A \approx 201.06\,\text{cm}^2\) b) \(A \approx 100.53\,\text{m}^2\)

- Model the walkway as the region between two concentric circles. - What is the outer radius? - Use the outer radius directly for the boundary length.

1. The flower bed's radius is \(3\,\text{ft}\), and the outer radius is \(4\,\text{ft}\). 2. The walkway area is \(\pi(4^2-3^2)\,\text{ft}^2=7\pi\,\text{ft}^2\approx22.0\,\text{ft}^2\). 3. The outer boundary is \(2\pi(4\,\text{ft})=8\pi\,\text{ft}\approx25.1\,\text{ft}\).

a) Approximately \(22.0\,\text{ft}^2\) b) \(8\pi\,\text{ft}\approx25.1\,\text{ft}\)

- How many circles fit across one side of the square? - Use the side length to find each circle's diameter. - Subtract the total circle area from the square's area. - Remember that there are four circles.

1. The square's area is \((8\,\text{ft})^2=64\,\text{ft}^2\). 2. Two circle diameters span the side of the square, so each diameter is \(\frac{8\,\text{ft}}{2}=4\,\text{ft}\). Each radius is \(2\,\text{ft}\). 3. The four circles have total area \(4\pi(2\,\text{ft})^2=16\pi\,\text{ft}^2\approx50.27\,\text{ft}^2\). 4. The grass area is \(64\,\text{ft}^2-16\pi\,\text{ft}^2\approx13.73\,\text{ft}^2\).

Approximately \(13.73\,\text{ft}^2\)

- Find the radius of the one large circle. - What is the diameter of each small circle when five diameters span \(20\,\text{in.}\)? - Compare the total area of the \(25\) small circles with the area of the large circle before subtracting from the square.

1. Each square sheet has area \(400\,\text{in.}^2\). 2. The largest circle has radius \(10\,\text{in.}\), so its area is \(100\pi\,\text{in.}^2\). 3. Each small circle has diameter \(4\,\text{in.}\) and radius \(2\,\text{in.}\). 4. The \(25\) small circles have total area \(25\pi(2\,\text{in.})^2=100\pi\,\text{in.}^2\). 5. The removed circle area is the same for both sheets, so each produces \(400-100\pi\,\text{in.}^2\) of scrap.

Both sheets produce the same amount of scrap: \(400-100\pi\,\text{in.}^2\).

- Find the tray area and the total area of all circular bases. - Express the coverage as a fraction of the tray area before converting to a percent. - Compare the total base areas in parts a and b.

1. The tray area is \(216\,\text{in.}^2\). 2. The six bases have total area \(6\pi(3\,\text{in.})^2=54\pi\,\text{in.}^2\). 3. The percent covered is \(\frac{54\pi}{216}\cdot100\%=25\pi\%\approx78.5\%\). 4. For the smaller glasses, the \(24\) bases also have total area \(24\pi(1.5\,\text{in.})^2=54\pi\,\text{in.}^2\). 5. Therefore, the covered percentage remains \(25\pi\%\approx78.5\%\).

a) \(25\pi\%\approx78.5\%\) b) The percent remains \(25\pi\%\approx78.5\%\) because the total circular-base area is unchanged.

- Identify the inner and outer radii first. - Model the walkway as an annulus. - For part b, subtract the two circumferences before evaluating. - Use the unrounded area for the cost.

1. The inner radius is \(10\,\text{ft}\), and the outer radius is \(15\,\text{ft}\). 2. The walkway area is \(\pi(15^2-10^2)\,\text{ft}^2=125\pi\,\text{ft}^2\approx392.70\,\text{ft}^2\). 3. The difference between the edge lengths is \(2\pi(15)-2\pi(10)=10\pi\,\text{ft}\approx31.42\,\text{ft}\). 4. Using the exact area, the cost is \(125\pi\cdot\$12.00\approx\$4{,}712.39\).

a) \(125\pi\,\text{ft}^2\approx392.70\,\text{ft}^2\) b) \(10\pi\,\text{ft}\approx31.42\,\text{ft}\) c) Approximately \(\$4{,}712.39\)

- Convert the island diameter to a radius. - Model the roadway as an annulus. - For the path difference, subtract the two edge circumferences before evaluating.

1. The island radius is \(25\,\text{ft}\), so its area is \(625\pi\,\text{ft}^2\approx1963.50\,\text{ft}^2\). 2. The outer radius is \(45\,\text{ft}\). 3. The roadway area is \(\pi(45^2-25^2)\,\text{ft}^2=1400\pi\,\text{ft}^2\approx4398.23\,\text{ft}^2\). 4. The path-length difference is \(2\pi(45)-2\pi(25)=40\pi\,\text{ft}\approx125.66\,\text{ft}\).

a) \(625\pi\,\text{ft}^2\approx1963.50\,\text{ft}^2\) b) \(1400\pi\,\text{ft}^2\approx4398.23\,\text{ft}^2\) c) \(40\pi\,\text{ft}\approx125.66\,\text{ft}\) longer

- Which part of the banner has enough measurements to find its area directly? - How do the areas of two parts that do not overlap relate to the whole? - Once you know the triangle's area, which measurement is still missing?

1. The parallelogram's area is \(8\cdot4.5=36\,\text{in.}^2\). 2. The triangle's area is the total less the parallelogram's area: \(66-36=30\,\text{in.}^2\). 3. For the triangle, \(30=\frac{1}{2}\cdot12\cdot h\), so \(30=6h\). 4. Therefore, \(h=5\,\text{in.}\).

The triangle's height is \(5\,\text{in.}\).

- Use the rectangle's area to find its width. - Subtract the rectangle's area from the combined area. - Use the garden's width as the triangle's base. - Solve the triangle area equation for its height.

1. Let \(w\) feet be the garden's width. From \(12w=120\), obtain \(w=10\). 2. The triangular flower bed has area \(150-120=30\,\text{ft}^2\). 3. Its base is the garden's width, \(10\,\text{ft}\). Write \(30=\frac{1}{2}\cdot10\cdot h\). 4. Simplify: \(30=5h\). 5. Divide by \(5\): \(h=6\).

The triangular flower bed is \(6\,\text{ft}\) high.

- Decide which component areas are added and which area is subtracted. - Write the area formulas for a rectangle, square, and triangle. - Keep the units consistent. - Sketching the pieces may help you organize the calculation.

1. The rectangle's area is \(5\cdot3=15\,\text{cm}^2\). 2. The square's area is \(2^2=4\,\text{cm}^2\). 3. The triangle's area is \(\frac{1}{2}\cdot1.2\cdot1.5=0.9\,\text{cm}^2\). 4. The remaining area is \(15+4-0.9=18.1\,\text{cm}^2\).

The remaining area is \(18.1\,\text{cm}^2\).

- Find the area of each shape separately. - Use the trapezoid area formula. - Use the parallelogram area formula. - Divide the parallelogram's area by the total area to find its percent.

1. The trapezoid's area is \(\frac{1}{2}(6+4)\cdot3=15\,\text{cm}^2\). 2. The parallelogram's area is \(4\cdot2.5=10\,\text{cm}^2\). 3. The total area is \(15+10=25\,\text{cm}^2\). 4. The parallelogram occupies \(\frac{10}{25}=0.4=40\%\) of the total area.

The total area is \(25\,\text{cm}^2\), and the parallelogram occupies \(40\%\) of it.

- Find the total area removed by the three squares. - Does a square's area depend on where it is located? - Find the parallelogram's area first. - How do you find the total area of three identical shapes?

1. The parallelogram's area is \(12\cdot8=96\,\text{cm}^2\). 2. Each square has area \((2.5)^2=6.25\,\text{cm}^2\), so the three squares have total area \(3\cdot6.25=18.75\,\text{cm}^2\). 3. The remaining area is \(96-18.75=77.25\,\text{cm}^2\). 4. Moving the cutouts does not change their total area, so the remaining area stays the same as long as the squares do not overlap and remain inside the plate.

a) \(77.25\,\text{cm}^2\) b) No. The same total area is removed regardless of the squares' locations.

- Use length times width for each rectangle. - Find the total area covered by all flower beds. - Subtract the covered area from the full lawn area. - Replace the fixed number of beds with \(n\).

1. The full lawn area is \(ab\). 2. The two flower beds have total area \(2xy\). 3. The remaining lawn area is \(A=ab-2xy\). 4. Substitute the values: \(A=25\cdot 15-2\cdot 4\cdot 2.5=375-20=355\,\text{m}^2\). 5. For \(n\) identical beds, the remaining area is \(A=ab-nxy\).

a) \(A=ab-2xy\) b) \(355\,\text{m}^2\) c) \(A=ab-nxy\)

- Identify what \(x^2\) represents. - Apply exponents before multiplication and subtraction. - Compare the area for a sample value of \(x\) with the area after doubling \(x\). - A square's area depends on the square of its side length.

1. Expression (1) is correct. The sheet area is \(LB\), and the four squares have total area \(4x^2\). 2. Substitute the values: \(A=20\cdot 15-4\cdot 3^2=300-36=264\,\text{cm}^2\). 3. The original punched-out area is \(4x^2\). Doubling the side length gives \(4(2x)^2=16x^2\), which is four times \(4x^2\), not twice as much. The claim is false.

a) Expression (1), \(A=LB-4x^2\) b) \(264\,\text{cm}^2\) c) The claim is false. Doubling the side length makes the punched-out area four times as large.

- Trace the entire boundary of the U-shaped figure. - The two top horizontal pieces together have length \(b-w\). - Find the area by subtracting the cutout from the original rectangle.

1. Subtract the rectangular cutout from the original rectangle: \(A=bh-wd\). 2. For the perimeter, include the bottom edge \(b\), the two outer vertical edges \(2h\), the two remaining top pieces with combined length \(b-w\), the two inner vertical edges \(2d\), and the bottom of the cutout \(w\). 3. Therefore, \(P=b+2h+(b-w)+2d+w=2b+2h+2d=2(b+h+d)\).

\(A=bh-wd\) \(P=2b+2h+2d\)

- The cutout has the same area in both positions. - List the boundary segments in each case. - Compare the boundary segments removed with the new boundary segments created.

1. In both figures, an area of \(x^2\) is removed from an area of \(s^2\), so each area is \(s^2-x^2\). 2. In Figure A, two outer segments of total length \(2x\) are replaced by two new inner segments of the same total length. Therefore, \(P_A=4s\). 3. In Figure B, one outer segment of length \(x\) is replaced by three boundary segments of total length \(3x\). The perimeter increases by \(2x\), so \(P_B=4s+2x\). 4. Figure B has the greater perimeter by \(2x\).

a) Both figures have area \(s^2-x^2\). b) \(P_A=4s\); \(P_B=4s+2x\) c) Figure B has the greater perimeter by \(2x\).

- Write the two heights first. - Use area equals width times height for each rectangle. - Add the two area expressions and set their sum equal to the combined area. - Use the width to calculate each individual area.

1. Let \(w\) inches be the common width. 2. The heights are \(8\,\text{in.}\) and \(16\,\text{in.}\). 3. Add the two area expressions: \(8w + 16w = 192\). 4. Combine like terms: \(24w = 192\). 5. Divide by \(24\): \(w = 8\). 6. The first area is \(8\,\text{in.} \times 8\,\text{in.} = 64\,\text{in.}^2\), and the second area is \(16\,\text{in.} \times 8\,\text{in.} = 128\,\text{in.}^2\).

The common width is \(8\,\text{in.}\). The areas are \(64\,\text{in.}^2\) and \(128\,\text{in.}^2\).

- Identify a horizontal segment that would create two familiar figures. - Find the rectangle's dimensions and the triangle's height. - Add the rectangle and triangle areas.

1. Use a horizontal segment from the upper end of the \(10\,\text{ft}\) side to split the figure. 2. The rectangle measures \(40\,\text{ft}\) by \(10\,\text{ft}\), so its area is \(40\cdot10=400\,\text{ft}^2\). 3. The triangle's height is \(20-10=10\,\text{ft}\), and its base is \(40\,\text{ft}\). Its area is \(\frac{1}{2}\cdot40\cdot10=200\,\text{ft}^2\). 4. The total area is \(400+200=600\,\text{ft}^2\).

Rectangle: \(400\,\text{ft}^2\) Right triangle: \(200\,\text{ft}^2\) Total: \(600\,\text{ft}^2\)

- Compare the full outside dimensions to find the two missing inner side lengths. - The perimeter includes all six boundary segments. - Split the L-shape into two rectangles. - Add the rectangle areas. - Divide the total area by the coverage per package and round up if necessary.

1. The missing vertical side is \(18-9=9\,\text{ft}\), and the missing horizontal side is \(24-12=12\,\text{ft}\). 2. The perimeter is \(24+18+12+9+12+9=84\,\text{ft}\). 3. Split the patio into a \(24\,\text{ft}\times9\,\text{ft}\) rectangle and a \(12\,\text{ft}\times9\,\text{ft}\) rectangle. The area is \(24\cdot9+12\cdot9=216+108=324\,\text{ft}^2\). 4. The number of packages is \(324\div18=18\). The cost is \(18\cdot\$34.90=\$628.20\).

a) \(84\,\text{ft}\) b) The area is \(324\,\text{ft}^2\). The homeowner needs \(18\) packages, costing \(\$628.20\).

- Which part of the enclosing rectangle actually belongs to the board? - How could you find its area without counting the empty opening? - For perimeter, trace the actual boundary, including the edges inside the opening.

1. The enclosing rectangle is \(6\,\text{in.}\times7\,\text{in.}\), so its area is \(6\cdot7=42\,\text{in.}^2\). 2. The cutout measures \(4\,\text{in.}\times3\,\text{in.}\), so its area is \(4\cdot3=12\,\text{in.}^2\). The C-shaped area is \(42-12=30\,\text{in.}^2\). 3. The four horizontal boundary segments total \(6+4+4+6=20\,\text{in.}\). 4. The two unlabeled right-side vertical segments together measure \(7-3=4\,\text{in.}\), so the vertical boundary segments total \(7+3+4=14\,\text{in.}\). 5. The perimeter is \(20+14=34\,\text{in.}\).

a) \(30\,\text{in.}^2\). b) \(34\,\text{in.}\).

- Split the L-shaped patio into rectangles. - You could also subtract a missing corner from one large rectangle. - The garden bed is not tiled, so what should you do with its area? - Find the patio area before subtracting the garden bed.

1. Split the L-shape into a \(6\,\text{ft}\times10\,\text{ft}\) rectangle and a \(6\,\text{ft}\times4\,\text{ft}\) rectangle. Their total area is \(6\cdot10+6\cdot4=60+24=84\,\text{ft}^2\). 2. The garden bed's area is \(2\cdot3=6\,\text{ft}^2\). 3. The tiled area is \(84-6=78\,\text{ft}^2\).

The area to be tiled is \(78\,\text{ft}^2\).

- Divide the T-shape into two rectangles. - Read the dimensions of each rectangle carefully from the diagram. - Multiply the total area by the cost per square meter.

1. Divide the T-shape into two rectangles. The top rectangle has area \(8\cdot3=24\,\text{m}^2\), and the lower rectangle has area \(4\cdot3=12\,\text{m}^2\). 2. The total area is \(24+12=36\,\text{m}^2\). 3. The total cost is \(36\cdot\$8.50=\$306.00\).

a) The lawn has area \(36\,\text{m}^2\). b) The sod costs \(\$306.00\).

- Divide the patio into two rectangles with known dimensions. - Find and add the areas of the two rectangles. - Multiply the total area by the price per square meter. - Add the one-time delivery fee.

1. Divide the patio into a \(5\,\text{m}\times2\,\text{m}\) rectangle and a \(3\,\text{m}\times6\,\text{m}\) rectangle. 2. Their areas are \(5\cdot2=10\,\text{m}^2\) and \(3\cdot6=18\,\text{m}^2\), so the patio area is \(10+18=28\,\text{m}^2\). 3. The pavers cost \(28\cdot\$35.00=\$980.00\). 4. Including delivery, the total cost is \(\$980.00+\$150.00=\$1130.00\).

a) The patio has area \(28\,\text{m}^2\). b) The total cost is \(\$1130.00\).

- Split the T-shape into a top rectangle and a lower rectangle. - Find each rectangle's area and add them. - Multiply the total area by the cost per square foot.

1. The top rectangle measures \(20\,\text{ft}\) by \(6\,\text{ft}\), so its area is \(20\cdot6=120\,\text{ft}^2\). 2. The lower rectangle measures \(8\,\text{ft}\) by \(10\,\text{ft}\), so its area is \(8\cdot10=80\,\text{ft}^2\). 3. The stage's total area is \(120+80=200\,\text{ft}^2\). 4. The carpet cost is \(200\cdot\$5=\$1000\).

a) The stage's area is \(200\,\text{ft}^2\). b) The carpet costs \(\$1000\).

- Find the area of the carpet before the opening is removed. - Find the area of the rectangular opening. - Subtract the opening's area from the full rectangle's area. - Keep the area units consistent.

1. The full rectangle's area is \(24\cdot18=432\,\text{ft}^2\). 2. The opening's area is \(14\cdot8=112\,\text{ft}^2\). 3. The remaining carpet area is \(432-112=320\,\text{ft}^2\).

The remaining carpet area is \(320\,\text{ft}^2\).

- View the cut-away regions as rectangles. - Use the square's side length to determine missing dimensions. - Pay attention to which visible \(3\,\text{cm}\) label is horizontal and derive the missing vertical length separately. - Add the areas of the two removed rectangles.

1. The original square has area \(12\cdot12=144\,\text{cm}^2\). 2. The lower removed rectangle extends \(12-7=5\,\text{cm}\) across. Its height is the part of the \(12\,\text{cm}\) side not accounted for by the visible \(5\,\text{cm}\) and \(4\,\text{cm}\) vertical segments, so its height is \(12-5-4=3\,\text{cm}\). Its area is \(5\cdot3=15\,\text{cm}^2\). 3. The other removed rectangle is \(12-7-3=2\,\text{cm}\) wide and \(5\,\text{cm}\) high, so its area is \(2\cdot5=10\,\text{cm}^2\). 4. The total cut-away area is \(15+10=25\,\text{cm}^2\).

The cut-away wood has an area of \(25\,\text{cm}^2\).

- Use symmetry to determine the width of the right side section. - Split the U-shape into three rectangles. - Find the total area, then divide by the area covered by one bag.

1. By symmetry, each side section is \(15\,\text{ft}\) wide. The opening is \(50-15-15=20\,\text{ft}\) wide, and the lower connecting section is \(40-30=10\,\text{ft}\) high. 2. Each side rectangle has area \(15\cdot40=600\,\text{ft}^2\). The lower middle rectangle has area \(20\cdot10=200\,\text{ft}^2\). 3. The total lawn area is \(600+600+200=1400\,\text{ft}^2\). 4. The number of bags is \(1400\div200=7\).

The lawn requires \(7\) bags of fertilizer.

- Can you find every missing horizontal and vertical length from the lengths shown? - How could familiar shapes help you measure the whole garden without overlap? - For the border, which segments are actually on the garden's edge?

1. The unlabeled lower-right vertical side is \(7-3=4\,\text{ft}\), and the upper horizontal side is \(8-3=5\,\text{ft}\). 2. Split the L-shape into rectangles measuring \(8\,\text{ft}\times4\,\text{ft}\) and \(5\,\text{ft}\times3\,\text{ft}\). 3. The area is \(8\cdot4+5\cdot3=47\,\text{ft}^2\). 4. Add all six outside edges for the border: \(8+4+3+3+5+7=30\,\text{ft}\).

a) \(47\,\text{ft}^2\). b) \(30\,\text{ft}\) of border.

- Split the T-shape into a top rectangle and a lower rectangle. - Use symmetry to find the two equal overhang lengths. - Add every outside edge to find the perimeter.

1. The top rectangle measures \(20\,\text{ft}\) by \(4\,\text{ft}\), so its area is \(20\cdot4=80\,\text{ft}^2\). 2. The lower rectangle measures \(12\,\text{ft}\) by \(8\,\text{ft}\), so its area is \(12\cdot8=96\,\text{ft}^2\). 3. The total area is \(80+96=176\,\text{ft}^2\). 4. Each horizontal overhang is \((20-12)\div2=4\,\text{ft}\). Adding the outside edges gives \(12+8+4+4+20+4+4+8=64\,\text{ft}\).

a) The patio's area is \(176\,\text{ft}^2\). b) The patio's perimeter is \(64\,\text{ft}\).

- Treat the U-shape as a large rectangle with a smaller rectangle removed. - Add every boundary segment to find the full perimeter. - Subtract the width of the opening from the perimeter.

1. View the plaza as a \(30\,\text{ft}\times18\,\text{ft}\) rectangle with a \(12\,\text{ft}\times9\,\text{ft}\) rectangular section removed. 2. The area is \(30\cdot18-12\cdot9=540-108=432\,\text{ft}^2\). 3. The full perimeter is \(30+18+9+9+12+9+9+18=114\,\text{ft}\). 4. Subtract the opening: \(114-6=108\,\text{ft}\).

a) The plaza's area is \(432\,\text{ft}^2\). b) \(108\,\text{ft}\) of railing is needed.

- Identify which part of the wall will not be painted. - Find the area of the wall and the area of the window. - Subtract to find the area that will be painted. - Use the paint coverage rate to find the number of liters needed. - Organize the calculation into separate steps.

1. The entire wall has area \(5\cdot3=15\,\text{m}^2\). 2. The window has area \(2\cdot1=2\,\text{m}^2\). 3. The area to be painted is \(15-2=13\,\text{m}^2\). 4. The amount of paint needed is \(13\div5=2.6\) liters.

\(2.6\) liters

- View the pool as one large rectangle with a smaller rectangle removed. - Determine the dimensions of the inner opening. - You could also split the U-shape into three rectangles. - Check how the total height is divided between the sections.

1. The enclosing rectangle has area \(30\cdot24=720\,\text{ft}^2\). 2. The rectangular opening is \(12\,\text{ft}\times15\,\text{ft}\), so its area is \(12\cdot15=180\,\text{ft}^2\). 3. The pool's area is \(720-180=540\,\text{ft}^2\).

The swimming pool's area is \(540\,\text{ft}^2\).

- Convert all lengths to centimeters first. - Try subtracting the opening from the outer rectangle. - Then verify by adding the areas of three rectangles.

1. Convert the measurements: \(80\,\text{mm}=8\,\text{cm}\), and \(20\,\text{mm}=2\,\text{cm}\). 2. Subtraction method: The outer rectangle has area \(10\cdot8=80\,\text{cm}^2\). The opening is \(10-2-2=6\,\text{cm}\) wide and \(5\,\text{cm}\) deep, so its area is \(6\cdot5=30\,\text{cm}^2\). The template's area is \(80-30=50\,\text{cm}^2\). 3. Addition method: The bottom rectangle is \(10\,\text{cm}\) wide and \(8-5=3\,\text{cm}\) high, so its area is \(30\,\text{cm}^2\). The two upper columns have total area \(2(2\cdot5)=20\,\text{cm}^2\). Thus, the area is \(30+20=50\,\text{cm}^2\).

The template's area is \(50\,\text{cm}^2\).

- Split the outer figure into rectangles or squares. - You could also start with a larger enclosing rectangle and subtract a missing corner. - Remember to subtract the square hole. - Find the outer area before accounting for the hole. - Identify the dimensions of each component.

1. The outer figure can be split into a \(6\,\text{cm}\times6\,\text{cm}\) square and a \(2\,\text{cm}\times3\,\text{cm}\) rectangle. Its outer area is \(6\cdot6+2\cdot3=36+6=42\,\text{cm}^2\). 2. The square hole has area \(2\cdot2=4\,\text{cm}^2\). 3. The area of the figure is \(42-4=38\,\text{cm}^2\).

The area is \(38\,\text{cm}^2\).

- Identify the dimensions of the outer rectangle. - Read the width and depth of the rectangular opening directly from the diagram. - Subtract the opening's area from the outer rectangle's area.

1. The outer rectangle has area \(8\cdot5=40\,\text{cm}^2\). 2. The opening is labeled \(4\,\text{cm}\) wide and \(3\,\text{cm}\) deep, so its area is \(4\cdot3=12\,\text{cm}^2\). 3. The U-shaped figure has area \(40-12=28\,\text{cm}^2\).

The area is \(28\,\text{cm}^2\).

- Split the figure into two rectangles with known dimensions. - Find missing side lengths by subtracting parallel lengths. - Add every boundary segment to find the full perimeter. - Subtract the gate width because no fence is needed there.

1. Split the figure into an \(18\,\text{ft}\times36\,\text{ft}\) rectangle and a \(27\,\text{ft}\times15\,\text{ft}\) rectangle. The area is \(18\cdot36+27\cdot15=648+405=1053\,\text{ft}^2\). 2. The missing horizontal side is \(45-18=27\,\text{ft}\), and the missing vertical side is \(36-15=21\,\text{ft}\). 3. The full perimeter is \(36+45+15+27+21+18=162\,\text{ft}\). 4. Subtract the gate width: \(162-9=153\,\text{ft}\).

a) \(1053\,\text{ft}^2\) b) \(153\,\text{ft}\) of fencing

- Complete each figure to a rectangle or square. - Identify the right triangle that must be subtracted. - Compare the two final areas.

1. Figure A fits inside a \(5\,\text{cm}\times5\,\text{cm}\) square. The missing upper-right triangle has legs \(2\,\text{cm}\) and \(2\,\text{cm}\), so Figure A has area \(25-\frac{1}{2}\cdot2\cdot2=23\,\text{cm}^2\). 2. Figure B fits inside a \(6\,\text{cm}\times4\,\text{cm}\) rectangle. The missing lower-left triangle has legs \(2\,\text{cm}\) and \(2\,\text{cm}\), so Figure B has area \(24-\frac{1}{2}\cdot2\cdot2=22\,\text{cm}^2\). 3. The difference is \(23-22=1\,\text{cm}^2\).

a) Figure A has area \(23\,\text{cm}^2\). b) Figure B has area \(22\,\text{cm}^2\). c) Figure A is larger by \(1\,\text{cm}^2\).

- Identify segments that would split each figure into familiar shapes. - Read each base and height from the grid. - Add the component areas for each figure.

1. Figure A consists of a \(4\,\text{cm}\times2\,\text{cm}\) rectangle and a triangle with base \(4\,\text{cm}\) and height \(2\,\text{cm}\). Its area is \(4\cdot2+\frac{1}{2}\cdot4\cdot2=8+4=12\,\text{cm}^2\). 2. Figure B consists of a \(3\,\text{cm}\times2\,\text{cm}\) rectangular shaft and a triangle with base \(4\,\text{cm}\) and height \(2\,\text{cm}\). Its area is \(3\cdot2+\frac{1}{2}\cdot4\cdot2=6+4=10\,\text{cm}^2\). 3. Figure C consists of a \(3\,\text{cm}\times3\,\text{cm}\) rectangle, a \(1\,\text{cm}\times1\,\text{cm}\) square, and a right triangle with legs \(2\,\text{cm}\) and \(1\,\text{cm}\). Its area is \(9+1+\frac{1}{2}\cdot2\cdot1=11\,\text{cm}^2\).

Figure A has area \(12\,\text{cm}^2\), Figure B has area \(10\,\text{cm}^2\), and Figure C has area \(11\,\text{cm}^2\).

- Look for a rectangle and a triangle. - Count grid intervals to find each base and height. - Add the two component areas.

1. Split the arrow into a rectangular shaft and a triangular arrowhead. 2. The shaft measures \(2\,\text{cm}\) by \(2\,\text{cm}\), so its area is \(2\cdot2=4\,\text{cm}^2\). 3. The arrowhead has base \(4\,\text{cm}\) and height \(3\,\text{cm}\), so its area is \(\frac{1}{2}\cdot4\cdot3=6\,\text{cm}^2\). 4. The total area is \(4+6=10\,\text{cm}^2\).

The arrow's area is \(10\,\text{cm}^2\).

- Split the ship into a trapezoid and a triangle. - Complete the crown to a rectangle and subtract the missing triangles. - Use the grid spacing to determine all dimensions.

1. The ship's hull is a trapezoid with bases \(4\,\text{cm}\) and \(6\,\text{cm}\) and height \(2\,\text{cm}\), so its area is \(\frac{1}{2}(4+6)\cdot2=10\,\text{cm}^2\). The sail is a right triangle with legs \(1\,\text{cm}\) and \(3\,\text{cm}\), so its area is \(\frac{1}{2}\cdot1\cdot3=1.5\,\text{cm}^2\). The ship's area is \(11.5\,\text{cm}^2\). 2. The crown fits inside a \(4\,\text{cm}\times3\,\text{cm}\) rectangle. Two triangles, each with base \(2\,\text{cm}\) and height \(2\,\text{cm}\), are missing. Its area is \(12-2\left(\frac{1}{2}\cdot2\cdot2\right)=8.0\,\text{cm}^2\).

a) \(11.5\,\text{cm}^2\) b) \(8.0\,\text{cm}^2\)

- Split Figure A into a central rectangle and two side triangles. - Split Figure B into two rectangles. - Add the component areas and compare the totals.

1. Figure A consists of a \(2\times3\) rectangle and two right triangles, each with legs \(1\) and \(3\). Its area is \(2\cdot3+2\left(\frac{1}{2}\cdot1\cdot3\right)=6+3=9\) square units. 2. Figure B consists of a \(2\times4\) rectangle and a \(1\times1\) square. Its area is \(2\cdot4+1\cdot1=9\) square units. 3. The two areas are equal.

a) Figure A has area \(9\) square units. b) Figure B has area \(9\) square units. c) The figures have equal areas.

- Identify a surrounding rectangle for each figure. - Find the areas of the outside right triangles. - Subtract the outside areas from the rectangle's area.

1. Figure D fits inside a \(4\times4\) square with area \(16\) square units. Four outside right triangles each have legs \(1\) and \(3\), so their total area is \(4\left(\frac{1}{2}\cdot1\cdot3\right)=6\) square units. Figure D has area \(16-6=10\) square units. 2. Figure E fits inside a \(5\times4\) rectangle with area \(20\) square units. The four outside triangles have total area \(2\left(\frac{1}{2}\cdot3\cdot2\right)+2\left(\frac{1}{2}\cdot2\cdot2\right)=10\) square units. Figure E has area \(20-10=10\) square units.

Figure D and Figure E each have area \(10\) square units.

- Identify the horizontal segment below the roof that separates the two shapes. - Find the rectangle's area and the triangle's area. - Add the two areas.

1. Split the figure into a lower rectangle and an upper triangle. 2. The rectangle measures \(4\times3\), so its area is \(4\cdot3=12\) square units. 3. The triangle has base \(4\) units and height \(2\) units, so its area is \(\frac{1}{2}\cdot4\cdot2=4\) square units. 4. The total area is \(12+4=16\) square units.

The figure has area \(16\) square units.

- Where might one straight line break the region into shapes with familiar area formulas? - What easy-to-measure rectangle could surround the whole figure? - Are both methods counting exactly the same covered region?

1. Draw a vertical segment from the inward top vertex straight down to the bottom side. This splits the figure into a \(2\times3\) rectangle of area \(6\) square units and a trapezoid. 2. The trapezoid has parallel sides \(1\) and \(3\) units long, separated by \(2\) units. Its area is \(\frac{1}{2}(1+3)\cdot2=4\) square units. The full area is \(6+4=10\) square units. 3. Alternatively, complete the figure to a \(4\times3\) rectangle of area \(12\) square units. 4. Subtract the missing right triangle, with legs \(2\) and \(2\) units, of area \(\frac{1}{2}\cdot2\cdot2=2\) square units. This gives \(12-2=10\) square units. 5. Both methods give the same result.

a) Decomposition gives \(10\) square units. b) Completion and subtraction give \(10\) square units. c) The results are equal.

- Identify the central rectangle, the top triangle, and the two lower triangles. - Read their dimensions from the grid. - Add all four component areas.

1. The central rectangle measures \(2\,\text{cm}\) by \(4\,\text{cm}\), so its area is \(2\cdot4=8\,\text{cm}^2\). 2. The top triangle has base \(2\,\text{cm}\) and height \(2\,\text{cm}\), so its area is \(\frac{1}{2}\cdot2\cdot2=2\,\text{cm}^2\). 3. Each lower fin is a right triangle with legs \(1\,\text{cm}\) and \(1\,\text{cm}\), so the two fins have total area \(2\left(\frac{1}{2}\cdot1\cdot1\right)=1\,\text{cm}^2\). 4. The total area is \(8+2+1=11\,\text{cm}^2\).

The rocket has area \(11\,\text{cm}^2\).

- Consider the horizontal diagonal connecting the leftmost and rightmost vertices. - Count grid intervals to find the shared base and each perpendicular height. - Add the two triangle areas.

1. Draw the horizontal diagonal joining the leftmost and rightmost vertices. It spans \(6\) grid units and divides the quadrilateral into two triangles. 2. The upper triangle has height \(2\) grid units, so its area is \(\frac{1}{2}\cdot6\cdot2=6\) square units. 3. The lower triangle also has height \(2\) grid units, so its area is \(6\) square units. 4. The total area is \(6+6=12\) square units.

The quadrilateral has area \(12\) square units.

- Split Figure a) into two nonoverlapping rectangles. - Use the grid to find Figure b)'s base and height. - Compare the two results.

1. Figure a) can be split into a \(4\times1\) rectangle and a nonoverlapping \(1\times4\) rectangle. Its area is \(4\cdot1+1\cdot4=8\) square units. 2. Figure b) is a triangle with base \(4\) units and height \(4\) units. Its area is \(\frac{1}{2}\cdot4\cdot4=8\) square units. 3. The figures have equal areas.

a) Figure a) has area \(8\) square units. b) Figure b) has area \(8\) square units. The two areas are equal.

- Identify the trapezoidal hull and triangular sail. - Read each base and height from the grid. - Add the two component areas.

1. The hull is a trapezoid with bases \(6\) and \(4\) and height \(2\). Its area is \(\frac{1}{2}(6+4)\cdot2=10\) square units. 2. The sail is a right triangle with base \(2\) and height \(4\). Its area is \(\frac{1}{2}\cdot2\cdot4=4\) square units. 3. The sailboat's total area is \(10+4=14\) square units.

The sailboat has area \(14\) square units.

- Split the figure into three rectangles. - Count grid intervals to find each rectangle's dimensions. - Add the three areas.

1. Split the figure into a left rectangle and two right rectangular arms. 2. The left rectangle measures \(4\times6\), so its area is \(4\cdot6=24\) square units. 3. Each right arm measures \(3\times2\), so each has area \(6\) square units. 4. The total area is \(24+6+6=36\) square units.

The figure has area \(36\) square units.

- Identify the horizontal segment that would create a rectangle and a triangle. - Count grid intervals to find each base and height. - Add the two areas.

1. The lower rectangle measures \(5\,\text{cm}\) by \(2\,\text{cm}\), so its area is \(5\cdot2=10\,\text{cm}^2\). 2. The upper triangle has base \(5\,\text{cm}\) and height \(3\,\text{cm}\), so its area is \(\frac{1}{2}\cdot5\cdot3=7.5\,\text{cm}^2\). 3. The total area is \(10+7.5=17.5\,\text{cm}^2\).

The figure has area \(17.5\,\text{cm}^2\).

- Identify a segment that would split the figure into two rectangles. - Add the two rectangle areas.

1. Split the figure into a lower \(3\,\text{cm}\times2\,\text{cm}\) rectangle and an upper \(2\,\text{cm}\times2\,\text{cm}\) rectangle. 2. Their areas are \(3\cdot2=6\,\text{cm}^2\) and \(2\cdot2=4\,\text{cm}^2\). 3. The total area is \(6+4=10\,\text{cm}^2\).

The figure has area \(10\,\text{cm}^2\).

- Split the figure into two rectangles. - Count grid intervals to find their dimensions.

1. Split the figure into a lower \(5\,\text{cm}\times1\,\text{cm}\) rectangle and an upper-right \(3\,\text{cm}\times2\,\text{cm}\) rectangle. 2. Their areas are \(5\cdot1=5\,\text{cm}^2\) and \(3\cdot2=6\,\text{cm}^2\). 3. The total area is \(5+6=11\,\text{cm}^2\).

The figure has area \(11\,\text{cm}^2\).

- Split the figure into rectangles and a triangle. - Pay special attention to the slanted edge.

1. Split the figure into a lower \(4\,\text{cm}\times2\,\text{cm}\) rectangle, an upper-right \(2\,\text{cm}\times2\,\text{cm}\) rectangle, and a right triangle with legs \(1\,\text{cm}\) and \(2\,\text{cm}\). 2. The component areas are \(8\,\text{cm}^2\), \(4\,\text{cm}^2\), and \(\frac{1}{2}\cdot1\cdot2=1\,\text{cm}^2\). 3. The total area is \(8+4+1=13\,\text{cm}^2\).

The figure has area \(13\,\text{cm}^2\).

- Identify the square in the center. - View the four points as congruent triangles. - Add the square and triangle areas.

1. The center is a square with side length \(2\,\text{cm}\), so its area is \(2\cdot2=4\,\text{cm}^2\). 2. Each of the four points is a triangle with base \(2\,\text{cm}\) and height \(2\,\text{cm}\), so each has area \(\frac{1}{2}\cdot2\cdot2=2\,\text{cm}^2\). 3. The total area is \(4+4\cdot2=12\,\text{cm}^2\).

The star has area \(12\,\text{cm}^2\).

- Split Figure a) into rectangles. - Use the base and height of Figure b). - Compare the two results.

1. Figure a) can be split into a \(3\,\text{cm}\times1\,\text{cm}\) rectangle and a \(1\,\text{cm}\times1\,\text{cm}\) square. Its area is \(3\cdot1+1\cdot1=4\,\text{cm}^2\). 2. Figure b) is a triangle with base \(4\,\text{cm}\) and height \(2\,\text{cm}\). Its area is \(\frac{1}{2}\cdot4\cdot2=4\,\text{cm}^2\). 3. The figures have equal areas.

Each figure has area \(4\,\text{cm}^2\), so they are equal in area.

- Split Figure a) into two rectangles. - Split Figure b) into a rectangle and a triangle. - Read all dimensions from the grid.

1. Figure a) consists of a lower \(4\times2\) rectangle and an upper \(2\times3\) rectangle. Its area is \(4\cdot2+2\cdot3=14\) square units. 2. Figure b) consists of a \(4\times3\) rectangle and a triangle with base \(4\) and height \(3\). Its area is \(4\cdot3+\frac{1}{2}\cdot4\cdot3=12+6=18\) square units.

a) Figure a) has area \(14\) square units, and Figure b) has area \(18\) square units. b) Figure b) can be split into a \(4\times3\) rectangle and a triangle with base \(4\) and height \(3\).

- Trace each figure's entire boundary to find the perimeter. - For area, split the figure into rectangles or subtract missing pieces from a larger rectangle. - Include the short segments around every inward corner. - Use the shaded reference square to identify one square unit.

1. Figure 1: Add the eight side lengths: \(P=5+2+1+3+3+2+3+3=22\) units. Decomposing the figure into rectangles gives \(A=3\cdot3+2\cdot5+1\cdot3=9+10+3=22\) square units. 2. Figure 2: Add the eight side lengths: \(P=3+1+1+1+1+1+1+3=12\) units. Counting by horizontal strips gives \(A=3+2+1=6\) square units.

1: \(P=22\) units, \(A=22\) square units 2: \(P=12\) units, \(A=6\) square units

- Split Figure f into a rectangle and a triangle. - Use the base and height of Figure g. - Compare the two areas.

1. Figure f consists of a \(4\times2\) rectangle and a triangle with base \(4\) and height \(2\). Its area is \(4\cdot2+\frac{1}{2}\cdot4\cdot2=8+4=12\) square units. 2. Figure g is a triangle with base \(6\) and height \(4\). Its area is \(\frac{1}{2}\cdot6\cdot4=12\) square units. 3. The figures have equal areas.

Both figures have area \(12\) square units, so they are equal in area.

- Subtraction may be easier than adding many small pieces. - What rectangle encloses the entire figure? - What shape is the notch cut into the top?

1. The enclosing rectangle is \(6\) units by \(4\) units, so its area is \(6\cdot4=24\) square units. 2. The triangular notch has a base of \(2\) units and a height of \(3\) units, so its area is \(\frac{1}{2}\cdot2\cdot3=3\) square units. 3. The figure's area is \(24-3=21\) square units.

The area is \(21\) square units.

- Split the figure into two rectangles. - Or complete it to a \(4\times4\) square and subtract the missing corner. - Add or subtract the appropriate areas.

1. Split the figure into a \(4\times2\) rectangle and a \(2\times2\) rectangle. 2. Their areas are \(4\cdot2=8\) and \(2\cdot2=4\) square units. 3. The total area is \(8+4=12\) square units.

The figure has area \(12\) square units.

- Begin with a complete \(5\times5\) square. - Identify the dimensions of the missing rectangle. - Subtract the missing area from the square's area.

1. The outer square has area \(5\cdot5=25\) square units. 2. The cutout is \(4\) units by \(3\) units, so its area is \(4\cdot3=12\) square units. 3. The C-shaped area is \(25-12=13\) square units.

The area is \(13\) square units.

- Find the area before the opening is removed. - Find the area of the square opening. - Subtract the opening's area from the full area.

1. The full panel's area is \(15\cdot10=150\,\text{cm}^2\). 2. The square opening's area is \(4^2=16\,\text{cm}^2\). 3. The remaining glass area is \(150-16=134\,\text{cm}^2\).

The area of the remaining glass is \(134\,\text{cm}^2\).

- For decomposition, identify horizontal segments that would create three rectangles. - Use the labeled arm height and the total height to determine the other arm's height. - For completion, identify the outer rectangle and the missing rectangle. - Compare the number and type of calculations in the two methods.

1. Decomposition: One horizontal arm is \(2\,\text{cm}\) high. The other is \(7-3-2=2\,\text{cm}\) high. The top and bottom rectangles therefore each measure \(10\,\text{cm}\times2\,\text{cm}\), and the left connecting rectangle measures \(4\,\text{cm}\times3\,\text{cm}\). The area is \(20+20+12=52\,\text{cm}^2\). 2. Completion: The surrounding rectangle measures \(10\,\text{cm}\times7\,\text{cm}\), so its area is \(70\,\text{cm}^2\). The missing rectangle measures \(6\,\text{cm}\times3\,\text{cm}\), so the figure's area is \(70-18=52\,\text{cm}^2\). 3. Both methods give the same area. One reasonable preference is the completion method because it uses two rectangle areas and one subtraction. The decomposition method is also a valid preference with a clear explanation.

The area is \(52\,\text{cm}^2\) by either method. Either method may be identified as easier if the choice is supported with a reasonable explanation.

- Which part of the lot will receive pavement? - How does the garden area affect the amount to be paved? - What area does the paving price apply to?

1. The full parking lot has area \(60\cdot40=2400\,\text{ft}^2\). 2. The landscaped area has area \(15\cdot10=150\,\text{ft}^2\). 3. The paved area is \(2400-150=2250\,\text{ft}^2\). 4. The paving cost is \(2250\cdot\$4.50=\$10{,}125.00\).

a) \(2250\,\text{ft}^2\). b) \(\$10{,}125.00\).

- Find the area before the hole is removed. - Find the area of the square. - Subtract the missing area.

1. The parallelogram's area is \(20\cdot15=300\,\text{cm}^2\). 2. The square hole's area is \(4^2=16\,\text{cm}^2\). 3. The remaining area is \(300-16=284\,\text{cm}^2\).

The remaining area is \(284\,\text{cm}^2\).

- Find the area of the full outer rectangle. - Find the area of the inner opening. - Subtract to keep only the frame area.

1. The outer rectangle's area is \(10\cdot8=80\,\text{in.}^2\). 2. The opening's area is \(6\cdot4=24\,\text{in.}^2\). 3. The frame's area is \(80-24=56\,\text{in.}^2\).

The area of the frame is \(56\,\text{in.}^2\).

- Start with one large enclosing rectangle. - Read the cutout's width and height from the diagram. - Subtract the cutout area from the outer area.

1. The enclosing rectangle has area \(36\cdot32=1152\,\text{in.}^2\). 2. The rectangular cutout is labeled \(24\,\text{in.}\) wide and \(16\,\text{in.}\) high, so its area is \(24\cdot16=384\,\text{in.}^2\). 3. The desktop's area is \(1152-384=768\,\text{in.}^2\).

The desktop's area is \(768\,\text{in.}^2\).

- View the bed as a square with a rectangle missing. - Read the missing rectangle's dimensions from the diagram. - Subtract that area from the square.

1. Begin with the enclosing \(10\,\text{ft}\times10\,\text{ft}\) square, which has area \(100\,\text{ft}^2\). 2. The missing rectangle is labeled \(4\,\text{ft}\) wide and \(6\,\text{ft}\) high, so its area is \(4\cdot6=24\,\text{ft}^2\). 3. The flower bed's area is \(100-24=76\,\text{ft}^2\).

The flower bed's area is \(76\,\text{ft}^2\).

- Split the floor plan into three rectangles. - Notice that the two side sections have different heights. - Add the three rectangle areas.

1. Split the floor into a bottom rectangle and two side rectangles. 2. The bottom rectangle has area \(24\cdot8=192\,\text{ft}^2\). 3. The left upper rectangle has area \(6\cdot8=48\,\text{ft}^2\), and the right upper rectangle has area \(6\cdot12=72\,\text{ft}^2\). 4. The total area is \(192+48+72=312\,\text{ft}^2\).

The storage room has floor area \(312\,\text{ft}^2\).

- Find the area of the full rectangle. - Find the area of the rectangular notch. - Subtract the notch from the full plate.

1. The full \(10\,\text{cm}\times10\,\text{cm}\) plate has area \(10\cdot10=100\,\text{cm}^2\). 2. The notch measures \(2\,\text{cm}\) by \(3\,\text{cm}\), so its area is \(2\cdot3=6\,\text{cm}^2\). 3. The remaining area is \(100-6=94\,\text{cm}^2\).

The remaining plate has area \(94\,\text{cm}^2\).

- Which simpler rectangles could make up each plot, or complete its outline? - How can you find each area without counting any part twice? - Compare the areas using the same square units.

1. Plot a) is a \(36\,\text{ft}\times30\,\text{ft}\) rectangle with a \(12\,\text{ft}\times12\,\text{ft}\) section removed. Its area is \(36\cdot30-12\cdot12=936\,\text{ft}^2\). 2. Plot b) consists of a \(30\,\text{ft}\times30\,\text{ft}\) rectangle and an \(18\,\text{ft}\times9\,\text{ft}\) extension. Its area is \(30\cdot30+18\cdot9=1062\,\text{ft}^2\). 3. Since \(1062>936\), plot b) has the greater area.

a) Plot a): \(936\,\text{ft}^2\). Plot b): \(1062\,\text{ft}^2\). b) Plot b) has the greater area.

- Which two rectangles make up the T-shape? - For the perimeter, the two unlabeled overhangs can be handled together by comparing the top width with the stem width. - Trace the entire outside boundary without assuming the two overhangs are equal.

1. Split the plate into a \(12\,\text{cm}\times3\,\text{cm}\) top rectangle and a \(4\,\text{cm}\times5\,\text{cm}\) stem. The area is \(12\cdot3+4\cdot5=36+20=56\,\text{cm}^2\). 2. For the perimeter, the vertical boundary lengths total \(3+5+5+3=16\,\text{cm}\). The two unlabeled horizontal overhangs together measure \(12-4=8\,\text{cm}\), so the horizontal boundary lengths total \(12+4+8=24\,\text{cm}\). Therefore, the perimeter is \(16+24=40\,\text{cm}\).

The area is \(56\,\text{cm}^2\), and the perimeter is \(40\,\text{cm}\).

- Find the area of the full rectangle. - Find the area of the smaller opening. - Subtract the removed area.

1. The full panel's area is \(80\cdot50=4000\,\text{cm}^2\). 2. The opening's area is \(30\cdot20=600\,\text{cm}^2\). 3. The remaining area is \(4000-600=3400\,\text{cm}^2\).

The remaining glass area is \(3400\,\text{cm}^2\).

- The path adds to both ends of the length and both ends of the width. - Find the area of the entire outer rectangle. - Subtract the area of the garden bed.

1. The outer rectangle is \(6+2=8\,\text{m}\) long and \(4+2=6\,\text{m}\) wide because the path adds \(1\,\text{m}\) on both sides of each dimension. 2. The outer area is \(8\cdot6=48\,\text{m}^2\). 3. The garden bed has area \(6\cdot4=24\,\text{m}^2\). 4. The paved area is \(48-24=24\,\text{m}^2\).

The paved path has area \(24\,\text{m}^2\).

- Split the T-shape into rectangles. - Find each rectangle's area and add. - Round the number of cans up because a partial can cannot be purchased.

1. Split the deck into a \(24\,\text{ft}\times9\,\text{ft}\) rectangle and a \(9\,\text{ft}\times12\,\text{ft}\) rectangle. 2. The total area is \(24\cdot9+9\cdot12=216+108=324\,\text{ft}^2\). 3. The number of cans is \(324\div135=2.4\). Since only whole cans can be purchased, \(3\) cans are needed.

The deck's area is \(324\,\text{ft}^2\), and \(3\) cans are needed.

- Find the area of the original square. - Split the finished part into two rectangles. - Subtract the part's area from the square's area.

1. The original board has area \(40\cdot40=1600\,\text{cm}^2\). 2. The finished part can be split into a \(40\,\text{cm}\times30\,\text{cm}\) rectangle and a \(30\,\text{cm}\times10\,\text{cm}\) rectangle. Its area is \(1200+300=1500\,\text{cm}^2\). 3. The waste area is \(1600-1500=100\,\text{cm}^2\).

The waste has area \(100\,\text{cm}^2\).

- View the frame as a full rectangle with a smaller rectangle removed. - Subtract the center area from the outer area.

1. The full plate's area is \(25\cdot20=500\,\text{cm}^2\). 2. The cutout's area is \(15\cdot10=150\,\text{cm}^2\). 3. The frame's area is \(500-150=350\,\text{cm}^2\).

The finished frame has an area of \(350\,\text{cm}^2\).

- Treat the pool as a rectangular opening in the outside area. - Subtract the pool area from the full rectangle. - Multiply the deck area by the price per square foot.

1. The outside area is \(36\cdot24=864\,\text{ft}^2\). 2. The pool area is \(24\cdot12=288\,\text{ft}^2\). 3. The deck area is \(864-288=576\,\text{ft}^2\). 4. The cost is \(576\cdot\$4.50=\$2592.00\).

The deck area is \(576\,\text{ft}^2\), and the total cost is \(\$2592.00\).

- Split the T-shape into two rectangles. - Use the labeled dimensions to identify the upper rectangle's width and height. - Add the two rectangle areas.

1. Split the roof into a lower \(40\,\text{ft}\times10\,\text{ft}\) rectangle and an upper \(8\,\text{ft}\times20\,\text{ft}\) rectangle. 2. Their areas are \(40\cdot10=400\,\text{ft}^2\) and \(8\cdot20=160\,\text{ft}^2\). 3. The total area is \(400+160=560\,\text{ft}^2\).

The roof has area \(560\,\text{ft}^2\).

- Find the area of the entire wall. - Find the combined area of all five windows. - Subtract the window area, then use the cost per square meter.

1. The entire wall has area \(20\cdot10=200\,\text{m}^2\). 2. One window has area \(2\cdot1.5=3\,\text{m}^2\), so the five windows have total area \(5\cdot3=15\,\text{m}^2\). 3. The area to be painted is \(200-15=185\,\text{m}^2\). 4. The total cost is \(185\cdot\$8.00=\$1480.00\).

The total painting cost is \(\$1480.00\).

- Find the area of the whole wall first. - Find the area of the doorway. - Subtract the doorway area from the wall area.

1. The entire wall has area \(3.5\cdot2.4=8.4\,\text{m}^2\). 2. The doorway has area \(0.8\cdot2=1.6\,\text{m}^2\). 3. The area to be painted is \(8.4-1.6=6.8\,\text{m}^2\).

The area to be painted is \(6.8\,\text{m}^2\).

- Split the plaza into two rectangles. - Read each rectangle's width and height from the diagram. - Add the two rectangle areas.

1. The left rectangle measures \(24\,\text{ft}\times12\,\text{ft}\), and the narrow right rectangle is labeled \(9\,\text{ft}\) wide and \(15\,\text{ft}\) high. 2. The total area is \(24\cdot12+9\cdot15=288+135=423\,\text{ft}^2\).

The plaza's total area is \(423\,\text{ft}^2\).

- Identify a segment that would split the L-shape into two rectangles. - Find the missing height by subtraction. - Add the two rectangle areas.

1. The lower rectangle measures \(15\,\text{cm}\) by \(5\,\text{cm}\), so its area is \(15\cdot5=75\,\text{cm}^2\). 2. The upper-left rectangle is \(5\,\text{cm}\) wide and \(15-5=10\,\text{cm}\) high, so its area is \(5\cdot10=50\,\text{cm}^2\). 3. The total area is \(75+50=125\,\text{cm}^2\).

The wooden piece has area \(125\,\text{cm}^2\).

- Split the figure into three narrow vertical rectangles. - Find the width of each step. - A horizontal decomposition also works.

1. Split the figure into three vertical rectangles. From left to right, their dimensions are \(9\,\text{ft}\times18\,\text{ft}\), \(9\,\text{ft}\times12\,\text{ft}\), and \(9\,\text{ft}\times6\,\text{ft}\). 2. The total area is \(9\cdot18+9\cdot12+9\cdot6=162+108+54=324\,\text{ft}^2\).

The terrace's total area is \(324\,\text{ft}^2\).

- Divide the shape into three horizontal rectangles. - Build each strip's width by adding the visible \(2\,\text{cm}\) horizontal step lengths. - Use the labeled vertical step lengths for the strip heights, then add the three areas.

1. Divide the shape into three horizontal rectangles. The labeled horizontal steps are each \(2\,\text{cm}\), so the three strip widths are \(2+2+2=6\,\text{cm}\), \(2+2=4\,\text{cm}\), and \(2\,\text{cm}\). 2. Their labeled heights are \(1\,\text{cm}\), \(2\,\text{cm}\), and \(2\,\text{cm}\). Their areas are \(6\cdot1=6\,\text{cm}^2\), \(4\cdot2=8\,\text{cm}^2\), and \(2\cdot2=4\,\text{cm}^2\). 3. The total area is \(6+8+4=18\,\text{cm}^2\).

The logo has area \(18\,\text{cm}^2\).

- Divide the L-shape into two rectangles. - Use the given dimensions to find any missing side length. - First determine how many liters fall on each square meter.

1. Divide the roof into two rectangles. Their areas are \(12\cdot15=180\,\text{m}^2\) and \((20-12)\cdot8=64\,\text{m}^2\). 2. The total roof area is \(180+64=244\,\text{m}^2\). 3. A rainfall depth of \(8\,\text{mm}\) gives \(8\,\text{L}\) per square meter, so the runoff is \(244\cdot8=1952\,\text{L}\).

A total of \(1952\,\text{L}\) of rainwater runs off the roof.

- Subtracting one inner rectangle may be easier than adding several pieces. - Find the dimensions of the enclosing rectangle. - Find the dimensions of the opening that will not be paved.

1. The enclosing rectangle has area \(24\cdot18=432\,\text{ft}^2\). 2. The inner opening has area \(12\cdot9=108\,\text{ft}^2\). 3. The patio area is \(432-108=324\,\text{ft}^2\). 4. The total cost is \(324\cdot\$5.50=\$1782.00\).

The total paving cost is \(\$1782.00\).

- Find each lot's area separately. - Split Lot B into two rectangles. - Subtract the smaller area from the larger area.

1. Lot A has area \(66\cdot54=3564\,\text{ft}^2\). 2. Split Lot B into a \(75\,\text{ft}\times36\,\text{ft}\) rectangle and a \(30\,\text{ft}\times(60-36)\,\text{ft}\) rectangle. Its area is \(75\cdot36+30\cdot24=2700+720=3420\,\text{ft}^2\). 3. The difference is \(3564-3420=144\,\text{ft}^2\), so Lot A is larger.

Lot A is larger by \(144\,\text{ft}^2\).

- Split the figure into two rectangles. - Determine the dimensions of both parts. - Multiply the total area by the price per square foot.

1. Split the patio into a \(36\,\text{ft}\times15\,\text{ft}\) rectangle and an \(18\,\text{ft}\times15\,\text{ft}\) rectangle. 2. The total area is \(36\cdot15+18\cdot15=540+270=810\,\text{ft}^2\). 3. The total cost is \(810\cdot\$4.50=\$3645.00\).

a) \(810\,\text{ft}^2\) b) \(\$3645.00\)

- Find the area of the large rectangle. - Subtract the areas of the three equal squares. - Multiply the paved area by the cost per square foot.

1. The full courtyard has area \(30\cdot16=480\,\text{ft}^2\). 2. The three \(4\,\text{ft}\times4\,\text{ft}\) cutouts have total area \(3(4\cdot4)=48\,\text{ft}^2\). 3. The paved area is \(480-48=432\,\text{ft}^2\). 4. The material cost is \(432\cdot\$7=\$3024\).

a) The paved area is \(432\,\text{ft}^2\). b) The materials cost \(\$3024\).

- Find the area of the original square. - Find the area removed from the center. - Subtract the opening from the original area.

1. The full square has area \(10\cdot10=100\,\text{cm}^2\). 2. The rectangular opening has area \(6\cdot8=48\,\text{cm}^2\). 3. The logo's area is \(100-48=52\,\text{cm}^2\).

The logo has area \(52\,\text{cm}^2\).

- Divide the figure into rectangles. - Tile boxes can only be purchased in whole numbers. - Use the coverage of one box to find how many boxes are needed.

1. A reasonable estimate for the floor area is about \(40\,\text{m}^2\). 2. Divide the floor into two rectangles. Their areas are \(8\cdot3=24\,\text{m}^2\) and \(3\cdot4=12\,\text{m}^2\). 3. The exact area is \(24+12=36\,\text{m}^2\). 4. The number of boxes is \(36\div5=7.2\), so \(8\) whole boxes are required. 5. The total cost is \(8\cdot\$89.50=\$716.00\).

A reasonable estimate is about \(40\,\text{m}^2\), and the exact area is \(36\,\text{m}^2\). Ms. Meyer must buy \(8\) boxes, costing \(\$716.00\) in total.

- Split the figure into two rectangles. - For another method, complete it to a larger rectangle. - Subtract the missing corner from the larger rectangle.

1. Horizontal decomposition: A \(7\,\text{cm}\times3\,\text{cm}\) lower rectangle and a \(3\,\text{cm}\times3\,\text{cm}\) upper square have total area \(7\cdot3+3\cdot3=30\,\text{cm}^2\). 2. Completion: A \(7\,\text{cm}\times6\,\text{cm}\) outer rectangle has area \(42\,\text{cm}^2\). Subtract the \(4\,\text{cm}\times3\,\text{cm}\) missing rectangle: \(42-12=30\,\text{cm}^2\).

Both methods give an area of \(30\,\text{cm}^2\).

- Split the figure into two rectangles. - Find the length that is not labeled directly. - Compare parallel sides to determine missing lengths.

1. The lower rectangle measures \(10\,\text{cm}\times4\,\text{cm}\), so its area is \(40\,\text{cm}^2\). 2. The upper rectangle is \(6\,\text{cm}\) wide. Its height is \(9-4=5\,\text{cm}\), so its area is \(6\cdot5=30\,\text{cm}^2\). 3. The total area is \(40+30=70\,\text{cm}^2\).

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

- Begin with the plate before the cutout is removed. - Find the area of the missing rectangle. - Subtraction is simpler than dividing the frame into many pieces.

1. The full plate's area is \(12\cdot8=96\,\text{cm}^2\). 2. The cutout's area is \(6\cdot4=24\,\text{cm}^2\). 3. The remaining area is \(96-24=72\,\text{cm}^2\).

The remaining area is \(72\,\text{cm}^2\).

- Find the gray area without counting the inner rectangle. - Subtract the flower bed from the full patio area. - Multiply the net area by the cost per square foot.

1. The entire patio area is \(30\cdot18=540\,\text{ft}^2\). 2. The flower bed area is \(9\cdot6=54\,\text{ft}^2\). 3. The paved area is \(540-54=486\,\text{ft}^2\). 4. The total cost is \(486\cdot\$2.80=\$1360.80\).

The pavers cost \(\$1360.80\).

- Identify a segment that would split the L-shape into two rectangles. - For the second method, identify the outer rectangle and the missing upper-right rectangle. - Check that both methods give the same result.

1. Decomposition: The lower rectangle has area \(48\cdot16=768\,\text{ft}^2\). The upper-left rectangle is \(20\,\text{ft}\) wide and \(42-16=26\,\text{ft}\) high, so its area is \(20\cdot26=520\,\text{ft}^2\). The total is \(768+520=1288\,\text{ft}^2\). 2. Completion: The outer rectangle has area \(48\cdot42=2016\,\text{ft}^2\). The missing rectangle measures \(48-20=28\,\text{ft}\) by \(26\,\text{ft}\), so its area is \(728\,\text{ft}^2\). The difference is \(2016-728=1288\,\text{ft}^2\).

The parking area has area \(1288\,\text{ft}^2\).

- View the patio as an enclosing rectangle with a rectangular opening removed. - Read the opening's width and height directly from the diagram. - Subtract the opening's area from the enclosing rectangle's area.

1. The enclosing rectangle has area \(36\cdot24=864\,\text{ft}^2\). 2. The top opening is labeled \(18\,\text{ft}\) wide and \(15\,\text{ft}\) high, so its area is \(18\cdot15=270\,\text{ft}^2\). 3. The patio area is \(864-270=594\,\text{ft}^2\).

The patio's total area is \(594\,\text{ft}^2\).

- Split the wall into a rectangle and a triangle. - The window area must be subtracted. - Use one-half of base times height for the gable.

1. The rectangular wall area is \(24\cdot12=288\,\text{ft}^2\). 2. The triangular gable area is \(\frac{1}{2}\cdot24\cdot9=108\,\text{ft}^2\). 3. The window area is \(6\cdot4.5=27\,\text{ft}^2\). 4. The area to paint is \(288+108-27=369\,\text{ft}^2\).

The area to be painted is \(369\,\text{ft}^2\).

- Start with the area of the full rectangle. - What is the perpendicular distance from the triangle's opposite vertex to the right side? - Subtract the triangle's area from the rectangle's area.

1. The full rectangle has area \(10\cdot6=60\,\text{cm}^2\). 2. The removed triangle has base \(6\,\text{cm}\) and perpendicular height \(10-6=4\,\text{cm}\). 3. Its area is \(\frac12\cdot6\cdot4=12\,\text{cm}^2\). 4. The remaining area is \(60-12=48\,\text{cm}^2\).

The logo has area \(48\,\text{cm}^2\).

- Split the figure into two simpler shapes. - Determine the dimensions of the two rectangles.

1. Split the figure into a \(3\,\text{cm}\times6\,\text{cm}\) rectangle and a \(3\,\text{cm}\times3\,\text{cm}\) rectangle. 2. The total area is \(3\cdot6+3\cdot3=18+9=27\,\text{cm}^2\). 3. Equivalently, subtract a \(3\,\text{cm}\times3\,\text{cm}\) square from a \(6\,\text{cm}\times6\,\text{cm}\) square: \(36-9=27\,\text{cm}^2\).

The area is \(27\,\text{cm}^2\).

- Find the areas of the two component shapes separately. - Subtract the rectangle's height from the total height to find the triangle's height.

1. The rectangle's area is \(8\cdot5=40\,\text{cm}^2\). 2. The triangle's height is \(8-5=3\,\text{cm}\), so its area is \(\frac{1}{2}\cdot8\cdot3=12\,\text{cm}^2\). 3. The total area is \(40+12=52\,\text{cm}^2\).

The area is \(52\,\text{cm}^2\).

- Find the area of the full sheet. - Subtract the area of the rectangular cutout.

1. The large rectangle has area \(10\cdot7=70\,\text{cm}^2\). 2. The cutout has area \(4\cdot3=12\,\text{cm}^2\). 3. The remaining area is \(70-12=58\,\text{cm}^2\).

The remaining metal has area \(58\,\text{cm}^2\).

- Split the figure into two rectangles. - Find missing side lengths by comparing the full dimensions. - The perimeter is the length of the entire outside boundary. - You can also subtract a missing corner from a large rectangle to find the area.

1. Split the figure into a \(12\,\text{ft}\times15\,\text{ft}\) rectangle and a \(9\,\text{ft}\times9\,\text{ft}\) rectangle. The area is \(12\cdot15+9\cdot9=180+81=261\,\text{ft}^2\). 2. The missing horizontal length is \(21-12=9\,\text{ft}\), and the missing vertical length is \(15-6=9\,\text{ft}\). 3. The perimeter is \(21+9+9+6+12+15=72\,\text{ft}\).

The area is \(261\,\text{ft}^2\), and the perimeter is \(72\,\text{ft}\).

- Split the figure into two rectangles. - Use the total height and lower-rectangle height to find the upper height. - Trace the entire outside boundary for the perimeter.

1. Split the figure into a lower \(24\,\text{ft}\times12\,\text{ft}\) rectangle and an upper rectangle. The upper rectangle is \(12\,\text{ft}\) wide and \(21-12=9\,\text{ft}\) high. 2. The area is \(24\cdot12+12\cdot9=288+108=396\,\text{ft}^2\). 3. The six outside edges are \(12\), \(9\), \(12\), \(12\), \(24\), and \(21\) feet, so the perimeter is \(12+9+12+12+24+21=90\,\text{ft}\).

The area is \(396\,\text{ft}^2\), and the perimeter is \(90\,\text{ft}\).

- Add the areas of the main rectangle and the tab. - For perimeter, handle the two unlabeled vertical pieces together by subtracting the tab height from the rectangle height. - Trace every remaining outside edge once.

1. The main rectangle has area \(15\cdot10=150\,\text{cm}^2\), and the tab has area \(5\cdot4=20\,\text{cm}^2\). The total area is \(170\,\text{cm}^2\). 2. For the perimeter, the two unlabeled vertical pieces beside the tab together measure \(10-4=6\,\text{cm}\). The remaining boundary consists of two \(15\,\text{cm}\) horizontal sides, two \(5\,\text{cm}\) tab edges, the \(4\,\text{cm}\) tab height, and the \(10\,\text{cm}\) left side. Thus the perimeter is \(6+15+5+4+5+15+10=60\,\text{cm}\).

The area is \(170\,\text{cm}^2\), and the perimeter is \(60\,\text{cm}\).

- Subtract the opening from the outer rectangle to find area. - Include the inner edges of the U when finding perimeter. - The two unlabeled bottom pieces can be handled together as the outer width minus the opening width.

1. The outer \(10\,\text{cm}\times8\,\text{cm}\) rectangle has area \(80\,\text{cm}^2\). The \(6\,\text{cm}\times5\,\text{cm}\) opening has area \(30\,\text{cm}^2\). 2. The colored area is \(80-30=50\,\text{cm}^2\). 3. For the perimeter, the two unlabeled bottom pieces together measure \(10-6=4\,\text{cm}\). Add the outer top \(10\), the two outer sides \(8+8\), the combined bottom pieces \(4\), the two inner sides \(5+5\), and the inner horizontal edge \(6\): \(10+8+8+4+5+5+6=46\,\text{cm}\).

The area is \(50\,\text{cm}^2\), and the perimeter is \(46\,\text{cm}\).

- Identify a line that would split the figure into two rectangles. - Determine the missing side lengths of the rectangles. - Add the two component areas.

1. The lower rectangle measures \(10\,\text{cm}\times3\,\text{cm}\), so its area is \(30\,\text{cm}^2\). 2. The upper-left rectangle is \(4\,\text{cm}\) wide and \(7-3=4\,\text{cm}\) high, so its area is \(16\,\text{cm}^2\). 3. The total area is \(30+16=46\,\text{cm}^2\).

The area is \(46\,\text{cm}^2\).

- Split the floor plan into two rectangles. - Find the missing height by subtraction. - Add the rectangle areas.

1. The lower rectangle measures \(40\,\text{ft}\) by \(12\,\text{ft}\), so its area is \(40\cdot12=480\,\text{ft}^2\). 2. The upper-left rectangle is \(16\,\text{ft}\) wide and \(30-12=18\,\text{ft}\) high, so its area is \(16\cdot18=288\,\text{ft}^2\). 3. The total floor area is \(480+288=768\,\text{ft}^2\).

The apartment has \(768\,\text{ft}^2\) of floor area.

- Is it easier to cut this stepped region into parts or to add a missing part around it? - What makes an edge part of the outside perimeter? - Check that no area is counted twice and that no outer edge is missed.

1. The figure consists of a \(12\,\text{cm}\times4\,\text{cm}\) base rectangle and two \(2\,\text{cm}\times3\,\text{cm}\) raised rectangles. 2. The area is \(12\cdot4+2(2\cdot3)=60\,\text{cm}^2\). 3. Adding every outside edge gives \(12+4+3+2+3+3+3+2+3+3+2+4=44\,\text{cm}\).

The area is \(60\,\text{cm}^2\), and the perimeter is \(44\,\text{cm}\).

- Subtract the two square notches from the original rectangle. - Compare the original top edge with the three edges created by each notch. - Account for both notches when finding perimeter.

1. The original rectangle measures \(40\,\text{in.}\times24\,\text{in.}\), so its area is \(40\cdot24=960\,\text{in.}^2\). 2. The two \(8\,\text{in.}\times8\,\text{in.}\) notches remove \(2(8\cdot8)=128\,\text{in.}^2\). 3. The remaining area is \(960-128=832\,\text{in.}^2\). 4. The original perimeter is \(2(40+24)=128\,\text{in.}\). Each notch increases the perimeter by \(16\,\text{in.}\), so the new perimeter is \(128+2(16)=160\,\text{in.}\).

The area is \(832\,\text{in.}^2\), and the perimeter is \(160\,\text{in.}\).

- Can you split the figure into two rectangles? - Add all the outside edge lengths to find the perimeter. - Make sure you do not omit or double-count an edge.

1. Split the figure into two rectangles. The lower rectangle has area \(10\cdot3=30\,\text{cm}^2\), and the upper rectangle has area \(4\cdot5=20\,\text{cm}^2\). 2. The total area is \(30+20=50\,\text{cm}^2\). 3. The two horizontal segments beside the stem have a combined length of \(10-4=6\,\text{cm}\). 4. The perimeter is \(10+2\cdot3+2\cdot5+4+6=36\,\text{cm}\).

The area is \(50\,\text{cm}^2\), and the perimeter is \(36\,\text{cm}\).

- Find the area that will actually be planted with grass. - Do not include the flower bed area. - Convert the final amount from grams to kilograms.

1. The park has area \(40\cdot25=1000\,\text{m}^2\). 2. The flower bed has area \(10\cdot5=50\,\text{m}^2\). 3. The grass area is \(1000-50=950\,\text{m}^2\). 4. The seed needed is \(950\cdot25=23{,}750\,\text{g}\). 5. Convert to kilograms: \(23{,}750\,\text{g}=23.75\,\text{kg}\).

\(23.75\,\text{kg}\) of grass seed is needed.

- Subtract the door area from the wall area. - Convert the tile side length to feet. - Divide the tiled area by one tile's area.

1. The wall area is \(12\cdot8=96\,\text{ft}^2\). 2. The door area is \(3\cdot7=21\,\text{ft}^2\), so the tiled area is \(96-21=75\,\text{ft}^2\). 3. Since \(6\,\text{in.}=0.5\,\text{ft}\), one tile has area \(0.5\cdot0.5=0.25\,\text{ft}^2\). 4. The number of tiles is \(75\div0.25=300\).

A total of \(300\) tiles are needed.

- Split the L-shape into two rectangles. - Determine whether you need to subtract to find a missing side length. - Find each rectangle's area and add the results.

1. Split the countertop into two nonoverlapping rectangles. 2. The lower rectangle has area \(8\cdot2=16\,\text{ft}^2\). 3. The upper-left rectangle is \(10-2=8\,\text{ft}\) tall, so its area is \(2\cdot8=16\,\text{ft}^2\). 4. The total area is \(16+16=32\,\text{ft}^2\).

The countertop has an area of \(32\,\text{ft}^2\).

- Split the T into a vertical rectangle and a horizontal rectangle. - Find each rectangle's area. - Add the two areas.

1. The vertical stem measures \(2\,\text{cm}\) by \(7\,\text{cm}\), so its area is \(2\cdot7=14\,\text{cm}^2\). 2. The top bar measures \(8\,\text{cm}\) by \(2\,\text{cm}\), so its area is \(8\cdot2=16\,\text{cm}^2\). 3. The total area is \(14+16=30\,\text{cm}^2\).

The area is \(30\,\text{cm}^2\).

- Divide the T-shape into two rectangles and find their areas. - Multiply the total area by the cost per square meter.

1. Divide the T-shape into two nonoverlapping rectangles. The vertical stem has area \(1\cdot3=3\,\text{m}^2\), and the top bar has area \(3\cdot1=3\,\text{m}^2\). 2. The total area is \(3+3=6\,\text{m}^2\). 3. The material cost is \(6\cdot\$45.00=\$270.00\).

The specialty glass costs \(\$270.00\).

- Find the area of the large rectangle first. - Find the area of one square cutout. - Subtract the total area of both cutouts from the rectangle's area.

1. The area of the full plate is \(30\cdot20=600\,\text{cm}^2\). 2. Each square cutout has area \(5\cdot5=25\,\text{cm}^2\). 3. The two cutouts have a total area of \(2\cdot25=50\,\text{cm}^2\). 4. The remaining area is \(600-50=550\,\text{cm}^2\).

The remaining area is \(550\,\text{cm}^2\).

- What fraction of the parallelogram is each outside triangle? - Use the given triangle's area to find the parallelogram's total area.

1. Because \(AE\) is half of \(AB\), \(\triangle ADE\) has one fourth the area of the parallelogram. Therefore, the parallelogram's area is \(4\cdot9=36\,\text{cm}^2\). 2. Since \(F\) is the midpoint of \(BC\), \(\triangle FCD\) also has one fourth the area of the parallelogram, or \(9\,\text{cm}^2\). 3. Subtract the two corner triangles: \(36-9-9=18\,\text{cm}^2\).

The area of quadrilateral \(EBFD\) is \(18\,\text{cm}^2\).

- Count the four corner triangles. - Determine what fraction of the parallelogram each corner triangle occupies. - Subtract their combined fraction from the whole.

1. At each corner, the small triangle has a base equal to half a side of the parallelogram and a corresponding height equal to half the parallelogram's height. Its area is therefore \(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}\) of the parallelogram's area. 2. The four corner triangles occupy \(4\cdot\frac{1}{8}=\frac{1}{2}\) of the total area. 3. The inner quadrilateral occupies the other half, so its area is \(\frac{1}{2}\cdot80=40\,\text{cm}^2\).

The inner quadrilateral has area \(40\,\text{cm}^2\).

- First use the midpoint to find the area of \(\triangle AEC\). - Convert the ratio \(AD:DC=2:1\) into the fraction \(\frac{AD}{AC}\). - Compare triangles that share the same height.

1. Since \(E\) is the midpoint of \(AB\), triangle \(AEC\) has half the base of triangle \(ABC\) and the same height. Thus, its area is \(\frac{1}{2}\cdot45=22.5\,\text{cm}^2\). 2. Since \(AD:DC=2:1\), \(AD=\frac{2}{3}AC\). Triangles \(ADE\) and \(AEC\) share the same height to line \(AC\), so \(\triangle ADE\) has \(\frac{2}{3}\) the area of \(\triangle AEC\). 3. The shaded area is \(\frac{2}{3}\cdot22.5=15\,\text{cm}^2\).

Triangle \(ADE\) has area \(15\,\text{cm}^2\).

- Compare the base lengths of triangles \(ABD\) and \(ABC\). - Determine what fraction of triangle \(ABC\) is triangle \(ABD\). - A segment from a vertex to the midpoint of the opposite side divides a triangle into two equal-area triangles. - Work from the large triangle to the smaller triangles one step at a time.

1. Since \(BD=\frac{1}{3}BC\), triangles \(ABD\) and \(ABC\) share the same height to line \(BC\), so \(A_{ABD}=\frac{1}{3}\cdot60=20\,\text{cm}^2\). 2. Since \(M\) is the midpoint of \(AD\), triangles \(ABM\) and \(BDM\) have equal bases on line \(AD\) and the same height from \(B\). Therefore, \(A_{ABM}=\frac{1}{2}\cdot20=10\,\text{cm}^2\).

Triangle \(ABM\) has area \(10\,\text{cm}^2\).

- It may be easier to subtract the areas around the shaded triangle from the area of the square. - Identify the three right triangles surrounding \(\triangle AEF\). - Use the midpoint information to find the leg lengths of those right triangles. - Recall the area formula for a right triangle.

1. The area of square \(ABCD\) is \(12\cdot12=144\,\text{cm}^2\). 2. The three unshaded corner triangles have areas \(A_{ABE}=\frac{1}{2}\cdot12\cdot6=36\,\text{cm}^2\), \(A_{ADF}=\frac{1}{2}\cdot12\cdot6=36\,\text{cm}^2\), and \(A_{ECF}=\frac{1}{2}\cdot6\cdot6=18\,\text{cm}^2\). 3. Subtract these areas from the area of the square: \(A_{AEF}=144-36-36-18=54\,\text{cm}^2\).

Triangle \(AEF\) has area \(54\,\text{cm}^2\).

- Express each area in terms of \(r\). - For a ring, subtract the inner circle's area from the outer circle's area. - How does a length scale factor affect area?

1. The inner circle has area \(A_1=\pi r^2\). The outer radius of the second region is \(2r\), so the ring area is \(A_2=\pi(2r)^2-\pi r^2=3\pi r^2\). 2. Therefore, \(A_1:A_2=1:3\). 3. With \(r=10\,\text{cm}\), the second ring has area \(\pi(20^2-10^2)\,\text{cm}^2=300\pi\,\text{cm}^2\). 4. The third ring has area \(\pi(30^2-20^2)\,\text{cm}^2=500\pi\,\text{cm}^2\), which is greater than \(300\pi\,\text{cm}^2\). 5. To quadruple a circle's area, \(\pi r_{\text{new}}^2=4\pi r^2\), so \(r_{\text{new}}=2r\).

a) \(1:3\) b) The second ring has area \(300\pi\,\text{cm}^2\), and the third has area \(500\pi\,\text{cm}^2\), so the third is larger. c) Multiply the radius by \(2\).

- The largest circle has diameter equal to the square's side length. - Subtract the circle area from the square area. - For part c, write both areas using a variable side length and simplify their ratio.

1. The square area is \(576\,\text{in.}^2\). 2. The circle radius is \(12\,\text{in.}\), so its area is \(144\pi\,\text{in.}^2\). 3. The scrap area is \(576-144\pi\,\text{in.}^2\approx123.61\,\text{in.}^2\). 4. The scrap percentage is \(100\left(1-\frac{\pi}{4}\right)\%\approx21.46\%\). 5. For side length \(s\), both the square area and inscribed-circle area contain a factor of \(s^2\), which cancels in the scrap fraction.

a) \(576-144\pi\,\text{in.}^2\approx123.61\,\text{in.}^2\) b) \(100\left(1-\frac{\pi}{4}\right)\%\approx21.46\%\) c) Both areas scale with \(s^2\), so the side-length factor cancels in the scrap fraction.

- Find the areas of the rectangular wall and triangular gable. - Subtract the areas of all three windows. - Account for two coats before finding the amount of paint. - Paint pails can only be purchased in whole numbers, so round the number of pails up when necessary.

1. The rectangular part of the wall has area \(10\cdot5=50\,\text{m}^2\). The gable height is \(8-5=3\,\text{m}\), so the triangular gable has area \(\frac{1}{2}\cdot10\cdot3=15\,\text{m}^2\). The entire front has area \(50+15=65\,\text{m}^2\). 2. The two square windows have total area \(2\cdot(2\cdot2)=8\,\text{m}^2\), and the rectangular window has area \(2\cdot1=2\,\text{m}^2\). The total window area is \(10\,\text{m}^2\). 3. One coat covers \(65-10=55\,\text{m}^2\). Two coats require coverage for \(2\cdot55=110\,\text{m}^2\). 4. The paint needed is \(110\div5=22\) liters. 5. Two pails contain only \(20\) liters, so \(3\) pails are required. The minimum cost is \(3\cdot\$45.00=\$135.00\).

The paint pails cost a minimum of \(\$135.00\).

- Convert every measurement to centimeters first. - Split the T into a top bar and a stem. - Alternatively, use an enclosing rectangle and subtract the missing corners. - A third possible approach is to split the figure into vertical strips.

1. Convert the mixed units: \(40\,\text{mm}=4\,\text{cm}\) and \(30\,\text{mm}=3\,\text{cm}\). 2. Method 1: Split the T into a \(12\,\text{cm}\times3\,\text{cm}\) top bar and a \(4\,\text{cm}\times6\,\text{cm}\) stem. The area is \(12\cdot3+4\cdot6=36+24=60\,\text{cm}^2\). 3. Method 2: Start with a \(12\,\text{cm}\times9\,\text{cm}\) enclosing rectangle. Subtract two \(4\,\text{cm}\times6\,\text{cm}\) corner rectangles: \(12\cdot9-2(4\cdot6)=108-48=60\,\text{cm}^2\).

The area is \(60\,\text{cm}^2\). One method adds the top bar and stem; another subtracts two missing corners from an enclosing rectangle.

- Enclose each figure in a rectangle. - Find and subtract the outside right triangles. - Subtract the two final areas to compare them.

1. The blue triangle fits inside a \(4\,\text{cm}\times3\,\text{cm}\) rectangle. The three outside triangles have areas \(2\,\text{cm}^2\), \(3\,\text{cm}^2\), and \(1.5\,\text{cm}^2\). Its area is \(12-(2+3+1.5)=5.5\,\text{cm}^2\). 2. The green quadrilateral fits inside a \(4\,\text{cm}\times3\,\text{cm}\) rectangle. The four outside triangles have areas \(1\,\text{cm}^2\), \(1\,\text{cm}^2\), \(2\,\text{cm}^2\), and \(2\,\text{cm}^2\). Its area is \(12-(1+1+2+2)=6\,\text{cm}^2\). 3. The difference is \(6-5.5=0.5\,\text{cm}^2\).

a) The blue triangle has area \(5.5\,\text{cm}^2\). b) The green quadrilateral has area \(6\,\text{cm}^2\). c) The green quadrilateral is larger by \(0.5\,\text{cm}^2\).

- Which simple region could you use as a starting point for the area? - Which regions are inside the garden, and which are outside it? - For perimeter, trace only edges exposed to the outside.

1. The main \(8\,\text{m}\times5\,\text{m}\) rectangle has area \(40\,\text{m}^2\). 2. One \(2\,\text{m}\times1\,\text{m}\) rectangle is added and two matching rectangles are removed, so the area is \(40+2-2-2=38\,\text{m}^2\). 3. Trace every segment of the outer boundary once. The total perimeter is \(32\,\text{m}\).

The garden bed has area \(38\,\text{m}^2\) and perimeter \(32\,\text{m}\).

- First compare the areas of triangles \(XST\) and \(XSZ\). - Use the ratio \(XT:TZ=1:3\) to determine what fraction of \(XZ\) is \(XT\). - Build the full triangle in two stages. - When a triangle's base is tripled while its height stays the same, how does its area change?

1. Since \(XT:TZ=1:3\), \(XT=\frac{1}{4}XZ\). Triangles \(XST\) and \(XSZ\) share the same height to line \(XZ\), so \(A_{XSZ}=4\cdot4=16\,\text{cm}^2\). 2. Since \(XS:SY=1:2\), \(XS=\frac{1}{3}XY\). Triangles \(XSZ\) and \(XYZ\) share the same height to line \(XY\), so \(A_{XYZ}=3\cdot16=48\,\text{cm}^2\).

Triangle \(XYZ\) has area \(48\,\text{cm}^2\).