unit 6 similar triangles-homework 3 proving triangles are similar
Gina Wilson(All Things Algebra) 2014
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Triangles are similar if they have two congruent angles or two sides of one are proportional to two sides of the other and the included angle are equal.
Correct responses:
9. ΔSUT ~ ΔWUV by AA similarity postulate
10. ΔKDH ~ ΔABD by SAS similarity
11. ΔKJL and ΔGJH are not similar
12. ΔSRT ~ ΔPRQ by SAS similarity
13. ΔAEB ~ ΔCED are similar by SAS similarity
14. ΔPQR and ΔPST are not similar
15. ΔMBR ~ ΔLPZ are similar by AA similarity postulate
16. ΔLNK ~ ΔJNM are similar by SAS similarity
9. The given properties from the diagram are;
∠STU = ∠WVU = 72°
∠STU ≅ ∠WVU by definition of congruency
∠SUT ≅ ∠WUV by vertical angles theorem
[tex]\displaystyle 10. \hspace{0.3 cm} \frac{\overline{AD}}{\overline{DH}} = \frac{21}{15} = \mathbf{ \frac{7}{5}}[/tex]
[tex]\displaystyle \frac{\overline{AB}}{\overline{KH}} = \frac{28}{20} = \frac{7}{5}[/tex]
Therefore, given that the two sides of triangle ΔKDH are proportional to two sides on ΔABD and the included angles are equal, therefore;
[tex]\displaystyle 11. \hspace{0.3 cm} \frac{\overline{LJ}}{\overline{JH}} = \frac{95}{72} \neq \frac{87}{68} = \mathbf{ \frac{\overline{KJ}}{\overline{GJ}}}[/tex]
ΔKJL and ΔGJH are not similar.
[tex]\displaystyle 12. \hspace{0.3 cm} \frac{\overline{PR}}{\overline{SR}} = \frac{49}{35} = \frac{7}{5} = \frac{42}{30} = \mathbf{\frac{\overline{QR}}{\overline{TR}}}[/tex]
∠PRQ ≅ ∠SRT
Therefore;
[tex]\displaystyle 13. \hspace{0.3 cm} \frac{\overline{BE}}{\overline{DE}} = \frac{42}{32} =\frac{21}{16} = \mathbf{\frac{\overline{AE}}{\overline{CE}}}[/tex]
∠CED ≅ ∠BEA
Therefore;
14. ∠PQR 101° ≠ 102° = ∠STP
Therefore;
ΔPQR and ΔPST are not similar.
15. Given that ΔMBR and ΔPLZ are isosceles triangles.
∠MBR and ∠LPZ are the base angles of the isosceles triangles ΔMBR and ΔPLZ respectively
∠MBR = ∠MRB by base angles of isosceles triangle ΔMBR
∠LPZ = ∠LZP
∠MBR ≅ ∠LPZ given
∠MBR = ∠LPZ definition of congruency
∠MRB = ∠LZP by transitive property
Therefore;
16. [tex]\overline{JN}[/tex] = √(30.6² - 27²) = 14.4
[tex]\displaystyle \frac{\overline{MN}}{\overline{KN}} = \frac{27}{15} = \frac{9}{5} = 1.8 = \frac{14.4}{8} = \mathbf{ \frac{\overline{JN}}{\overline{NL}}}[/tex]
∠LNK = ∠JNM = 90° by vertical angles theorem.
Learn more about similar triangles here:
https://brainly.com/question/2410410