Question

ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20° and `"HP"/"ED" = 4/5`. Then construct ΔRHP and ΔNED

Answer 1

In ∆NED, ∠D = 30°

and ∠N = 20°   ......(i) [Given]

∴ ∠E = 130°    .....(ii) [Remaining angle of a triangle]

∆RHP ∼ ∆NED

∴ `"RH"/"NE" = "HP"/"ED" = "PR"/"DN"`  .....[Corresponding sides of similar triangles]

∴ `"RH"/7 = 4/5`   ......[Given]

∴ RH = `(4 xx 7)/5` = 5.6 cm

Also, ∠R = ∠N, ∠H = ∠E, ∠P = ∠D  ......(iii) [Corresponding angles of similar triangles]

∴ ∠R = 20°, ∠H = 130°, ∠P = 30°   ......[From (i),(ii) and (iii)]

Steps of Construction:

	∆NED	∆RHP
i.	Draw seg NE of 7 cm	Draw seg RH of 5.6 cm
ii.	Draw a ray NA and EB such that ∠ANE = 20° and ∠BEN = 130°.	Draw a ray RC and HD such that ∠CRH = 20° and ∠DHR = 130°.
iii.	Name the point of intersection of rays D.	Name the point of intersection of rays P.