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ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20°, (HP)/(ED) = 4/5, then construct ΔRHP and ∆NED.

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प्रश्न

ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20°, `(HP)/(ED) = 4/5`, then construct ΔRHP and ∆NED.

ज्यामितीय चित्र
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उत्तर


Analysis:

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`

∴ RH = 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.
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