Sadhana Bhadoriya

Ex – 6.2 Que No. 3 Q.3) In Fig. 6.18, if LM || CB and LN || CD, prove that  AM/AB = AN/AD. Solution:-

Ex – 6.3 Que No. 16 Q.16) If AD and PM are medians of triangles ABC and PQR, respectively where △ABC ~ △PQR, prove that Solution:-

Ex – 6.3 Que No. 15 Q.15) A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Solution:-

Ex – 6.3 Que No. 14 Q.14) Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that △ABC ~ △PQR. Solution:-

Ex – 6.3 Que No. 13 Q.13) D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD Solution:-

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:-