Question

20)
a) similar; sss and sas similarity
b) not similar
c) similar; aa similarity
d) similar; sas similarity
find the missing length. the triangles in each pair are similar.

1. △abc ~ △alk

a) not similar
b) similar; sss similarity
c) similar; aa similarity
d) similar; sas similarity

1. △abc ~ △agf

23)

Explanation:

Problem 20

Step1: Analyze side ratios

For triangles \( \triangle QUV \) and \( \triangle QRS \), check side ratios. \( \frac{QU}{QR}=\frac{6}{6 + 18}=\frac{6}{24}=\frac{1}{4} \), \( \frac{QV}{QS}=\frac{7}{21}=\frac{1}{3} \). Wait, no, maybe I misread. Wait, \( QU = 6 \), \( UR = 18 - 6 = 12 \)? Wait, no, the segment from \( Q \) to \( R \) is 18? Wait, the diagram: \( Q \) to \( U \) is 6, \( U \) to \( R \) is 18 - 6? Wait, no, the length from \( Q \) to \( R \) is 18? Wait, the first triangle: \( Q \) to \( U \) is 6, \( U \) to \( R \) is 18 - 6 = 12? Wait, no, the base \( QS \) is 21, \( QV \) is 7, \( VS \) is 21 - 7 = 14. Then \( \frac{QU}{QR}=\frac{6}{6 + 12}=\frac{6}{18}=\frac{1}{3} \), \( \frac{QV}{QS}=\frac{7}{21}=\frac{1}{3} \), and \( \angle Q \) is common. So by SAS similarity (two sides proportional, included angle equal), the triangles are similar.

Step2: Match with options

Option D says "similar; SAS similarity", which matches.

Step1: Analyze angles

In \( \triangle ABC \) and \( \triangle ALK \), \( \angle A \) is common. \( \angle ABC = 90^\circ \) (since \( \angle C = 47^\circ \), so \( \angle A = 180 - 90 - 47 = 43^\circ \)? Wait, no, \( \angle ALK \) has \( 48^\circ \), wait, no. Wait, \( \triangle ABC \) and \( \triangle ALK \): \( \angle A \) is common. \( \angle ABC = 90^\circ \) (since \( \angle C = 47^\circ \), so \( \angle A = 180 - 90 - 47 = 43^\circ \)? Wait, no, maybe \( \angle B \) and \( \angle L \) are right angles? Wait, the diagram: \( B \) and \( L \) are right angles? Then \( \angle A \) is common, so by AA similarity (two angles equal: right angle and \( \angle A \)), the triangles are similar.

Step2: Match with options

Option C says "similar; AA similarity", which matches.

Step1: Set up proportion for similar triangles

Since \( \triangle ABC \sim \triangle AGF \), the ratios of corresponding sides are equal. So \( \frac{AC}{AF}=\frac{BC}{GF}=\frac{AB}{AG} \). Wait, \( AB = 50 + 15 = 65 \)? No, \( AG = 15 \), \( AB = 50 \)? Wait, no, \( A \) is the common vertex. So \( \frac{AG}{AB}=\frac{AF}{AC} \)? Wait, \( AG = 15 \), \( AB = 50 \), \( AF = 18 \), let \( AC = x \) (the missing length, which is \( C A \), so \( CA = x \), \( AF = 18 \), so \( \frac{AG}{AB}=\frac{AF}{CA} \)? Wait, no, corresponding sides: \( \triangle AGF \) and \( \triangle ABC \), so \( \frac{AG}{AB}=\frac{AF}{AC} \). So \( \frac{15}{50}=\frac{18}{x} \).

Step2: Solve for x

Cross - multiply: \( 15x = 50\times18 \), \( 15x = 900 \), \( x=\frac{900}{15}=60 \). Wait, no, wait, maybe I mixed up. Wait, \( AG = 15 \), \( AB = 50 \), \( AF = 18 \), and \( CA \) is the side we need. So \( \frac{AG}{AB}=\frac{AF}{CA} \) → \( \frac{15}{50}=\frac{18}{CA} \) → \( CA=\frac{50\times18}{15}=60 \). Wait, but the missing length is \( C A \), which is 60? Wait, no, the segment \( C A \) is the base? Wait, the diagram: \( C \) to \( A \) is the horizontal side? Wait, maybe \( \frac{AG}{AB}=\frac{AF}{AC} \), so \( AC=\frac{AB\times AF}{AG}=\frac{50\times18}{15}=60 \).

Answer:

D. similar; SAS similarity

Problem 21