Question
$\triangle RHP \sim \triangle NED$, In $\triangle NED , NE =7 cm , \angle D =30^{\circ}, \angle N =20^{\circ}$ and $\frac{ HP }{ ED }=\frac{4}{5}$. Then construct $\triangle RHP$ and $\triangle N E D$
✓
Answer
In $\triangle N E D, \angle D=30^{\circ}$
and $\angle N =20^{\circ}$ (i) [Given]
$\therefore \angle E=130^{\circ}$
....(ii) [Remaining angle of a triangle]
$\triangle RHP \sim \triangle NED$
$\therefore \frac{ RH }{ NE }=\frac{ HP }{ ED }=\frac{ PR }{ DN } \ldots$....[Corresponding sides of similar triangles]
$\therefore \frac{ RH }{7}=\frac{4}{5} \quad \ldots . . .[$ Given]
$\therefore RH =\frac{4 \times 7}{5}=5.6 cm$
Also, $\angle R =\angle N , \angle H =\angle E , \angle P =\angle D$ (iii) [Corresponding angles of similar triangles]
$\therefore \angle R =20^{\circ}, \angle H =130^{\circ}, \angle P =30^{\circ}$ [From (i),(ii) and (iii)]
Steps of Construction:
| /_\NED | Delta RHP |
| Draw seg NE of 7cm | Draw seg RH of 5.6 cm |
| 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^(@). |
| Name the point of intersection of r | Name the point of intersection of rays P. |
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