A) \[DE=12cm,\,\,\,\angle F=50{}^\circ \]
B) \[DE=12cm,\,\,\,\angle F=100{}^\circ \]
C) \[EF=12cm,\,\,\,\angle D=100{}^\circ \]
D) \[EF=12cm,\,\,\,\angle D=30{}^\circ \]
Correct Answer: B
Solution :
[b] Given, \[\Delta ABC-\Delta DEF,\] |
then \[\angle A=\angle D=30{}^\circ \] |
\[\angle C=\angle E=50{}^\circ \] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle B=\angle F=180{}^\circ -(30{}^\circ +50{}^\circ )=100{}^\circ \] |
Also, \[AB=5\,cm,\] \[AC=8\,cm\] and \[DF=7.5\,cm\] |
\[\therefore \,\,\,\,\,\,\,\,\frac{AB}{DF}=\frac{AC}{DE}\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\frac{5}{7.5}=\frac{8}{DE}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,DE=\frac{8\times 7.5}{5}=12\,cm\] |
Hence. \[DE=12cm,\,\,\,\angle F=100{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec