Sadhana Bhadoriya

Ex – 6.3 Que No. 12 Q.12) Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of △PQR (see Fig. 6.41). Show that △ABC ~ △PQR. Solution:-

Ex – 6.3 Que No. 11 Q.11) In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that △ABD ~ △ECF. Solution:-

Ex – 6.3 Que No. 10 Q.10) CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of △ABC and △EFG respectively. If △ABC ~ △FEG, show that:    i. CD/GH = AC/FG ii. △DCB ~ △HGE iii. △DCA ~ △HGF  i. CD/GH = AC/FG Solution:- ii. △DCB ~ △HGE

Ex – 6.3 Que No. 9 Q.9) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: i. △ ABC ~ △ AMP Solution:- ii. CA/PA = BC/MP Solution:- We know that, if two triangles are similar, then the ratio of their corresponding sides are proportional.

Ex – 6.3 Que No. 8 Q.8) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ABE ~ △CFB. Solution:-

Ex – 6.3 Que No. 7 Q.7) In Fig. 6.38, altitudes AD and CE of D ABC intersect each other at the point P. Show that: (i). △AEP ~ △CDP Solution:- (ii). △ABD ~ △CBE Solution:- (iii). △AEP ~ △ADB Solution:- (iv). △PDC ~ △BEC Solution:-

Ex – 6.3 Que No. 6 Q.6) In Fig. 6.37, if △ABE ≅ △ACD, shows that △ADE ~ △ABC. Solution:-

Ex – 6.3 Que No. 5 Q.5) S and T are points on the PR and QR sides of the △PQR, such that ∠P = ∠RTS. Show that △RPQ ~ △RTS. Solution:-

Ex – 6.3 Que No. 4 Q.4) In Fig. 6.36, QR/ QS = QT/ PR = and ∠1 = ∠2. Show that △PQS ~ △TQR. Solution:-

Ex – 6.3 Que No. 3 Q.3) Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD. Solution:-