Free math worksheets

- Focus on the angle markings rather than the apparent side lengths. - How many corresponding angle pairs are needed for the angle-angle criterion?

The marked angles show two pairs of corresponding congruent angles. Therefore, the triangles are similar by the angle-angle similarity criterion.

Yes. The triangles are similar by AA.

- One pair of congruent angles is already marked. - Look for an angle that belongs to both triangles.

1. The diagram gives \(\angle C\cong\angle E\). 2. The triangles also share the angle at \(A\), because \(B\) and \(D\) lie on the same ray from \(A\), and \(C\) and \(E\) lie on the other ray. 3. Therefore, \(\triangle ABC\sim\triangle ADE\) by AA similarity.

AA similarity.

- Use the stated parallel segments rather than estimating angles from the drawing. - Treat \(AB\) and \(AC\) as transversals of the parallel segments. - Identify one corresponding-angle pair on each side of the triangle.

Because \(DE\parallel BC\), corresponding angles give \(\angle ADE\cong\angle ABC\) and \(\angle AED\cong\angle ACB\). Therefore, triangle \(ADE\sim\triangle ABC\) by AA.

\(\angle ADE\cong\angle ABC\) and \(\angle AED\cong\angle ACB\), so \(\triangle ADE\sim\triangle ABC\) by AA.

- Use the angle markings to match corresponding vertices before using side lengths. - Find the scale factor from a pair of corresponding sides whose lengths are known. - Apply the same scale factor to the side corresponding to \(8\).

The angle markings match \(A\) with \(D\) and \(B\) with \(E\), so \(AB\) corresponds to \(DE\) and \(AC\) corresponds to \(DF\). The scale factor from triangle \(ABC\) to triangle \(DEF\) is \(\frac{9}{6}=\frac{3}{2}\). Therefore, \(x=8\cdot\frac{3}{2}=12\).

\(x=12\)

- Match vertices by their angle markings before comparing side lengths. - Keep the vertex order consistent in the similarity statement. - Use one known pair of corresponding sides for the scale factor.

1. The one-arc angles give \(A\leftrightarrow Q\), and the two-arc angles give \(B\leftrightarrow R\). Therefore, \(C\leftrightarrow P\). 2. The correct statement is \(\triangle ABC\sim\triangle QRP\). 3. Corresponding sides \(AB=4\) and \(QR=6\), so the scale factor is \(\frac{6}{4}=\frac{3}{2}\).

\(\triangle ABC\sim\triangle QRP\), with scale factor \(\frac{3}{2}\).

- Recall exactly how many corresponding angle pairs the AA criterion requires. - Use the fact that the three angles in a triangle sum to \(180^\circ\). - Try constructing two different triangles that share one angle but not a second angle.

The statement is not correct. AA requires two pairs of corresponding congruent angles, not just one pair. For example, one triangle could have angles \(60^\circ,50^\circ,70^\circ\), while another could have angles \(60^\circ,40^\circ,80^\circ\). Both contain a \(60^\circ\) angle, but the other angles do not match, so the triangles are not similar.

No. One equal angle is not enough for AA. For example, \((60^\circ,50^\circ,70^\circ)\) and \((60^\circ,40^\circ,80^\circ)\) are not similar triangles.

- Identify the right angle in each height-shadow triangle. - Parallel sun rays create another pair of equal angles. - Match height with height and shadow length with shadow length. - Set up one proportion using corresponding sides.

The person and flagpole are both vertical, so each height-shadow triangle has a right angle. Parallel sunlight creates the same acute angle with the ground in both triangles. Therefore, the triangles are similar by AA. Using corresponding height-to-shadow ratios, \(\frac{6}{4}=\frac{h}{18}\), so \(h=27\,\text{ft}\).

The flagpole is \(27\,\text{ft}\) tall.

- Look at the pair of angles formed at the intersection \(O\). - After establishing similarity, identify which side corresponds to \(AO\). - Use the same scale factor on the side corresponding to \(OB\).

1. The diagram gives \(\angle A\cong\angle C\). 2. Angles \(\angle AOB\) and \(\angle COD\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle AOB\sim\triangle COD\) by AA. 4. The scale factor from \(\triangle AOB\) to \(\triangle COD\) is \(\frac{CO}{AO}=\frac{4}{6}=\frac{2}{3}\). 5. Since \(OB=9\) corresponds to \(OD=x\), \(x=\frac{2}{3}\cdot9=6\).

a) The triangles are similar by AA. b) \(x=6\)

- One angle pair is already known to correspond. - Express each triangle's third angle using the \(180^\circ\) angle sum. - The second matching angle does not have to use the same named position in both triangles. - Check every possible correspondence of the two remaining angles.

Since \(\angle A=\angle D=50^\circ\), one angle pair already matches. Triangle \(ABC\) has \(\angle C=180^\circ-50^\circ-(2x+10)^\circ=(120-2x)^\circ\). Triangle \(DEF\) has \(\angle F=180^\circ-50^\circ-3x^\circ=(130-3x)^\circ\). One possibility is \(\angle B=\angle E\): \(2x+10=3x\), giving \(x=10\). The other possible correspondence is \(\angle B=\angle F\): \(2x+10=130-3x\), giving \(x=24\). Both values produce valid positive triangle angles, so both guarantee AA similarity, with different vertex correspondences.

\(x=10\) or \(x=24\)

- Use the parallel segments to establish AA similarity before setting up proportions. - Combine the two pieces on \(\overline{AB}\) to find the whole side. - Determine the scale factor from triangle \(ADE\) to triangle \(ABC\). - Apply that same factor to the other corresponding sides.

Because \(DE\parallel BC\), triangle \(ADE\sim\triangle ABC\) by AA. From the diagram, \(AD=6\), \(DB=3\), so \(AB=9\). The scale factor from the smaller triangle to the larger triangle is \(\frac{AB}{AD}=\frac{9}{6}=\frac{3}{2}\). Thus \(AC=8\cdot\frac{3}{2}=12\), so \(EC=12-8=4\). Also, \(BC=10\cdot\frac{3}{2}=15\).

\(EC=4\) and \(BC=15\)