A) \[16+8\sqrt{15}\,cm\]
B) \[16+4\sqrt{17}\,cm\]
C) \[16+2\sqrt{11}\,cm\]
D) None of these
Correct Answer: B
Solution :
[b] Since, ABCD is an isosceles trapezium we have, non-parallel sides are equal. i.e. AD = BC and AG = HB = 2 Now, in \[\Delta \,AGD\] \[A{{D}^{2}}=D{{G}^{2}}+A{{G}^{2}}\] \[\Rightarrow \] \[AD=\sqrt{{{8}^{2}}+{{2}^{2}}}\] \[=\sqrt{68}=2\sqrt{17}\] \[\therefore \] \[CB=2\sqrt{17}\] (\[\therefore \] EF=DG=CH) Thus, the perimeter of the trapezium \[=AB+2\,(AD)+DC\] \[=16+4\sqrt{17}\,cm\] |
You need to login to perform this action.
You will be redirected in
3 sec