Area and Perimeter — Free Pre-Algebra Review Games.

Perimeter = 2(l + w) = 2(8 + 5) = 26.

Area = length * width = 6 * 4 = 24.

Area of a square = side^2 = 9^2 = 81.

Perimeter = 4 * side = 4 * 7 = 28.

Area = (1/2) * base * height = (1/2) * 10 * 6 = 30.

Circumference = 2 * pi * r = 2 * 3.14 * 7 = 43.96.

Area = pi * r^2 = 3.14 * 25 = 78.5.

Area = l * w, so l = 48/6 = 8.

Area of parallelogram = base * height = 12 * 5 = 60.

Perimeter = 5 + 12 + 13 = 30.

Area = (1/2)(b1 + b2)(h) = (1/2)(6+10)(4) = (1/2)(16)(4) = 32.

Radius = 10. Area = pi * 10^2 = 3.14 * 100 = 314.

Area = 10*4 + 6*3 = 40 + 18 = 58.

Radius = 6. Area of semicircle = (1/2)*pi*r^2 = (1/2)*3.14*36 = 56.52.

P = 2(l+w), 30 = 2(9+w), 15 = 9+w, w = 6. Area = 9*6 = 54.

The perimeter formula \(P = 2(l + w)\) gives \(2(12+5) = 34\), since perimeter is the total distance around all four sides. The choice \(17\) is wrong because it only adds the length and width once instead of doubling the sum. Remember that perimeter always requires doubling the sum of adjacent sides for a rectangle.

The area of a square is \(A = s^2\), so \(11^2 = 121\) square units. The choice \(44\) is incorrect because it comes from multiplying the side by \(4\), which actually gives the perimeter, not the area. Always square the side length for area and multiply by \(4\) for perimeter of a square.

Perimeter of a square uses \(P = 4s\), so \(4(15) = 60\). The choice \(225\) is wrong because that is \(15^2\), the area formula, not the perimeter formula. Keep the area and perimeter formulas for squares distinct in your mind to avoid mixing them up.

Rectangle area is \(A = l \times w\), so \(9 \times 7 = 63\) square units. The choice \(32\) is incorrect because it results from adding \(9+7\) and doubling, which gives perimeter, not area. Always multiply length and width for area, never add them.

Triangle area uses \(A = \frac{1}{2}bh\), so \(\frac{1}{2}(8)(5) = 20\) square units. The choice \(40\) is wrong because it forgets to multiply by \(\frac{1}{2}\), giving the area of a rectangle instead. Always remember the \(\frac{1}{2}\) factor that distinguishes triangle area from rectangle area.

Circumference is the distance around a circle and is given by \(C = 2\pi r\), using the radius and the constant \(\pi\). The choice \(C = \pi r^2\) is wrong because that formula calculates area, not the distance around the circle. Keep circumference and area formulas separate since one uses \(r\) linearly and the other squares \(r\).

The area of a circle is found using \(A = \pi r^2\), since area depends on the square of the radius. The choice \(A = 2\pi r\) is incorrect because that expression calculates circumference, the distance around the circle. Memorize that area formulas involve squaring a dimension while circumference formulas do not.

Perimeter of any polygon is the sum of all its side lengths, so \(4+5+6=15\). The choice \(20\) is wrong because it does not match the sum of the given sides. For any triangle, perimeter simply requires adding the three side lengths together with no special formula needed.

The correct formula for triangle area is \(A = \frac{1}{2}bh\), since a triangle occupies half the area of a rectangle with the same base and height. The choice \(A = bh\) is wrong because that formula computes the area of a rectangle or parallelogram, not a triangle. Always include the factor of \(\frac{1}{2}\) when finding triangle area.

Using \(P = 2(l+w)\), we get \(2(14+3) = 34\). The choice \(42\) is incorrect because it comes from multiplying \(14 \times 3\), which gives the area, not the perimeter. Double-check whether a problem asks for the sum-based perimeter or the product-based area.

Squaring the side length gives \(12^2 = 144\) square units, which is the area. The choice \(48\) is wrong because it results from \(4 \times 12\), which is the perimeter formula, not area. For squares, remember area squares the side while perimeter multiplies it by four.

The radius is always half the diameter, so \(r = \frac{10}{2} = 5\). The choice \(10\) is incorrect because that value is the diameter itself, not half of it. Always divide the diameter by \(2\) before using radius-based formulas like area or circumference.

Area combines two length dimensions multiplied together, so the unit becomes squared, giving square feet. The choice "Feet" is wrong because that unit only measures a single linear dimension, not a two-dimensional region. Always square the linear unit when reporting area, and cube it when reporting volume.

Using \(A = \pi r^2\), we compute \(3.14 \times 81 = 254.34\) square units. The choice \(56.52\) is wrong because that value is the circumference, \(2\pi r\), not the area. Always square the radius before multiplying by \(\pi\) when finding area.

Since \(C = \pi d\), we get \(\frac{22}{7} \times 14 = 44\). The choice \(154\) is incorrect because that value results from squaring and multiplying by \(\pi\), which is how area is calculated, not circumference. When the diameter is given directly, use \(C = \pi d\) instead of converting to radius first.

Since \(A = l \times w\), dividing \(60 \div 10 = 6\) gives the width. The choice \(15\) is wrong because it comes from an incorrect division that does not match the given area and length. When one dimension and the area are known, divide the area by the known side to find the missing one.

Since \(P = 4s\), dividing \(36 \div 4 = 9\) gives the side length. The choice \(18\) is incorrect because it comes from dividing by \(2\) instead of \(4\), ignoring that a square has four equal sides. Always divide the perimeter by \(4\) to isolate the side length of a square.

Using \(A = \frac{1}{2}bh\), we solve \(45 = \frac{1}{2}(9)(h)\), giving \(h = 10\). The choice \(5\) is wrong because it forgets to double the area before dividing by the base, missing the \(\frac{1}{2}\) factor in the formula. When solving for height, multiply the area by \(2\) before dividing by the base.

From \(P = 2(l+w)\), we get \(40 = 2(l+8)\), so \(l+8 = 20\) and \(l = 12\). The choice \(20\) is incorrect because it stops at \(l+w\) without subtracting the given width. Always isolate the length by first dividing the perimeter by \(2\), then subtracting the known width.

Dividing the area by \(\pi\) gives \(r^2 = \frac{78.5}{3.14} = 25\), so \(r = \sqrt{25} = 5\). The choice \(25\) is wrong because it stops after finding \(r^2\) without taking the square root to reach the actual radius. Remember that solving for radius from area requires an extra square root step after dividing by \(\pi\).

Applying \(A = \frac{1}{2}bh\) gives \(\frac{1}{2}(14)(3) = 21\) square units. The choice \(42\) is wrong because it skips the \(\frac{1}{2}\) multiplier, treating the triangle like a rectangle. Always halve the product of base and height for any triangle area calculation.

The side length is \(\sqrt{64} = 8\), and the perimeter is \(4 \times 8 = 32\). The choice \(16\) is incorrect because it doubles the side rather than multiplying it by \(4\) as required for perimeter. When given area, first find the side using a square root, then apply the perimeter formula.

Setting \(2(x + (x+3)) = 26\) gives \(4x + 6 = 26\), so \(4x = 20\) and \(x = 5\). The choice \(8\) is wrong because it represents the width, \(x+3\), rather than the value of \(x\) itself. Always solve the perimeter equation fully for the requested variable, not for an intermediate expression.

Using \(C = 2\pi r\), we compute \(2 \times 3.14 \times 3.5 = 21.98\). The choice \(10.99\) is incorrect because it uses only \(\pi r\) instead of doubling for the full circumference formula. Always include the factor of \(2\) when applying the circumference formula with radius.

Using \(A = \frac{1}{2}bh\), we get \(\frac{1}{2}(20)(15) = 150\) square units. The choice \(300\) is wrong because it omits the \(\frac{1}{2}\) factor and instead gives the area of a rectangle with the same dimensions. Always remember that a triangle's area is exactly half that of a rectangle sharing the same base and height.

Fencing corresponds to perimeter, so \(P = 2(15+9) = 48\) units. The choice \(135\) is incorrect because that value is the area, \(15 \times 9\), which measures surface, not boundary length. Word problems about fencing or borders always call for the perimeter formula.

Since \(C = \pi d\), we compute \(3.14 \times 16 = 50.24\). The choice \(200.96\) is wrong because it results from mistakenly squaring the diameter, which is used for area-related calculations, not circumference. When the diameter is directly given, apply \(C = \pi d\) without needing to find the radius first.

For a right triangle, the two legs act as base and height, so \(A = \frac{1}{2}(6)(8) = 24\). The choice \(10\) is incorrect because that value is the hypotenuse length from the Pythagorean theorem, not the area. When a triangle is right-angled, its legs can be used directly as base and height in the area formula.

Since \(P = 4s\) is linear in \(s\), doubling \(s\) also doubles the perimeter. The choice "It quadruples" is wrong because that describes how area changes with doubling, since area depends on \(s^2\). Remember that perimeter scales linearly with side length while area scales with its square.

Since \(A = l \times w\) is directly proportional to length, tripling the length triples the area. The choice "It quadruples" is incorrect because that would only occur if both dimensions changed, not just one. When only one dimension of a rectangle changes, the area changes by that same factor.

Since area depends on \(r^2\), the ratio becomes \(\left(\frac{3}{6}\right)^2 = \frac{1}{4}\). The choice \(1:2\) is wrong because it treats the ratio as linear in radius rather than squared, ignoring how area scales. Always square the radius ratio when comparing circle areas, not just the radii themselves.

Perimeter is the sum of all sides, so \(7+7+10 = 24\). The choice \(17\) is incorrect because it only adds two of the three sides, ignoring the base of length \(10\). Always add every side of a triangle, regardless of whether it is isosceles, scalene, or equilateral.

The outer dimensions become \(24 \times 14 = 336\) square feet, and subtracting the pool's area of \(200\) leaves a walkway area of \(136\) square feet. The choice \(200\) is wrong because that is simply the pool's own area, not the surrounding walkway. For border or walkway problems, always subtract the inner shape's area from the larger outer shape's area.

The square's area is \(10^2 = 100\), and the circle's area is \(3.14 \times 25 = 78.5\), so subtracting gives \(100 - 78.5 = 21.5\). The choice \(78.5\) is wrong because that value is just the circle's area, not the leftover region outside it. When a circle is inscribed in a square, always subtract the circle's area from the square's area to find the remaining space.

Setting \(2(w + 2w) = 54\) gives \(6w = 54\), so \(w = 9\) and length \(= 18\), making the area \(9 \times 18 = 162\). The choice \(81\) is wrong because it only uses \(9^2\), mistakenly treating the shape as a square rather than a rectangle with different length and width. When given a ratio between length and width, solve for both variables before computing area.

Setting \(w(w+4) = 96\) gives \(w^2 + 4w - 96 = 0\), which factors as \((w-8)(w+12)=0\), so \(w = 8\) since width cannot be negative. The choice \(12\) is wrong because it corresponds to the negative root's magnitude, which is not a valid width. When area problems produce a quadratic equation, discard any negative solutions since lengths must be positive.

The rectangle contributes \(8 \times 6 = 48\) square units, and the semicircle with radius \(3\) contributes \(0.5 \times 3.14 \times 9 = 14.13\) square units, giving a total of \(62.13\). The choice \(48\) is wrong because it only accounts for the rectangular portion and ignores the semicircular addition. For composite shapes, always add the areas of each individual piece together.

Solving \(62.8 = 2(3.14)r\) gives \(r = 10\), and then \(A = 3.14 \times 10^2 = 314\) square meters. The choice \(62.8\) is wrong because it simply restates the circumference rather than converting it into an area. When circumference is given but area is needed, first solve for the radius before applying the area formula.

Since both shapes share an area of \(64\), dividing \(64 \div 4 = 16\) gives the rectangle's length. The choice \(8\) is wrong because that is the square's side length, not the length of the differently proportioned rectangle. When two shapes share equal areas but different dimensions, use the area equation to solve for the unknown side.

The full perimeter is \(2(12+9) = 42\) feet, and subtracting the \(3\)-foot doorway gap leaves \(39\) feet of baseboard needed. The choice \(42\) is wrong because it fails to account for the doorway opening where no baseboard is installed. In real-world perimeter problems, always subtract any gaps such as doorways from the total perimeter.

Solving \(154 = \frac{22}{7}r^2\) gives \(r^2 = 49\), so \(r = 7\), and then \(C = 2 \times \frac{22}{7} \times 7 = 44\). The choice \(154\) is wrong because it merely restates the original area rather than converting it through the radius into circumference. When switching between area and circumference, always solve for the radius as an intermediate step.

The garden's area is \(20^2 = 400\), and the pond's area is \(3.14 \times 25 = 78.5\), so subtracting gives \(400 - 78.5 = 321.5\) square units. The choice \(400\) is wrong because it represents the entire garden without removing the space occupied by the pond. For a region with a shape removed from it, always subtract the inner shape's area from the total outer area.

Setting \(2(w + 3w) = 60\) gives \(8w = 60\), so \(w = 7.5\) and length \(= 22.5\), making the area \(7.5 \times 22.5 = 168.75\). The choice \(225\) is wrong because it comes from squaring \(15\), a value not derived correctly from the given perimeter and ratio conditions. When solving ratio-based perimeter problems, carefully substitute the resulting dimensions into the area formula rather than guessing a round number.

Each side is \(36 \div 3 = 12\), so the area is \(\frac{1.73}{4}(144) \approx 62.28\) square units. The choice \(144\) is wrong because that value is simply the side squared without applying the equilateral triangle's specific area coefficient. For equilateral triangles, always find the side length first, then apply the special area formula involving \(\sqrt{3}\).

Since area is the product of length and width, scaling both by \(3\) multiplies the area by \(3 \times 3 = 9\). The choice \(3\) is wrong because that would only be correct if a single dimension were scaled, not both. Whenever every dimension of a shape is scaled by the same factor, area increases by that factor squared.