Similarity — Free Geometry Review Games.

Similar figures have the same shape but can differ in size.

AA similarity: two pairs of equal angles guarantee similar triangles.

Scale factor 2 means each dimension is multiplied by 2.

B's side = 12/3 = 4.

AA (Angle-Angle) similarity only needs two pairs of congruent angles.

All ratios are 1:2 (3/6=4/8=5/10=1/2), so similar with scale factor 1:2.

Area ratio is the square of the scale factor: 9:25.

Corresponding sides of similar figures are in proportion.

6/8 = x/32, x = 6*32/8 = 24 ft.

10/DE = 2/3, DE = 30/2 = 15.

Scale factor = 20/30 = 2/3. Area ratio = (2/3)^2 = 4/9.

Scale factor = 39/13 = 3. Shortest side = 5*3 = 15.

By the side-splitter theorem: AD/DB = AE/EC, 3/6 = 4/EC, EC = 8.

Volume ratio = (scale)^3. Cube root of 8/27 = 2/3.

Scale = ST/PQ = 20/8 = 5/2. SU = PR * 5/2 = 10 * 5/2 = 25.

The scale factor is the ratio of corresponding side lengths. If $\triangle ABC \sim \triangle DEF$ with \(AB = 4\) and \(DE = 6\), the scale factor is \(\frac{4}{6} = \frac{2}{3}\). The area ratio would be \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\), not the scale factor itself. Angles in similar figures are congruent, so they cannot form a meaningful ratio.

SAS Similarity requires two pairs of corresponding sides to be in the same ratio AND the angles between (included by) those sides to be congruent. Choice B describes SAS Congruence, which requires equal side lengths rather than proportional ones. Choice C describes AA Similarity. Choice D is not a recognized similarity postulate.

When all three ratios of corresponding sides are equal, the triangles are similar by SSS (Side-Side-Side) Similarity. AA Similarity requires only two pairs of congruent angles. SAS Similarity requires two proportional sides with a congruent included angle. AAS is a congruence theorem that requires equal side lengths, not proportional ones.

Corresponding angles in similar polygons are always congruent — equal in measure. This is one of the two defining properties of similar polygons: corresponding angles are equal and corresponding sides are proportional. Angles do not scale; only lengths are multiplied by the similarity ratio.

When two polygons are similar with scale factor \(\frac{2}{5}\), every pair of corresponding sides satisfies that ratio and all corresponding angles are congruent. So \(\frac{AB}{PQ} = \frac{CD}{RS} = \frac{2}{5}\) and \(\angle B = \angle Q\). Choice D is incorrect because the area ratio equals \(\left(\frac{2}{5}\right)^2 = \frac{4}{25}\), not \(\frac{2}{5}\). Choice B is wrong because corresponding angles in similar figures are congruent, not supplementary.

Perimeter is a linear measure — the sum of side lengths. Since each side scales by factor \(k\), the entire perimeter also scales by \(k\). By contrast, area scales by \(k^2\) and volume scales by \(k^3\). A common mistake is to confuse the area ratio \(k^2\) with the perimeter ratio \(k\).

By AA Similarity, two triangles with two pairs of congruent angles are similar. Since the three angle measures match, AA applies directly — no side information is needed. However, similar triangles are not necessarily congruent; they have the same shape but can differ in size. Congruence additionally requires corresponding sides to be equal.

The scale factor from $\triangle ABC$ to $\triangle PQR$ is \(\frac{PQ}{AB} = \frac{6}{9} = \frac{2}{3}\). Applying this to \(BC\): \(QR = 12 \times \frac{2}{3} = 8\). Choice D (\(18\)) results from inverting the scale factor and computing \(12 \times \frac{3}{2} = 18\), which would make $\triangle PQR$ larger — inconsistent with \(PQ < AB\).

Since \(\overline{DE} \parallel \overline{BC}\), by AA Similarity $\triangle ADE \sim \triangle ABC$. The scale factor is \(\frac{AD}{AB} = \frac{3}{9} = \frac{1}{3}\). Therefore \(DE = BC \times \frac{1}{3} = 12 \times \frac{1}{3} = 4\). Choice B (\(6\)) results from incorrectly using \(\frac{AD}{DB} = \frac{3}{6} = \frac{1}{2}\) as the scale factor instead of \(\frac{AD}{AB}\).

The ratio of perimeters equals the scale factor: \(\frac{24}{36} = \frac{2}{3}\). So corresponding sides satisfy \(\frac{10}{x} = \frac{2}{3}\), giving \(x = \frac{10 \times 3}{2} = 15\). Choice B (\(12\)) results from adding the difference of the perimeters (\(12\)) to the smaller side, which has no geometric justification.

All corresponding linear measures in similar triangles — including altitudes, medians, and angle bisectors — share the same scale factor as the sides. So if the scale factor is \(\frac{3}{5}\), the altitudes are also in the ratio \(\frac{3}{5}\). Choice B (\(\frac{9}{25}\)) is the ratio of areas, and Choice C (\(\frac{27}{125}\)) would apply to volumes of similar 3D solids. Choice D inverts the ratio.

By the Angle Bisector Theorem, \(\frac{QT}{TR} = \frac{PQ}{PR} = \frac{10}{15} = \frac{2}{3}\). Let \(QT = 2k\) and \(TR = 3k\). Since \(QT + TR = 20\), we get \(5k = 20\), so \(k = 4\) and \(QT = 8\). Choice B (\(10\)) assumes the bisector cuts \(\overline{QR}\) in half, which only occurs when \(PQ = PR\). Choice C (\(12\)) results from inverting the ratio: \(\frac{QT}{TR} = \frac{PR}{PQ} = \frac{3}{2}\).

For any two similar figures, the ratio of areas equals the square of the scale factor. Here the scale factor is \(\frac{5}{8}\), so the area ratio is \(\left(\frac{5}{8}\right)^2 = \frac{25}{64}\). Choice B (\(\frac{5}{8}\)) is the linear scale factor, not the area ratio. Choice C (\(\frac{125}{512}\)) is the cube of the ratio, which applies to volumes of similar 3D figures. Choice D simplifies to \(\frac{5}{8}\), the same error as B.

Since \(\frac{AD}{DB} = \frac{2}{3}\), let \(AD = 2k\) and \(DB = 3k\), so \(AB = 5k\). The scale factor from $\triangle ADE$ to $\triangle ABC$ is \(\frac{AD}{AB} = \frac{2k}{5k} = \frac{2}{5}\). Therefore \(DE = BC \times \frac{2}{5} = 15 \times \frac{2}{5} = 6\). Choice C (\(10\)) results from incorrectly using \(\frac{AD}{DB} = \frac{2}{3}\) as the scale factor, giving \(15 \times \frac{2}{3} = 10\).

Since \(D\) is on \(\overline{AB}\), we have \(AB = AD + DB = 4 + 8 = 12\). The scale factor from $\triangle ADE$ to $\triangle ABC$ is \(\frac{AD}{AB} = \frac{4}{12} = \frac{1}{3}\). Therefore \(\frac{DE}{BC} = \frac{1}{3}\), giving \(BC = 3 \times DE = 3 \times 5 = 15\). Choice B (\(10\)) results from incorrectly using \(\frac{AD}{DB} = \frac{4}{8} = \frac{1}{2}\) as the scale factor.

Cross-multiplying gives \(5(x + 2) = 8(x - 1)\). Expanding: \(5x + 10 = 8x - 8\). Solving: \(18 = 3x\), so \(x = 6\). Verify: \(\frac{6+2}{8} = \frac{8}{8} = 1\) and \(\frac{6-1}{5} = \frac{5}{5} = 1\). Choice B (\(3\)) comes from forgetting to distribute on the right side: \(5(x+2) = 8x - 1\) leads to an incorrect equation.

AA Similarity states that two triangles are similar if two pairs of corresponding angles are congruent. Here \(\angle A = \angle D\) and \(\angle B = \angle E\), so by AA, $\triangle ABC \sim \triangle DEF$. The third angle pair is automatically equal: \(\angle C = \angle F = 180^\circ - 47^\circ - 83^\circ = 50^\circ\). No side information is needed. AAS is a congruence postulate requiring at least one pair of congruent sides.

Since $\triangle TPQ \sim \triangle TRS$, corresponding sides are proportional: \(\frac{RS}{PQ} = \frac{TR}{TP} = \frac{10}{6} = \frac{5}{3}\). Therefore \(RS = 9 \times \frac{5}{3} = 15\). Choice B (\(\frac{27}{5}\)) results from inverting the ratio, using \(\frac{TP}{TR} = \frac{6}{10} = \frac{3}{5}\) instead, which would give the length of a segment in the smaller triangle.

By the Angle Bisector Theorem, \(\frac{BD}{DC} = \frac{AB}{AC} = \frac{12}{8} = \frac{3}{2}\). Let \(BD = 3k\) and \(DC = 2k\). Since \(BD + DC = 15\): \(5k = 15\), so \(k = 3\) and \(BD = 9\). Choice C (\(7.5\)) assumes the bisector divides \(\overline{BC}\) into equal halves — this only occurs when \(AB = AC\) (isosceles triangle). Choice B (\(6\)) is \(DC\), not \(BD\).

The scale factor (smaller to larger) equals the ratio of perimeters: \(\frac{18}{30} = \frac{3}{5}\). The area ratio equals the square of the scale factor: \(\left(\frac{3}{5}\right)^2 = \frac{9}{25}\). Smaller area \(= 225 \times \frac{9}{25} = \frac{2025}{25} = 81\). Choice B (\(135\)) results from incorrectly applying the linear ratio to the area: \(225 \times \frac{3}{5} = 135\).

When the altitude is drawn to the hypotenuse, each leg is the geometric mean of the hypotenuse and the adjacent segment: \(AC^2 = AH \times AB = 4 \times (4 + 9) = 4 \times 13 = 52\). So \(AC = \sqrt{52} = 2\sqrt{13}\). Choice B (\(6\)) is the altitude \(CH\), found from \(CH^2 = AH \times HB = 4 \times 9 = 36\). Choice C (\(3\sqrt{13}\)) is leg \(BC\), since \(BC^2 = HB \times AB = 9 \times 13 = 117\).

Since \(\overline{AB} \parallel \overline{CD}\), triangles $\triangle ABE$ and $\triangle CDE$ are similar by AA Similarity (alternate interior angles are congruent and vertical angles at \(E\) are congruent). The scale factor from $\triangle CDE$ to $\triangle ABE$ is \(\frac{AB}{CD} = \frac{12}{8} = \frac{3}{2}\). Therefore \(\frac{AE}{EC} = \frac{3}{2}\). Choice C (\(\frac{2}{3}\)) is the reciprocal — it gives \(\frac{EC}{AE}\), not \(\frac{AE}{EC}\).

The perimeter of $\triangle ABC = 6 + 8 + 10 = 24$, so the scale factor is \(\frac{48}{24} = 2\). Since \(6^2 + 8^2 = 100 = 10^2\), triangle $ABC$ is a right triangle with legs \(6\) and \(8\). Its area is \(\frac{1}{2} \times 6 \times 8 = 24\). The area ratio is \(k^2 = 4\), so the area of $\triangle DEF = 24 \times 4 = 96$. Choice B (\(48\)) incorrectly scales the area by \(k = 2\) instead of \(k^2 = 4\).

By the Triangle Midsegment Theorem, the segment joining the midpoints of two sides is half the length of the third side: \(DE = \frac{1}{2} BC\). Setting up the equation: \(2x - 3 = \frac{1}{2}(3x + 4)\). Multiplying both sides by \(2\): \(4x - 6 = 3x + 4\), so \(x = 10\). Verify: \(DE = 17\) and \(BC = 34\), so \(DE = \frac{1}{2} BC\). Choice B (\(7\)) results from incorrectly setting \(2(2x-3) = 3x+4\) and solving \(4x - 6 = 3x + 4\) but then making an arithmetic error.

Perimeter scales by the same factor as side lengths. The sides of $\triangle DEF$ are \(6 \times \frac{3}{2} = 9\), \(8 \times \frac{3}{2} = 12\), and \(10 \times \frac{3}{2} = 15\), giving perimeter \(9 + 12 + 15 = 36\). Equivalently, \(24 \times \frac{3}{2} = 36\). Choice B (\(54\)) incorrectly applies the area ratio: \(24 \times \left(\frac{3}{2}\right)^2 = 24 \times \frac{9}{4} = 54\). Area, not perimeter, scales by \(k^2\).

Similarity requires that corresponding angles match exactly and that corresponding sides share a constant ratio, which preserves shape while allowing size to change. The choice "the polygons have the same perimeter" is wrong because equal perimeters say nothing about angle measures or the shape of the figure. On the exam, always check both the angle-congruence and side-proportionality conditions before declaring polygons similar.

The AA Postulate states that if two angles of one triangle are congruent to two angles of another, the third angles must also be congruent (angle sum \(180^\circ\)), making the triangles similar. "Two pairs of corresponding sides are proportional" describes part of the SSS similarity criterion, not AA, so it is not the rule being tested here. Remember that with triangles you only need two angle matches, since the third angle is automatically determined.

Dividing any pair of corresponding sides, such as \(\frac{6}{4}\), gives the scale factor \(\frac{3}{2}\) from the smaller to the larger triangle. The choice \(\frac{2}{3}\) is wrong because that ratio describes the scale factor from $\triangle DEF$ to $\triangle ABC$, the reverse direction. Always be careful about which triangle is the starting figure when stating a scale factor.

Cross-multiplying gives \(6x = 36\), so \(x = 6\), which satisfies the proportion exactly. The choice \(4\) is incorrect because substituting it gives \(\frac{4}{4} = 1\), which does not equal \(\frac{9}{6} = 1.5\). When solving proportions, always cross-multiply and verify the result by substitution.

The Triangle Proportionality Theorem states that a line parallel to one side of a triangle cuts the other two sides into segments whose ratios are equal, meaning the division is proportional, not necessarily equal in length. "Into equal segments" is wrong because the segments are only equal if the parallel line happens to be a midsegment. This theorem is the foundation for solving many missing-side problems involving parallel lines inside triangles.

For similar polygons, the ratio of perimeters always equals the scale factor of the sides, since perimeter is a linear sum of proportional sides. The choice \(4:9\) is incorrect because that ratio applies to areas, which scale with the square of the scale factor, not to perimeters. Remember: perimeter ratios match the scale factor directly, while area ratios equal the scale factor squared.

The scale factor from $\triangle DEF$ to $\triangle GHI$ is \(\frac{10}{5} = 2\), so multiplying \(EF = 7\) by \(2\) gives \(HI = 14\). The choice \(9\) is wrong because that value is simply \(FD\), an unrelated side rather than the correctly scaled counterpart of \(EF\). When triangles are similar, always match corresponding sides carefully before applying the scale factor.

Triangle 1's missing angle is \(180^\circ - 50^\circ - 70^\circ = 60^\circ\) and Triangle 2's missing angle is \(180^\circ - 70^\circ - 60^\circ = 50^\circ\), so both triangles actually share the identical angle set \(\{50^\circ, 60^\circ, 70^\circ\}\), making them similar by AA. The choice "No, because the given angles are not identical pairs" overlooks that computing the third angle in each triangle reveals a full match. Always calculate the missing angle using the triangle angle sum before concluding two triangles are not similar.

Because the sun's rays create similar right triangles for the person and the flagpole, the ratio of height to shadow is constant: \(\frac{5.5}{4} = \frac{h}{24}\), giving \(h = 33\) feet. The choice \(30\) ft is wrong because it results from an arithmetic slip rather than correctly cross-multiplying the proportion. Indirect measurement problems like this rely on AA similarity between the observer's shadow triangle and the object's shadow triangle.

By the Triangle Proportionality Theorem, \(\frac{AD}{DB} = \frac{AE}{EC}\), so \(\frac{6}{9} = \frac{8}{EC}\) gives \(EC = 12\). The choice \(9\) is incorrect because it merely repeats the value of \(DB\) rather than solving the proportion for \(EC\). Whenever a segment is parallel to a triangle's side, set up the ratio using the two divided sides, not unrelated given values.

Since \(\overline{DE} \parallel \overline{BC}\), \(\frac{AD}{DB} = \frac{AE}{EC}\), and substituting \(EC = 15 - AE\) gives \(\frac{4}{6} = \frac{AE}{15-AE}\), which solves to \(AE = 6\). The choice \(9\) is wrong because it comes from incorrectly assuming \(AE\) equals \(EC\) rather than solving the cross-multiplied equation. This type of problem requires expressing one unknown segment in terms of the total side length before applying the proportion.

Because \(\overline{DE} \parallel \overline{BC}\), $\triangle ADE \sim \triangle ABC$ by AA (shared angle \(A\) plus congruent corresponding angles from the parallel lines), so \(\frac{AD}{AB} = \frac{DE}{BC}\) gives \(\frac{3}{12} = \frac{5}{BC}\), and \(BC = 20\). The choice \(15\) is wrong because it results from misapplying the ratio as \(\frac{AD}{DE}\) instead of correctly pairing \(\frac{AD}{AB}\) with \(\frac{DE}{BC}\). When a segment inside a triangle is parallel to the base, always identify the smaller similar triangle first, then match corresponding sides.

Since perimeter ratios equal the scale factor for similar triangles, \(\frac{24}{36} = \frac{2}{3}\), and solving \(\frac{8}{x} = \frac{2}{3}\) gives \(x = 12\). The choice \(16\) is wrong because it doubles the given side rather than applying the actual \(2:3\) scale factor derived from the perimeters. Always convert a perimeter ratio into its simplest scale factor before scaling individual sides.

The scale factor of a dilation is found by dividing the image length by the original length, so \(\frac{15}{5} = 3\). The choice \(\frac{1}{3}\) is incorrect because that value would represent the scale factor for a reduction from the larger triangle back to the smaller one, not the enlargement described. Always divide the new length by the original length, in that order, to find the correct dilation scale factor.

The combined scale factor from \(\triangle A\) to \(\triangle C\) is the product of the two individual scale factors, \(2 \times 3 = 6\), so the corresponding side is \(4 \times 6 = 24\). The choice \(12\) is wrong because it applies only the scale factor from \(\triangle A\) to \(\triangle B\) and forgets to chain in the second similarity relationship to \(\triangle C\). When similarity relationships are chained, multiply the scale factors together rather than adding or using just one of them.

By the Geometric Mean (Altitude) Relationship, the altitude to the hypotenuse equals the geometric mean of the two segments it creates, so the altitude \(= \sqrt{4 \times 9} = \sqrt{36} = 6\). The choice \(36\) is wrong because it is the product of the segments rather than its square root, which is the actual geometric mean. Whenever an altitude is drawn to the hypotenuse of a right triangle, it creates two smaller triangles similar to the original and to each other, giving rise to this geometric mean relationship.

By the Angle Bisector Theorem, \(\frac{BD}{DC} = \frac{AB}{AC} = \frac{10}{14} = \frac{5}{7}\), and since \(BD + DC = 24\), solving \(BD = 24 \times \frac{5}{12}\) gives \(BD = 10\). The choice \(14\) is incorrect because that value equals \(AC\), not the correctly proportioned segment \(BD\) derived from the theorem. The Angle Bisector Theorem always splits the opposite side in the same ratio as the two adjacent sides of the triangle.

Cross-multiplying gives \(x(x+8) = (x+2)(x+3)\), which expands to \(x^2 + 8x = x^2 + 5x + 6\), and the \(x^2\) terms cancel to leave \(3x = 6\), so \(x = 2\). The choice \(4\) is wrong because substituting it into the original proportion gives \(\frac{4}{6} \neq \frac{7}{12}\), so it does not satisfy the equation. Even when a proportion looks quadratic at first glance, always expand fully before assuming the equation cannot be solved with simple linear steps.

The law of reflection creates two similar right triangles sharing equal angles at the mirror, so \(\frac{\text{pole height}}{30} = \frac{5}{2}\), giving a pole height of \(75\) feet. The choice \(12\) ft is wrong because it results from dividing instead of setting up the correct cross-multiplied proportion between the two similar triangles. The mirror method is a classic application of AA similarity, since the angle of incidence equals the angle of reflection at the mirror's surface.

For similar figures, area ratios equal the square of the scale factor, so with a \(1:2\) side ratio the area ratio is \(1:4\), meaning the smaller triangle's area is \(96 \div 4 = 24\) square units. The choice \(48\) is wrong because it treats the area ratio the same as the side ratio, forgetting to square the scale factor. Always remember that while perimeters scale linearly with the scale factor, areas scale with its square.