Free math worksheets

- Split the figure into two nonoverlapping rectangles. - Use the labeled lengths to find any rectangle side that is not labeled directly. - Add the two rectangle areas.

1. The lower rectangle is \(8\,\text{ft}\times3\,\text{ft}\), so its area is \(24\,\text{ft}^2\). 2. The upper-left rectangle has width \(8-3=5\,\text{ft}\) and height \(3\,\text{ft}\), so its area is \(15\,\text{ft}^2\). 3. The total area is \(24+15=39\,\text{ft}^2\).

The area is \(39\,\text{ft}^2\).

- Use the grid as the measuring unit instead of introducing a formula first. - Count the complete shaded squares row by row so none are skipped or counted twice.

1. Count the shaded unit squares by horizontal rows: there are \(4\) in the bottom row, \(3\) in the middle row, and \(2\) in the top row. 2. The total area is \(4+3+2=9\,\text{cm}^2\).

The area is \(9\,\text{cm}^2\).

- Treat the opening as area that has been removed from a larger rectangle. - Find the outer area and the opening area separately. - Subtract rather than add.

1. The outer rectangle has area \(12\cdot9=108\,\text{in}^2\). 2. The opening has area \(7\cdot4=28\,\text{in}^2\). 3. Subtract the opening: \(108-28=80\,\text{in}^2\).

The frame contains \(80\,\text{in}^2\) of metal.

- Identify the area of the complete circular plate first. - The square is a hole, so its area must be removed. - Keep \(\pi\) in the final exact answer.

1. The circle has radius \(5\,\text{cm}\), so its area is \(25\pi\,\text{cm}^2\). 2. The square hole has side length \(6\,\text{cm}\), so its area is \(36\,\text{cm}^2\). 3. The remaining area is \((25\pi-36)\,\text{cm}^2\).

The remaining area is \((25\pi-36)\,\text{cm}^2\).

- Compare the remaining area with the area of the full outer rectangle. - The missing corner is a rectangle. - Use the two side lengths of the missing rectangle to relate its area to \(x\).

1. Find the area of the full rectangle: \(14 \cdot 10 = 140\,\text{cm}^2\). 2. Find the area of the cutout: \(140 - 116 = 24\,\text{cm}^2\). 3. The cutout is a rectangle with side lengths \(x\) and \(4\,\text{cm}\), so: \(4x = 24\). 4. Solve for \(x\): \(x = 6\).

\(x = 6\,\text{cm}\)

- Identify which part of the region belongs to both rectangles. - When two areas are added, an overlap is counted once for each rectangle. - Adjust the sum so that every part of the plus shape is counted exactly once.

1. The vertical rectangle has area \(4\cdot10=40\,\text{ft}^2\), and the horizontal rectangle has area \(12\cdot3=36\,\text{ft}^2\). 2. Maya's sum counts the \(4\,\text{ft}\times3\,\text{ft}\) overlap twice. 3. Subtract one copy of the overlap: \(40+36-4\cdot3=64\,\text{ft}^2\).

Maya counted the overlap twice. The correct area is \(64\,\text{ft}^2\).

- Use the interior divider to view the patio as two nonoverlapping pieces. - For the triangular extension, use the side shared with the rectangle as its base and the labeled perpendicular segment as its height. - After finding the total area, a fraction of a box still requires buying a whole box.

1. The rectangular section has area \(12\cdot8=96\,\text{ft}^2\). 2. The triangular extension has base \(8\,\text{ft}\) and perpendicular height \(6\,\text{ft}\), so its area is \(\frac{1}{2}\cdot8\cdot6=24\,\text{ft}^2\). 3. The patio area is \(96+24=120\,\text{ft}^2\). 4. Since \(120\div18=6.\overline{6}\), seven whole boxes are needed. 5. The cost is \(7\cdot\$27.50=\$192.50\).

Seven boxes are needed, for a total cost of \(\$192.50\).

- Look for a simple bounding figure that contains the whole polygon. - Identify the small region inside the bounding figure but outside the garden. - Use the grid to read the horizontal and vertical lengths needed for that region's area.

1. Enclose the garden in the rectangle from \((0, 0)\) to \((8, 6)\). Its area is \(8\cdot6=48\,\text{m}^2\). 2. The part of that rectangle outside the garden is a right triangle in the upper-right corner with legs \(3\,\text{m}\) and \(3\,\text{m}\). 3. The triangle's area is \(\frac{1}{2}\cdot3\cdot3=4.5\,\text{m}^2\). 4. The garden's area is \(48-4.5=43.5\,\text{m}^2\).

The garden's area is \(43.5\,\text{m}^2\).

- Find the area of the whole trapezoid first. - Then find the area of the triangular opening. - Compare the opening's area to the trapezoid's area to get the percentage removed.

1. Find the area of the trapezoid: \(A=\frac{1}{2}(14+8)\cdot 6=\frac{1}{2}\cdot 22\cdot 6=66\,\text{cm}^2\). 2. Find the area of the triangular opening: \(A=\frac{1}{2}\cdot 4\cdot 3=6\,\text{cm}^2\). 3. Subtract to find the remaining metal area: \(66-6=60\,\text{cm}^2\). 4. Find the percentage removed: \(\frac{6}{66}\cdot 100\% \approx 9.1\%\).

Area remaining: \(60\,\text{cm}^2\) Percentage removed: \(9.1\%\)

- Express the side length of the inner square in terms of the uniform border width. - The frame area is the outer square's area minus the inner square's area. - After solving the resulting equation, reject any value that cannot represent a border inside an \(18\)-inch square.

1. The inner square has side length \(18-2x\), so the frame area is \(18^2-(18-2x)^2\). 2. Set this equal to the given frame area: \(324-(18-2x)^2=128\). 3. Then \((18-2x)^2=196\). Because the inner side length must be positive, \(18-2x=14\). 4. Solving gives \(x=2\,\text{in}\).

\(x=2\,\text{in}\).

- Divide the plotted region into strips between consecutive measurement lines. - Each strip can be modeled as a trapezoid whose parallel sides are the two neighboring measured widths. - Add the areas of all the strips to obtain the estimate for the whole region.

1. The four trapezoids all have width \(20\,\text{m}\), with endpoint heights \(0\) and \(30\), \(30\) and \(50\), \(50\) and \(40\), and \(40\) and \(0\) meters. 2. Their areas are \(20\cdot\frac{0+30}{2}=300\,\text{m}^2\), \(20\cdot\frac{30+50}{2}=800\,\text{m}^2\), \(20\cdot\frac{50+40}{2}=900\,\text{m}^2\), and \(20\cdot\frac{40+0}{2}=400\,\text{m}^2\). 3. The estimated total area is \(300+800+900+400=2400\,\text{m}^2\).

The trapezoidal estimate of the wetland's area is \(2400\,\text{m}^2\).

- For the subtraction method, start with the full bounding rectangle and remove each corner cutout once. - For the strip method, use the cutout dimensions to determine the width of each horizontal layer of the plate. - The two methods partition the same region differently, so their final totals should match.

1. By subtraction, the outer rectangle has area \(16\cdot12=192\,\text{cm}^2\). The cutouts have areas \(6\cdot5=30\,\text{cm}^2\) and \(4\cdot3=12\,\text{cm}^2\), so the plate area is \(192-30-12=150\,\text{cm}^2\). 2. By horizontal strips, the bottom strip is \(10\,\text{cm}\times5\,\text{cm}\), the middle strip is \(16\,\text{cm}\times4\,\text{cm}\), and the top strip is \(12\,\text{cm}\times3\,\text{cm}\). 3. Their total area is \(10\cdot5+16\cdot4+12\cdot3=50+64+36=150\,\text{cm}^2\). 4. Both decompositions give the same area, \(150\,\text{cm}^2\).

The plate has area \(150\,\text{cm}^2\). Both subtraction and horizontal-strip decomposition give this result.

- Use the center of the regular hexagon to decompose it into congruent triangles. - Recall the exact area relationship for an equilateral triangle, or derive it from a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle. - The circular region is a hole, so remove its area from the polygon's area.

1. A regular hexagon with side length \(6\,\text{cm}\) can be divided into six equilateral triangles of side length \(6\,\text{cm}\). 2. Each triangle has area \(\frac{\sqrt{3}}{4}\cdot6^2=9\sqrt{3}\,\text{cm}^2\), so the hexagon's area is \(54\sqrt{3}\,\text{cm}^2\). 3. The circular hole has radius \(2\,\text{cm}\), so its area is \(4\pi\,\text{cm}^2\). 4. The remaining metal area is \((54\sqrt{3}-4\pi)\,\text{cm}^2\).

The exact remaining area is \((54\sqrt{3}-4\pi)\,\text{cm}^2\).