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Pre-Calculus Homework Solutions 162

Pre-Calculus Homework Solutions 162 - NM 8 Since LMK MNK...

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Since LMK MNK and K K , LMK MNK by AA Similarity. M LK K ¬ M NK K 2( K 1 P 6 ) 9 KP ¬ 2( KP ) 9 KP 3( 2 K 5 P ) ¬ 3( K 9 P ) 25(9) ¬ 9( KP ) 2 25 ¬ ( KP ) 2 5 ¬ KP KM KP PM KP 2( KP ) 3( KP ) 3(5) or 15 PM 2( KP ) 2(5) or 10 Since L QRM and K QPM , LKM RPM by AA Similarity. L R K P ¬ K PM M R 25 P ¬ 1 1 5 0 25(10) ¬ 15( RP ) 2 1 5 5 0 ¬ RP 5 3 0 ¬ RP Since KNM PQM and K QPM , KNM PQM by AA Similarity. K PQ N ¬ K PM M P 9 Q ¬ 1 1 5 0 9(10) ¬ 15( PQ ) 90 ¬ 15( PQ ) 6 ¬ PQ RP RQ PQ 5 3 0 RQ 6 5 3 0 6 RQ 3 3 2 RQ m MQP m QPM m PMQ ¬ 180 90 m QPM m PMQ ¬ 180 m QPM m PMQ ¬ 90 Since RMP is a right angle, m RMQ m PMQ 90 m QPM m PMQ m RMQ m PMQ m QPM m RMQ , so QPM RMQ . Therefore, RQM MQP by AA Similarity. M RQ Q ¬ M QP Q ¬ M 6 Q ( MQ ) 2 ¬ 3 3 2 6 ( MQ ) 2 ¬ 64 MQ ¬ 8 R M M P ¬ Q Q M P R 1 M 0 ¬ 8 6 6( RM ) ¬ 10(8) RM ¬ 8 6 0 or 4 3 0 Since KNM PQM , N QM M ¬ K PQ N N 8 M ¬ 9 6 6( NM ) ¬ 8(9) NM ¬ 7 6 2 or 12 Since L QRM and LNM RQM , QRM NLM by AA Similarity. Q N R L ¬ R LM M ¬ 3 3 2 ( LM ) ¬ 16 4 3 0 LM ¬ 16 4 3 0 3 3 2 LM ¬ 20 30. X IJ J H YJ J and J J , so IJH XJY by SAS Similarity. m JXY 180 m WXJ 180 130 50 m JIH m JXY by corr. 50 m YIZ m JIH by vert. 50 m JYX m YIZ + m WZG 50 20 70 because exterior angle sum of remote interior angles m JHI m JYX 70 by corr. m J m JXY m JYX ¬ 180 m J 50 70 ¬ 180 m J ¬ 60 m JHG m JHI ¬ 180 Linear pair m JHG 70 ¬ 180 m JHG ¬ 110 31. RST is a right angle, so
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