Area and Volume — Free Geometry Review Games.

V = lwh = 5*3*4 = 60.

SA = 6s^2 = 6*9 = 54.

V = s^3 = 4^3 = 64.

Area = (1/2)*apothem*perimeter.

V = pi*r^2*h for a cylinder.

V = 3.14*9*10 = 282.6.

SA = 2*pi*r^2 + 2*pi*r*h = 2*3.14*4 + 2*3.14*2*7 = 25.12 + 87.92 = 113.04.

V = (1/3)*pi*r^2*h = (1/3)*3.14*9*12 = 113.04.

V = (4/3)*pi*r^3 = (4/3)*3.14*27 = 113.04.

V = (1/3)*base area*height = (1/3)*36*10 = 120.

V = B*h, h = 200/25 = 8.

Volume scales as r^3, so doubling r multiplies volume by 2^3 = 8.

V_cone = (1/3)*pi*r^2*h, which is 1/3 of the cylinder's volume.

Perimeter = 6*6 = 36. Area = (1/2)*5.2*36 = 93.6.

V = (2/3)*pi*r^3 = (2/3)*3.14*125 = 261.67.

The area of a triangle is \(A=\frac{1}{2}bh=\frac{1}{2}(10)(6)=30\), applying the standard triangle-area formula. The distractor \(60\) comes from forgetting the \(\frac{1}{2}\) factor and just multiplying base by height. Always remember a triangle's area is half of the base times the height, unlike a parallelogram.

The trapezoid area formula \(A=\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(5+9)(4)=28\) averages the two parallel sides before multiplying by height. The distractor \(56\) results from skipping the \(\frac{1}{2}\) factor entirely. Always average the two bases first, then multiply by the perpendicular height.

A parallelogram's area is \(A=bh=(7)(3)=21\), using base times perpendicular height directly. The distractor \(10.5\) mistakenly halves the product as if it were a triangle. Unlike triangles, parallelograms use the full base times height with no fractional factor.

Surface area sums the areas of all six rectangular faces, which come in three congruent pairs, giving \(SA=2(lw+lh+wh)\). The distractor $lwh$ is actually the volume formula, not surface area. Remember that surface area totals the areas of all faces, while volume measures the enclosed space.

Volume of any prism is base area times height, so \(V=Bh=(12)(5)=60\). The distractor \(17\) mistakenly adds the base area and height instead of multiplying them. This base-times-height rule applies to any prism regardless of the shape of its cross-section.

A cone's volume is one-third that of a cylinder with the same base and height, giving \(V=\frac{1}{3}\pi r^2 h\). The distractor \(\pi r^2 h\) is the cylinder volume formula, missing the one-third scaling factor. Remember that pyramids and cones both use a one-third factor compared to their prism or cylinder counterparts.

A pyramid's volume is one-third of a prism with the same base and height, so \(V=\frac{1}{3}Bh\). The distractor \(Bh\) is actually the prism volume formula, overstating the pyramid's volume threefold. This one-third relationship between pyramids and prisms mirrors the relationship between cones and cylinders.

The surface area of a sphere is \(SA=4\pi r^2\), a standard geometric formula derived from calculus but memorized for geometry courses. The distractor \(\frac{4}{3}\pi r^3\) is actually the volume formula for a sphere, not the surface area. Keep surface area (square units) and volume (cubic units) formulas distinct by checking the exponent on \(r\).

A square's area is \(A=s^2=9^2=81\), squaring the side length. The distractor \(36\) incorrectly multiplies the side by \(4\) as if computing perimeter instead of area. Remember that area formulas involve squaring a length, while perimeter formulas involve summing lengths.

Circle area is \(A=\pi r^2=3.14(25)=78.5\), squaring the radius before multiplying by \(\pi\). The distractor \(31.4\) instead computes the circumference formula \(2\pi r\), confusing area with perimeter. Always square the radius for area, since area is measured in square units.

A rectangular prism has six rectangular faces: top, bottom, and four sides, matching the six terms combined in its surface area formula. The distractor \(8\) actually counts the vertices of the prism, not its faces. Remembering Euler's relationship among faces, edges, and vertices can help verify counts on any polyhedron.

Volume of a rectangular prism is $V=lwh=(8)(3)(2)=48$, multiplying all three dimensions together. The distractor \(13\) mistakenly adds the dimensions instead of multiplying them. Always multiply all three linear dimensions to get a volume in cubic units.

The lateral surface unrolls into a rectangle with width equal to the circumference \(2\pi r\) and length \(h\), giving \(2\pi r h\). The distractor \(2\pi r^2\) actually represents the combined area of the two circular bases, not the lateral surface. Total surface area of a cylinder combines this lateral term with the two circular base areas.

The triangular base area is \(\frac{1}{2}(6)(4)=12\), and multiplying by the prism length gives \(V=Bh=12(10)=120\). The distractor \(240\) comes from forgetting to halve the triangle's base times height before multiplying by prism length. Always compute the cross-sectional area first, then multiply by the length of the prism.

Surface area is \(2(lw+lh+wh)=2(12+20+15)=2(47)=94\), summing all three pairs of face areas before doubling. The distractor \(47\) forgets to double the sum, only counting one of each pair of congruent faces. Always account for the fact that every face on a rectangular prism has a matching congruent face on the opposite side.

The area of a regular polygon is \(A=\frac{1}{2}(\text{perimeter})(\text{apothem})=\frac{1}{2}(24)(3.46)=41.52\), using perimeter \(6(4)=24\). The distractor \(20.76\) forgets to use the full perimeter, effectively halving the perimeter twice. Always compute the full perimeter first before applying the apothem formula.

Using \(A=\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(20)(6)=60\), the bases are averaged and multiplied by the height. The distractor \(120\) skips the \(\frac{1}{2}\) factor and just multiplies the sum of bases by height. Always average the two parallel bases before multiplying by the perpendicular height.

Total surface area of a cone is \(SA=\pi r^2+\pi r l=3.14(3)(3+5)=75.36\), combining the circular base with the lateral surface. The distractor \(47.1\) only computes the lateral surface \(\pi r l\) and omits the base area \(\pi r^2\). Remember that total surface area of a cone always includes both the base circle and the curved lateral surface.

The base area is \(B=6^2=36\), and pyramid volume is \(V=\frac{1}{3}Bh=\frac{1}{3}(36)(9)=108\). The distractor \(324\) forgets the one-third factor and computes the prism volume instead. Always apply the one-third scaling factor whenever finding a pyramid's volume from its base area and height.

The radius is half the diameter, so \(r=4\), and \(V=\frac{1}{3}\pi r^2 h=\frac{1}{3}(3.14)(16)(9)=150.72\). The distractor \(452.16\) mistakenly uses the diameter itself as the radius in the formula. Always convert a given diameter to its radius before applying any circle or cone volume formula.

Total surface area equals two triangular base areas plus the lateral area: \(2(12)+(5+5+6)(10)=24+160=184\). The distractor \(160\) only includes the lateral surface area and forgets the two triangular end faces. Always add both congruent bases to the unrolled lateral surface when finding total surface area of any prism.

The prism contributes \(V=4(4)(6)=96\) and the pyramid contributes \(V=\frac{1}{3}(16)(3)=16\), giving a total of \(112\). The distractor \(96\) only accounts for the prism and ignores the pyramid on top. For composite solids, always find and add each individual solid's volume separately using its own appropriate formula.

The area of a rhombus is \(A=\frac{1}{2}d_1d_2=\frac{1}{2}(6)(8)=24\), using half the product of the diagonals. The distractor \(48\) forgets to include the \(\frac{1}{2}\) factor in the formula. Remember that a rhombus's area uses its diagonals, not its side lengths, unlike most other quadrilaterals.

Since volume is $V=lwh$, doubling only one linear factor scales the entire product by exactly \(2\). The distractor "It quadruples" would only apply if two dimensions were doubled simultaneously. Whenever a single linear dimension changes by a scale factor \(k\), volume changes by that same factor \(k\), not \(k^2\) or \(k^3\).

Since \(V=s^3=64\), the side length is \(s=4\), and surface area is \(SA=6s^2=6(16)=96\). The distractor \(64\) mistakenly reuses the volume value as the surface area without recalculating. Always solve for the side length first using a cube root before computing any other cube property.

The radius is \(r=6\), and \(V=\frac{4}{3}\pi r^3=\frac{4}{3}(3.14)(216)=904.32\). The distractor \(452.16\) results from halving the correct volume, likely from mistakenly using the diameter as if it were the radius. Always halve a given diameter to obtain the radius before applying the sphere volume formula.

The perimeter is \(5(5)=25\), so \(A=\frac{1}{2}(25)(3.44)=43\) using the standard regular-polygon area formula. The distractor \(21.5\) forgets to multiply by the full perimeter, effectively using only one side length. Always calculate the full perimeter of a regular polygon before applying the apothem-based area formula.

Solving \(2\pi r=18.84\) gives \(r=3\), so \(V=\pi r^2h=3.14(9)(5)=141.3\). The distractor \(94.2\) mistakenly uses the circumference value directly as \(\pi r^2\) instead of first solving for the radius. Always isolate the radius from any given circumference before substituting into area or volume formulas.

Total surface area is the base area plus four triangular faces: \(36+4\left(\frac{1}{2}(6)(5)\right)=36+60=96\). The distractor \(60\) only includes the lateral triangular faces and omits the square base. Always add the base area to the sum of the lateral triangle areas when finding a pyramid's total surface area.

Capacity equals volume, so \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}(3.14)(9)(6)=56.52\). The distractor \(169.56\) forgets the one-third factor and instead computes the volume of a cylinder with the same dimensions. Remember that real-world capacity problems for cones still require the one-third scaling factor in the volume formula.

The radius is \(r=5\), so \(SA=4\pi r^2=4(3.14)(25)=314\). The distractor \(157\) omits the factor of \(4\) in the sphere surface area formula. Always remember that a sphere's surface area is exactly four times the area of a great circle with the same radius.

Volume of any prism is base area times height, so \(V=Bh=40.2(10)=402\). The distractor \(201\) results from mistakenly halving the correct product, as if applying a pyramid formula instead. Prisms use base area times height directly with no fractional scaling, unlike pyramids and cones.

Since \(V=\pi r^2h\), doubling the radius scales volume by \(2^2=4\) while halving the height scales it by \(\frac{1}{2}\), giving a net factor of \(4\times\frac{1}{2}=2\). The distractor "It stays the same" wrongly assumes the radius and height changes cancel evenly, ignoring that radius is squared in the formula. Always remember that radius changes have a squared effect on cylinder volume, while height changes have only a linear effect.

Solving \(150.72=\frac{1}{3}(3.14)(16)h\) gives \(16.75h\approx150.72\), so \(h=9\). The distractor \(12\) results from forgetting the one-third factor and dividing by the full cylinder coefficient instead. When solving backward for height, always isolate \(h\) carefully by first simplifying all the constant and radius terms together.

Solving \(904.32=\frac{4}{3}(3.14)r^3\) gives \(r^3=216\), so \(r=6\) after taking the cube root. The distractor \(12\) mistakenly treats the result as if it were the diameter rather than solving directly for the radius. When reversing a volume formula, always take the correct root (cube root for volume, square root for area) to isolate the linear dimension.

Since volume depends on the product of three linear dimensions, scaling each by \(3\) multiplies volume by \(3^3=27\). The distractor \(9\) mistakenly applies only a squared scale factor, as would be correct for area rather than volume. Always remember that uniform linear scaling by factor \(k\) changes volume by \(k^3\), not \(k\) or \(k^2\).

The slant height forms the hypotenuse of a right triangle with legs \(r\) and \(h\), so \(l=\sqrt{6^2+8^2}=\sqrt{100}=10\). The distractor \(14\) mistakenly adds the radius and height instead of applying the Pythagorean theorem. Always use the Pythagorean theorem to relate a cone's radius, height, and slant height.

The cylinder contributes \(\pi r^2h=3.14(9)(10)=282.6\) and the hemisphere contributes \(\frac{2}{3}\pi r^3=\frac{2}{3}(3.14)(27)=56.52\), giving a total of \(339.12\). The distractor \(282.6\) only accounts for the cylinder and ignores the hemisphere on top. For composite solids, always identify each individual piece and use the appropriate formula for each before summing.

Solving \(6s^2=150\) gives \(s^2=25\), so \(s=5\) after taking the square root. The distractor \(25\) mistakenly stops at \(s^2\) without taking the final square root to find the actual side length. When reversing a surface area formula, always finish by taking the appropriate root to isolate the linear dimension.

With base area \(B=100\), solving \(200=\frac{1}{3}(100)h\) gives \(h=\frac{600}{100}=6\). The distractor \(12\) forgets to account for the one-third factor when isolating \(h\), effectively treating the pyramid like a prism. Always keep the one-third factor explicit when solving pyramid volume equations for an unknown dimension.

For similar solids, the ratio of volumes equals the cube of the ratio of corresponding linear dimensions, so taking the cube root of \(27:8\) gives \(3:2\). The distractor \(9:4\) mistakenly takes the square root of the volume ratio, which would be appropriate for a surface area ratio instead. Remember that linear ratios relate to volume ratios by a cube, and to surface area ratios by a square.

The apothem is \(\frac{8\sqrt{3}}{2}\approx6.93\), and with perimeter \(48\), the area is \(\frac{1}{2}(48)(6.93)\approx166.3\). The distractor \(192.0\) mistakenly uses the side length itself in place of the apothem in the area formula. Always compute the apothem from the side length using its geometric relationship before applying the regular-polygon area formula.

Since the prism's volume is \(Bh\) and the pyramid's volume is \(\frac{1}{3}Bh\) with the same base and height, the ratio simplifies directly to \(1:3\) regardless of the actual base and height values. The distractor \(1:2\) incorrectly assumes the pyramid holds half the prism's volume, but the true relationship uses one-third, not one-half. This one-third relationship between a pyramid and a prism sharing the same base and height is a key geometric fact worth memorizing.

An inscribed sphere has a diameter equal to the cube's side, so its radius is \(5\), giving \(V=\frac{4}{3}\pi r^3=\frac{4}{3}(3.14)(125)\approx523.33\). The distractor \(261.67\) results from mistakenly halving the correct volume, as if the radius were only \(2.5\). Always recognize that an inscribed sphere's diameter equals the cube's side length, not its radius.

The full cone's volume is \(\frac{1}{3}(3.14)(25)(12)=314\), and since the liquid forms a similar cone scaled by \(\frac{1}{2}\) in every linear dimension, its volume scales by \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\), giving \(314\times\frac{1}{8}=39.25\). The distractor \(157\) mistakenly assumes the liquid volume is simply half the total volume, ignoring that partial-height cones scale cubically, not linearly. When a liquid fills a cone only partway up its height, always use the cube of the height ratio to find the corresponding volume fraction.