Saved Bookmarks
| 1. |
Solution of Ch-Triangle Ex-6.3 Q14 is wrong in your ncert solution.Correct this solution |
| Answer» Given :∆ABC and ∆PQR, AD and PM are mediansBD= BC/2 and QR = QR/2So, AB/PQ = AC/PR=AD/PMTo prove:∆abc ~∆pqrConstruction:ad extended upto e and PM extended upto L.join be,CE,ql and elAd=de and PM=mlBD=DC and QM=mrTherefore,abce and pqrl are ||gramAb=CE and AC=be PQ= elPr=qlIn ∆abe and ∆pqlAb/PQ=BC/QR=ad/pmAb/PQ=BC/2×2/QR=2ad/2pmAb/PQ=bd\\qm=ae/pl∆abe~∆pql by SSS ruleangle bae=angle qpl (equation first)Similarly ∆ace~∆prlTherefore angle cae=RPL (equation second)Adding equation first and secondAngle bae + cae=angle qpl+RPLAngle bac = Apr (equation third)In ∆abc and ∆pqrAngle bac =qpr from equation thirdAB/PQ=BC/QR (given)Therefore ∆ABC ~∆PQR (SAS)Proved | |