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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.
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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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