Free math worksheets

- Find the missing acute angle in each right triangle. - Compare the three angle measures in the two triangles. - Which triangle-similarity criterion uses two pairs of congruent angles?

1. Each triangle has a \(90^\circ\) angle. 2. The third angle in the first triangle is \(180^\circ-90^\circ-32^\circ=58^\circ\). 3. The third angle in the second triangle is \(180^\circ-90^\circ-58^\circ=32^\circ\). 4. Both triangles have angles \(90^\circ\), \(32^\circ\), and \(58^\circ\), so they are similar by AA.

The triangles are similar by AA because both have angle measures \(90^\circ\), \(32^\circ\), and \(58^\circ\).

- Find the missing interior angle in each triangle. - Compare the complete sets of angle measures. - Which similarity criterion applies when two pairs of corresponding angles are congruent?

1. The missing angle of the first triangle is \(180^\circ-40^\circ-80^\circ=60^\circ\). 2. The missing angle of the second triangle is \(180^\circ-80^\circ-60^\circ=40^\circ\). 3. Both triangles have angle measures \(40^\circ\), \(60^\circ\), and \(80^\circ\), so they are similar by AA.

Yes. Both triangles have angle measures \(40^\circ\), \(60^\circ\), and \(80^\circ\), so they are similar by AA.

- Use the \(180^\circ\) triangle angle sum. - What does matching angle measures tell you about similarity? - Does angle information alone determine a triangle’s size?

1. For Triangle 1, \(\gamma_1 = 180^\circ - (45^\circ + 75^\circ) = 60^\circ\). 2. For Triangle 2, \(\beta_2 = 180^\circ - (45^\circ + 60^\circ) = 75^\circ\). 3. Both triangles have angle measures \(45^\circ\), \(60^\circ\), and \(75^\circ\), so they have the same shape and are similar by AA. 4. No side length is given, so the triangles may have different scale factors. Therefore, they are not necessarily congruent.

a) \(\gamma_1 = 60^\circ\) and \(\beta_2 = 75^\circ\) b) The triangles have the same angle measures, so they have the same shape and are similar. c) No. They may be different sizes because no corresponding side lengths are specified.

- Imagine enlarging or reducing a triangle without changing its angles. - What is the difference between similar and congruent figures? - Can two equilateral triangles have different side lengths?

1. Paul’s claim is false. 2. Three matching angle measures determine the shape of a triangle but not its size. Such triangles are similar by AA. 3. For example, two equilateral triangles both have angles measuring \(60^\circ\), \(60^\circ\), and \(60^\circ\). One could have side length \(2\,\text{cm}\), while the other has side length \(5\,\text{cm}\). 4. The triangles have the same shape but different sizes, so they are not congruent. AAA is a similarity condition, not a congruence criterion.

Paul is incorrect. Three pairs of congruent angles guarantee similarity, not congruence. The triangles can have the same shape but different side lengths.

- What is true about the base angles of an isosceles triangle? - What is the sum of the interior angles of a triangle? - What angle condition is sufficient to prove two triangles similar?

1. In isosceles triangle \(ABC\), both base angles measure \(72^\circ\). 2. The vertex angle is \(180^\circ-72^\circ-72^\circ=36^\circ\). Thus, the angle measures are \(36^\circ,72^\circ,72^\circ\). 3. In triangle \(DEF\), the missing angle is \(180^\circ-36^\circ-72^\circ=72^\circ\). 4. Both triangles have the same three angle measures, so they are similar by AA.

Yes. Both triangles have interior angle measures \(36^\circ,72^\circ,72^\circ\), so they are similar by AA.

- What is the sum of the interior angles of a triangle? - Find all three angle measures in each triangle. - Which similarity criterion uses only angle information?

1. In triangle \(ABC\), \(\angle C=90^\circ\) and \(\angle A=40^\circ\), so \(\angle B=180^\circ-90^\circ-40^\circ=50^\circ\). 2. In triangle \(DEF\), one angle is \(90^\circ\) and one acute angle is \(50^\circ\), so the third angle is \(180^\circ-90^\circ-50^\circ=40^\circ\). 3. Both triangles have angle measures \(90^\circ,40^\circ,50^\circ\). Therefore, they are similar by AA.

The triangles are similar by AA because both have angle measures \(90^\circ,40^\circ,50^\circ\).

- What conditions can prove two triangles similar? - What is the third angle in a right triangle with a \(40^\circ\) angle? - For Sophia’s claim, compare the two legs and the angle between them. - Match the shorter leg to the shorter leg and the longer leg to the longer leg.

1. Lucas’s claim is true. A right triangle already has a \(90^\circ\) angle. If another angle is \(40^\circ\), the third angle is \(180^\circ-90^\circ-40^\circ=50^\circ\). Every such triangle has angle measures \(90^\circ,40^\circ,50^\circ\), so the triangles are similar by AA. 2. Sophia’s claim is true. In every such triangle, the included angle between the legs is \(90^\circ\), and the ratio of the two adjacent leg lengths is \(2\) to \(1\). Therefore, any two such triangles are similar by SAS.

Both claims are true. Lucas’s claim follows from AA because every triangle described has angles \(90^\circ,40^\circ,50^\circ\). Sophia’s claim follows from SAS because the included angle is \(90^\circ\) and the adjacent leg-length ratio is always \(2\) to \(1\).

- First find the third angle of the reference triangle. - Use the triangle angle sum for each comparison triangle. - In the isosceles triangle, the two base angles are congruent. - Compare the complete angle sets before deciding similarity.

1. The third angle of the reference triangle is \(180^\circ-48^\circ-72^\circ=60^\circ\), so its angles are \(48^\circ\), \(60^\circ\), and \(72^\circ\). 2. For a), the missing angle is \(180^\circ-72^\circ-60^\circ=48^\circ\). The triangle is similar to the reference triangle by AA. 3. For b), both base angles are \(48^\circ\), so the vertex angle is \(180^\circ-2\cdot48^\circ=84^\circ\). Its angles do not match the reference triangle, so it is not similar. 4. For c), the missing angle is \(180^\circ-48^\circ-60^\circ=72^\circ\). The triangle is similar to the reference triangle by AA.

a) Missing angle: \(48^\circ\). Similar. b) Missing angles: \(48^\circ\) and \(84^\circ\). Not similar. c) Missing angle: \(72^\circ\). Similar.

- Use only the two leg lengths and the angle between them. - Compare the rise-to-run ratios for the two ramps. - Check the angle included between each rise and run before naming the criterion.

1. The two legs of Ramp A have ratio \(\frac{3}{15}=\frac{1}{5}\). 2. The corresponding legs of Ramp B have ratio \(\frac{4}{20}=\frac{1}{5}\). 3. The included angle between each rise and run is \(90^\circ\). 4. Two pairs of corresponding sides are proportional and their included angles are congruent, so the ramp triangles are similar by SAS.

Yes. The ramp triangles are similar by SAS because \(\frac{3}{15}=\frac{4}{20}\) and the included angles are both \(90^\circ\).

- Find the third angle in each triangle. - Think about the side lengths after an equilateral triangle is cut in half. - What happens to a fraction when its numerator and denominator are multiplied by the same number?

1. Each triangle has angles of \(90^\circ\), \(30^\circ\), and \(60^\circ\), so the triangles are similar by AA. 2. Let the equilateral triangle have side length \(c\). Cutting it in half creates a right triangle whose hypotenuse is \(c\) and whose side opposite the \(30^\circ\) angle is \(\frac{c}{2}\). 3. Thus, \(\frac{\text{opposite leg}}{\text{hypotenuse}}=\frac{c/2}{c}=\frac{1}{2}\). 4. Doubling all side lengths multiplies both parts of the ratio by \(2\), so \(\frac{2a}{2c}=\frac{a}{c}\).

a) The triangles are similar by AA because both have angles of \(90^\circ\), \(30^\circ\), and \(60^\circ\). b) The ratio is \(\frac{1}{2}\). c) Scaling multiplies both the numerator and denominator by the same factor, so the ratio does not change.

- Which angle belongs to both triangles? - What do the two right-angle marks tell you? - Which similarity criterion uses two pairs of congruent angles?

1. Triangles \(ADC\) and \(ABC\) share \(\angle A\). 2. The right-angle marks show that \(\angle ADC=90^\circ\) and \(\angle ACB=90^\circ\). 3. Therefore, the triangles have two pairs of congruent corresponding angles, so \(\triangle ADC\sim\triangle ACB\) by AA.

\(\triangle ADC\sim\triangle ACB\) by AA because they share \(\angle A\) and each has a right angle.

- What is the sum of the interior angles of a triangle? - Which angle condition proves triangles similar? - What properties are preserved when a figure is enlarged by a dilation?

1. The missing angle in Triangle 1 is \(180^\circ-42^\circ-75^\circ=63^\circ\). 2. The missing angle in Triangle 2 is \(180^\circ-75^\circ-63^\circ=42^\circ\). 3. Both triangles have angle measures \(42^\circ,63^\circ,75^\circ\), so they are similar by AA. 4. Doubling every side length is a dilation with scale factor \(2\). A dilation preserves angle measures, so the new triangle has the same interior angles as the original.

a) Yes. Both triangles have angle measures \(42^\circ,63^\circ,75^\circ\), so they are similar by AA. b) All interior angle measures remain unchanged because a dilation preserves shape and angle measure.

- Match the shortest, middle, and longest sides in the two triangles. - What must be true about all three corresponding side-length ratios for SSS similarity? - Compare one ratio for each pair of corresponding sides.

1. Match the sides from shortest to longest: \(AB=6\,\text{cm}\) with \(DE=4\,\text{cm}\), \(BC=9\,\text{cm}\) with \(EF=6\,\text{cm}\), and \(AC=12\,\text{cm}\) with \(DF=8\,\text{cm}\). 2. The ratios are \(\frac{6}{4}=1.5\), \(\frac{9}{6}=1.5\), and \(\frac{12}{8}=1.5\). 3. All three corresponding side-length ratios are equal, so the triangles are similar by SSS.

Yes. \(\triangle ABC\sim\triangle DEF\) by SSS because every corresponding side-length ratio equals \(1.5\).

- What is the sum of the interior angles of a triangle? - Find the unlabeled angle in each triangle. - Which angle condition is sufficient to prove triangle similarity?

1. In triangle \(PQR\), \(\angle R=180^\circ-52^\circ-68^\circ=60^\circ\). 2. In triangle \(STU\), \(\angle S=180^\circ-68^\circ-60^\circ=52^\circ\). 3. Both triangles have angle measures \(52^\circ\), \(60^\circ\), and \(68^\circ\), so they are similar by AA.

\(\angle R=60^\circ\) and \(\angle S=52^\circ\). Yes, \(\triangle PQR\sim\triangle STU\) by AA.

- Which pair of angles is marked congruent? - What is true about opposite angles formed by intersecting lines? - Which triangle similarity criterion uses two angle pairs?

1. It is given that \(\angle ABE\cong\angle CDE\). 2. Angles \(\angle AEB\) and \(\angle CED\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle ABE\sim\triangle CDE\) by AA.

The triangles are similar by AA because \(\angle ABE\cong\angle CDE\) and \(\angle AEB\cong\angle CED\) as vertical angles.

- Which two angle pairs are marked congruent in the diagram? - How many congruent angle pairs are needed for AA similarity? - Match the vertices of the two triangles in corresponding order.

1. The right-angle marks show that \(\angle ACE\cong\angle EKF\). 2. The matching angle marks show that \(\angle CAE\cong\angle FEK\). 3. Therefore, \(\triangle ACE\sim\triangle EKF\) by AA.

\(\triangle ACE\sim\triangle EKF\) by AA because the diagram shows two pairs of congruent corresponding angles.

- Which smaller triangle shares the angle at \(B\) with the large triangle? - What congruent angle pair is marked in the diagram? - Use the order of corresponding vertices when naming the similar triangles.

1. Compare triangles \(BKP\) and \(BAC\). 2. Because \(K\) lies on \(BA\) and \(P\) lies on \(BC\), \(\angle KBP\cong\angle ABC\). 3. The matching angle marks show that \(\angle BKP\cong\angle BAC\). 4. Therefore, \(\triangle BKP\sim\triangle BAC\) by AA.

\(\triangle BKP\sim\triangle BAC\) by AA.

- Read the two side lengths adjacent to the marked angle in each panel. - Compare the ratios of corresponding side pairs. - Is the marked congruent angle included between those side pairs?

1. Compare the pairs of sides adjacent to the marked angles: \(\frac{12}{8}=1.5\) and \(\frac{15}{10}=1.5\). 2. The two pairs of corresponding sides are proportional. 3. The included angles are congruent because both are marked \(38^\circ\). 4. Therefore, \(\triangle ABC\sim\triangle DEF\) by SAS similarity.

Yes. \(\triangle ABC\sim\triangle DEF\) by SAS similarity because the adjacent side pairs are proportional and the included angles are congruent.

- What do the matching tick marks tell you about each triangle? - Use the marked vertex angle and the triangle angle sum to find a base angle. - Which similarity criterion applies once two angle measures match?

1. The tick marks show that each triangle is isosceles, and each vertex angle is marked \(44^\circ\). 2. In each triangle, each base angle measures \(\frac{180^\circ-44^\circ}{2}=68^\circ\). 3. Both triangles therefore have angle measures \(44^\circ\), \(68^\circ\), and \(68^\circ\), so they are similar by AA.

Yes. Both triangles have angle measures \(44^\circ\), \(68^\circ\), and \(68^\circ\), so they are similar by AA.

- Which angle is shared because \(D\) lies on \(AC\)? - Which second pair of congruent angles is given?

1. Because \(D\) lies on \(AC\), \(\angle BAD\cong\angle CAB\). 2. It is given that \(\angle ABD\cong\angle ACB\). 3. Therefore, \(\triangle ABD\sim\triangle ACB\) by AA.

\(\triangle ABD\sim\triangle ACB\) by AA.

- What do the matching tick marks tell you about each triangle? - Find the two acute angle measures in an isosceles right triangle. - Which similarity criterion uses angle measures?

1. Each triangle is a right triangle with two congruent legs, as shown by the tick marks. 2. In each isosceles right triangle, the two acute angles are congruent and have measure \(\frac{180^\circ-90^\circ}{2}=45^\circ\). 3. Both triangles have angle measures \(45^\circ\), \(45^\circ\), and \(90^\circ\), so they are similar by AA.

Yes. The triangles are similar by AA because both are \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangles.

- Match the shortest sides, the middle-length sides, and the longest sides. - Check whether one constant scale factor relates all three pairs.

1. Match the sides from shortest to shortest, middle to middle, and longest to longest. 2. The corresponding ratios are \(\frac{6}{4}=1.5\), \(\frac{9}{6}=1.5\), and \(\frac{11}{8}=1.375\). 3. The ratios are not all equal, so the triangles are not similar.

No. The triangles are not similar because the corresponding side-length ratios are \(1.5\), \(1.5\), and \(1.375\), which are not all equal.

- What angle is formed by a segment perpendicular to a line? - What does an angle bisector do to an angle? - Which similarity criterion uses two pairs of congruent angles?

1. Because \(BB_1\) and \(CC_1\) are perpendicular to the angle bisector, \(\angle AB_1B\) and \(\angle AC_1C\) are right angles. 2. The angle bisector divides \(\angle A\) into two congruent angles, so \(\angle BAB_1\cong\angle CAC_1\). 3. Therefore, \(\triangle ABB_1\sim\triangle ACC_1\) by AA.

\(\triangle ABB_1\sim\triangle ACC_1\) by AA because each triangle has a right angle and the acute angles at \(A\) are congruent.

- Compare all three pairs of side lengths rather than only one pair. - A single common multiplier for all three pairs is the condition needed for SSS similarity. - Use the side matches to determine which vertices correspond.

1. Compare corresponding-looking side ratios: \(\frac{DE}{AB}=\frac{16}{12}=\frac{4}{3}\), \(\frac{DF}{AC}=\frac{12}{9}=\frac{4}{3}\), and \(\frac{EF}{BC}=\frac{8}{6}=\frac{4}{3}\). 2. All three pairs of sides are proportional, so the triangles are similar by SSS. 3. The side matches are \(AB\leftrightarrow DE\), \(BC\leftrightarrow EF\), and \(AC\leftrightarrow DF\), so \(A\leftrightarrow D\), \(B\leftrightarrow E\), and \(C\leftrightarrow F\). 4. Therefore, \(\triangle ABC\sim\triangle DEF\), with scale factor \(\frac{4}{3}\) from the first triangle to the second.

Yes. The triangles are similar by SSS: \(\triangle ABC\sim\triangle DEF\). The scale factor from \(\triangle ABC\) to \(\triangle DEF\) is \(\frac{4}{3}\).

- Focus on the two sides that meet at each marked angle. - Compare the two pairs of those adjacent sides. - Check whether the congruent angle is the included angle between the proportional sides.

1. The marked included angles are congruent: \(\angle A=\angle D=50^\circ\). 2. The sides around those angles are proportional: \(\frac{DE}{AB}=\frac{10}{6}=\frac{5}{3}\) and \(\frac{DF}{AC}=\frac{15}{9}=\frac{5}{3}\). 3. Therefore, the triangles are similar by SAS, with correspondence \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\). 4. The scale factor from \(\triangle ABC\) to \(\triangle DEF\) is \(\frac{5}{3}\).

Yes. \(\triangle ABC\sim\triangle DEF\) by SAS, with scale factor \(\frac{5}{3}\).

- Compare the two sides that meet at the \(42^\circ\) angle in each triangle. - Check whether those two pairs of sides have the same scale factor. - For part b, use the vertex correspondence established in part a before choosing the corresponding side.

1. The sides around the included angle satisfy \(\frac{PQ}{XY}=\frac{6}{10}=\frac{3}{5}\) and \(\frac{PR}{XZ}=\frac{9}{15}=\frac{3}{5}\). 2. The included angles satisfy \(\angle QPR=\angle YXZ=42^\circ\). 3. Therefore, \(\triangle PQR\sim\triangle XYZ\) by SAS, with \(P\leftrightarrow X\), \(Q\leftrightarrow Y\), and \(R\leftrightarrow Z\). 4. The scale factor from \(\triangle PQR\) to \(\triangle XYZ\) is \(\frac{10}{6}=\frac{5}{3}\). 5. Thus, \(YZ=8.4\cdot\frac{5}{3}=14\,\text{cm}\).

a) \(\triangle PQR\sim\triangle XYZ\) by SAS. b) \(YZ=14\,\text{cm}\)

- Find the vertex angle of the first triangle. - Use the vertex angle of the second triangle to find its two base angles. - How is the area scale factor related to the length scale factor for similar figures?

1. In the first triangle, both base angles are \(70^\circ\), so the vertex angle is \(180^\circ-70^\circ-70^\circ=40^\circ\). 2. In the second triangle, the two base angles share the remaining \(180^\circ-40^\circ=140^\circ\), so each base angle is \(\frac{140^\circ}{2}=70^\circ\). 3. Both triangles have angle measures \(70^\circ,70^\circ,40^\circ\), so they are similar by AA. 4. The area scale factor is \(k^2=2.25\), so the length scale factor is \(k=\sqrt{2.25}=1.5\). 5. The base of the larger triangle is \(1.5\cdot5\,\text{cm}=7.5\,\text{cm}\).

a) Yes. Both triangles have angle measures \(70^\circ,70^\circ,40^\circ\). b) The base of the larger triangle is \(7.5\,\text{cm}\).

- Which missing side lengths are needed before all ratios can be compared? - Use the Pythagorean theorem in each right triangle. - How can side lengths show that one triangle is a scaled copy of another?

1. In triangle \(A\), the hypotenuse is \(c_1=\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{cm}\). 2. In triangle \(B\), the missing leg is \(b_2=\sqrt{20^2-12^2}=\sqrt{256}=16\,\text{cm}\). 3. Compare corresponding sides: \(\frac{a_2}{a_1}=\frac{12}{6}=2\), \(\frac{b_2}{b_1}=\frac{16}{8}=2\), and \(\frac{c_2}{c_1}=\frac{20}{10}=2\). 4. All three corresponding side-length ratios are equal, so the triangles are similar by SSS similarity.

Yes. Triangle \(A\) has side lengths \(6\,\text{cm}\), \(8\,\text{cm}\), and \(10\,\text{cm}\), and triangle \(B\) has side lengths \(12\,\text{cm}\), \(16\,\text{cm}\), and \(20\,\text{cm}\). Every side of triangle \(B\) is twice the corresponding side of triangle \(A\).

- Use \(A=\frac{1}{2}ab\) to choose two different pairs of positive leg lengths with the required area. - What must be true about corresponding side-length ratios for the triangles to be similar? - Can right triangles with equal areas have different shapes?

1. For a right triangle with leg lengths \(a\) and \(b\), the area is \(A=\frac{1}{2}ab\). 2. A right triangle with legs \(3\,\text{cm}\) and \(4\,\text{cm}\) has area \(\frac{1}{2}\cdot3\cdot4=6\,\text{cm}^2\). 3. A right triangle with legs \(2\,\text{cm}\) and \(6\,\text{cm}\) also has area \(\frac{1}{2}\cdot2\cdot6=6\,\text{cm}^2\). 4. The leg ratios are \(\frac{3}{4}\) and \(\frac{2}{6}=\frac{1}{3}\), so the triangles are not similar. Equal area does not guarantee similarity.

No. For example, right triangles with legs \(3\,\text{cm}\) and \(4\,\text{cm}\), and with legs \(2\,\text{cm}\) and \(6\,\text{cm}\), both have area \(6\,\text{cm}^2\) but are not similar because their leg-length ratios differ.

- Use the right-angle marks in the diagram. - Express the other acute angle of triangle \(ADC\) in terms of \(\angle A\). - How do the two angles at \(C\) fit together in the original right triangle? - Compare the complete angle sets of the two smaller triangles.

1. Both smaller triangles have a right angle at \(D\). 2. Let \(\angle A=\alpha\). In triangle \(ADC\), \(\angle ACD=90^\circ-\alpha\). 3. Since \(\angle ACB=90^\circ\), the remaining angle \(\angle DCB=\alpha\). Thus triangle \(CDB\) has angles \(90^\circ\), \(\alpha\), and \(90^\circ-\alpha\). 4. Therefore, \(\triangle ADC\sim\triangle CDB\) by AA. 5. When \(\alpha=35^\circ\), triangle \(ADC\) has angles \(35^\circ\), \(55^\circ\), and \(90^\circ\), while triangle \(CDB\) has angles \(55^\circ\), \(35^\circ\), and \(90^\circ\).

a) \(\triangle ADC\sim\triangle CDB\) by AA. b) \(\triangle ADC\): \(35^\circ,55^\circ,90^\circ\) \(\triangle CDB\): \(55^\circ,35^\circ,90^\circ\)

- What must be true about all corresponding side lengths for two triangles to be similar by SSS? - Find the three side-length ratios. - Does adding the same amount to each side preserve proportionality?

1. The side lengths of \(T_2\) are \(8\,\text{cm}\), \(10\,\text{cm}\), and \(12\,\text{cm}\). 2. Compare corresponding side-length ratios: \(\frac{8}{6}=\frac{4}{3}\), \(\frac{10}{8}=\frac{5}{4}\), and \(\frac{12}{10}=\frac{6}{5}\). 3. The three ratios are not equal, so the corresponding sides are not proportional. Therefore, the triangles are not similar by SSS.

No. The corresponding side-length ratios \(\frac{4}{3}\), \(\frac{5}{4}\), and \(\frac{6}{5}\) are not equal, so the triangles are not similar.

- What conditions define similar polygons? - Is there a triangle similarity criterion based only on angle measures? - Can two quadrilaterals have the same angle measures but different side-length proportions?

1. Similar polygons have congruent corresponding angles and proportional corresponding side lengths. 2. For triangles, the claim is true. If two corresponding angle pairs are congruent, then the triangles are similar by the AA similarity criterion. 3. For trapezoids, the claim is false. Consider two isosceles trapezoids that both have angle measures \(45^\circ,45^\circ,135^\circ,135^\circ\). One can have bases \(6\,\text{cm}\) and \(2\,\text{cm}\) with height \(2\,\text{cm}\), while another has bases \(8\,\text{cm}\) and \(4\,\text{cm}\) with height \(2\,\text{cm}\). Their corresponding base ratios are \(\frac{8}{6}\) and \(\frac{4}{2}\), which are not equal. Therefore, the trapezoids are not similar even though their corresponding angles are congruent.

a) The claim is true for triangles because two pairs of congruent corresponding angles establish similarity by AA. b) The claim is false for trapezoids. Two trapezoids can have congruent corresponding angles without having proportional corresponding side lengths.

- Substitute \(2a\) and \(2b\) into the tangent ratio. - Simplify the new ratio. - What happens to angles when a figure is scaled uniformly?

1. After doubling both legs, \(\tan(\alpha_{\text{new}})=\frac{2a}{2b}=\frac{a}{b}\). 2. The tangent ratio is unchanged, so the acute angle is unchanged: \(\alpha_{\text{new}}=\alpha\). 3. Geometrically, multiplying both legs by the same scale factor produces a triangle similar to the original. Corresponding angles in similar triangles are congruent.

Angle \(\alpha\) does not change. The ratio \(\frac{2a}{2b}\) equals \(\frac{a}{b}\), and the new triangle is a scaled copy of the original.

- What angle relationships hold in an isosceles triangle? - Could the given angle be either a vertex angle or a base angle? - What must be true about corresponding angles for triangles to be similar? - Find the other angles in both possible cases.

1. In an isosceles triangle, the \(50^\circ\) angle could be the vertex angle or a base angle. 2. If \(50^\circ\) is the vertex angle, each base angle is \(\frac{180^\circ-50^\circ}{2}=65^\circ\). The angle measures are \(50^\circ,65^\circ,65^\circ\). 3. If \(50^\circ\) is a base angle, the other base angle is also \(50^\circ\), and the vertex angle is \(180^\circ-2\cdot50^\circ=80^\circ\). The angle measures are \(50^\circ,50^\circ,80^\circ\). 4. Because two different angle sets are possible, the triangles are not guaranteed to be similar.

The \(50^\circ\) angle might be the vertex angle, giving angles \(50^\circ,65^\circ,65^\circ\), or a base angle, giving angles \(50^\circ,50^\circ,80^\circ\). These triangles are not similar, so the given information is insufficient.

- Use the shared angle at \(A\) and the angle relationships created by the parallel segments. - Use the two labeled pieces on \(AB\) to find the whole side. - Match corresponding sides of the similar triangles before setting up a proportion.

1. The triangles share \(\angle A\). Since \(DE\parallel BC\), a second pair of corresponding angles is congruent, so \(\triangle ADE\sim\triangle ABC\) by AA. 2. From the diagram, \(AB=AD+DB=4+2=6\,\text{cm}\). 3. Corresponding sides satisfy \(\frac{DE}{BC}=\frac{AD}{AB}\), so \(\frac{DE}{7.5}=\frac{4}{6}\). 4. Therefore, \(DE=7.5\cdot\frac{4}{6}=5\,\text{cm}\).

a) \(\triangle ADE\sim\triangle ABC\) by AA. b) \(DE=5\,\text{cm}\)

- Which angles are right angles because \(CD\) is an altitude? - Express the acute angles in terms of one angle in the original right triangle. - Which similarity criterion follows from two pairs of congruent angles?

1. Because \(CD\perp AB\), \(\angle ADC\) and \(\angle CDB\) are both right angles. 2. Let \(\angle CAD=\alpha\). Since \(\triangle ACD\) is a right triangle, \(\angle ACD=90^\circ-\alpha\). 3. In \(\triangle ABC\), \(\angle CBA=90^\circ-\alpha\). Because \(D\) lies on \(AB\), \(\angle CBD=\angle CBA\). 4. Thus, \(\angle ACD\cong\angle CBD\), and the triangles also have congruent right angles. Therefore, \(\triangle ACD\sim\triangle CBD\) by AA.

\(\triangle ACD\sim\triangle CBD\) by AA because both have a right angle and \(\angle ACD\cong\angle CBD\).

- How are an exterior angle and its adjacent interior angle related? - How are the base angles of an isosceles triangle related?

1. In the left triangle, the interior angle adjacent to the \(120^\circ\) exterior angle is \(180^\circ-120^\circ=60^\circ\). 2. Because the left triangle is isosceles, its other base angle is also \(60^\circ\), so its vertex angle is \(60^\circ\). 3. In the right isosceles triangle, the vertex angle is \(60^\circ\). Each base angle is \(\frac{180^\circ-60^\circ}{2}=60^\circ\). 4. Both triangles have three \(60^\circ\) angles, so they are similar by AA.

Yes. Both triangles are equilateral, so they are similar by AA.

- What do the matching angle marks at \(A\) tell you? - Replace \(\angle ABM\) with the corresponding angle that uses point \(C\) on the same line. - Which two inscribed angles intercept arc \(AC\)?

1. The matching angle marks at \(A\) show that \(\angle BAM\cong\angle KAC\). 2. Because \(M\) lies on \(BC\), \(\angle ABM=\angle ABC\). 3. Angles \(\angle ABC\) and \(\angle AKC\) are inscribed angles that intercept the same arc \(AC\), so they are congruent. 4. Therefore, \(\triangle ABM\sim\triangle AKC\) by AA.

\(\triangle ABM\sim\triangle AKC\) by AA because the angle bisector gives one congruent angle pair and the inscribed angle theorem gives \(\angle ABM\cong\angle AKC\).

- The parallel segments create two triangles with a shared vertex \(S\). - Compare their corresponding angles before comparing their side lengths. - Use the side ratio and the ray directions to interpret the similarity as a dilation.

1. Lines \(AC\) and \(BD\) meet at \(S\). 2. Since \(AB\parallel CD\), triangles \(SAB\) and \(SCD\) are similar by AA. 3. Therefore, \(\frac{SC}{SA}=\frac{SD}{SB}=\frac{CD}{AB}=\frac{9}{6}=\frac{3}{2}\). 4. Because each image point lies on the same ray from \(S\) as its preimage, the scale factor is positive. 5. A dilation centered at \(S\) with scale factor \(\frac{3}{2}\) sends \(A\) to \(C\) and \(B\) to \(D\). 6. Thus, \(S=AC\cap BD\) is the required center of dilation.

The dilation center is \(S=AC\cap BD\), and the scale factor is \(\frac{3}{2}\).

- Identify exactly which two sides meet at each stated angle. - SAS requires more than two proportional side pairs and any congruent angle. - Ask whether the stated angle is the angle between the two side pairs being compared.

1. The two given side pairs are proportional because \(\frac{8}{12}=\frac{6}{9}=\frac{2}{3}\). 2. However, \(\angle B\) is not the included angle between \(AB\) and \(AC\), and \(\angle E\) is not the included angle between \(DE\) and \(DF\). 3. SAS similarity requires the congruent angle to lie between the two proportional side pairs. 4. Here the information has an SSA-type structure. Geometrically, after fixing one side and the \(30^\circ\) ray, a circle representing the other given side can intersect that ray in two different positions, so the data need not determine one triangle shape. 5. Therefore, the classmate has not proved the triangles similar.

No. The \(30^\circ\) angles are not included between the two proportional side pairs, so SAS does not apply. The given information does not guarantee similarity.

- Use the triangle angle sum to find both missing angles. - Match corresponding vertices by equal angle measures. - Is the \(6.5\,\text{cm}\) side opposite the same angle in both triangles?

1. In \(\triangle ABC\), \(\beta = 180^\circ - 50^\circ - 70^\circ = 60^\circ\). 2. In \(\triangle DEF\), \(\epsilon = 180^\circ - 60^\circ - 70^\circ = 50^\circ\). 3. Both triangles have angle measures \(50^\circ\), \(60^\circ\), and \(70^\circ\), so they are similar by AA. 4. The side \(b = 6.5\,\text{cm}\) is opposite the \(60^\circ\) angle, while side \(e = 6.5\,\text{cm}\) is opposite the \(50^\circ\) angle. These are not corresponding sides. 5. Because distinct angles must have distinct opposite side lengths, the common value \(6.5\,\text{cm}\) cannot represent corresponding sides in congruent triangles. Therefore, the triangles are not congruent.

The triangles are not congruent. They are similar, but the equal-length sides lie opposite different angle measures and are not corresponding sides.

- Rewrite the product equation as an equation of ratios. - Which angle is shared by the two triangles? - Which similarity criterion combines two proportional side pairs with the included angle?

1. Rewrite the given equation by dividing both sides by \(CB\cdot BK\): \(\frac{AB}{CB}=\frac{BP}{BK}\). 2. The included angle at \(B\) is common to both triangles: \(\angle ABC\cong\angle PBK\). 3. Two pairs of sides that include the congruent angle are proportional, so \(\triangle ABC\sim\triangle PBK\) by SAS.

\(\triangle ABC\sim\triangle PBK\) by SAS because \(\frac{AB}{CB}=\frac{BP}{BK}\) and the triangles share the included angle at \(B\).

- Avoid choosing the parallel radii along the line of centers; that choice makes the endpoint line coincide with the center line. - Use the two parallel radii to identify a pair of similar triangles. - Show that the intersection is the unique external division point of \(\overline{O_1O_2}\) in the ratio \(R:r\).

1. The two radii satisfy \(O_1A\parallel O_2B\), point in the same direction, and have lengths \(R\) and \(r\). 2. Because \(R\ne r\) and the radius direction is not parallel to \(O_1O_2\), line \(AB\) is not parallel to line \(O_1O_2\), so the lines meet at \(S\). 3. Triangles \(SO_1A\) and \(SO_2B\) are similar because \(O_1A\parallel O_2B\) and the other corresponding sides lie on the same two intersecting lines. 4. Thus \(\frac{SO_1}{SO_2}=\frac{O_1A}{O_2B}=\frac{R}{r}\). 5. Because the parallel radii point in the same direction and \(R\ne r\), point \(S\) lies outside segment \(\overline{O_1O_2}\). It is the unique external point on the line of centers whose distances to \(O_1\) and \(O_2\) have ratio \(R:r\). 6. That external division point depends only on \(O_1\), \(O_2\), \(R\), and \(r\), not on the chosen valid radius direction. Therefore, every valid choice produces the same point \(S\).

Choose same-direction parallel radii that are not parallel to the line of centers, and intersect the line through their endpoints with line \(O_1O_2\). The intersection \(S\) satisfies \(SO_1:SO_2=R:r\) and lies outside \(\overline{O_1O_2}\). Since there is exactly one external division point with that ratio, every valid radius direction gives the same external center of similarity.

- Do not match vertices by the order in which the side lengths are listed. - First determine which three side pairs share one scale factor. - A median corresponds according to its vertex and the opposite side, so use the vertex correspondence you established.

1. Match the side lengths by a common factor: \(QR=3AB\), \(RP=3BC\), and \(PQ=3CA\). 2. Therefore, all three corresponding side pairs are proportional, so the triangles are similar by SSS. 3. Since \(AB\leftrightarrow QR\), \(BC\leftrightarrow RP\), and \(CA\leftrightarrow PQ\), the vertex correspondence is \(A\leftrightarrow Q\), \(B\leftrightarrow R\), \(C\leftrightarrow P\). Thus, \(\triangle ABC\sim\triangle QRP\). 4. The scale factor from \(\triangle ABC\) to \(\triangle QRP\) is \(3\). A median from \(A\) to \(BC\) therefore corresponds to the median from \(Q\) to \(RP\). 5. Its length is \(3\cdot5.5=16.5\).

a) \(\triangle ABC\sim\triangle QRP\) by SSS. b) The corresponding median is the median from \(Q\) to \(RP\), and its length is \(16.5\).