question_answer

ABCD is an isosceles trapezium with AB = 10 cm and CD = 6 cm. If EF = 8 cm. What is the perimeter of ABCD?

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 \[=AB+2\,(AD)+DC\] \[=16+4\sqrt{17}\,cm\]