question_answer

For a trapezium parallel sides (in metres) are 11 and 25 and non - parallel sides (in metres) are 15 and 13. The area (in sq. m.) of the trapezium is:

A) 84

B) 512

C) 432

D) 216

E) None of these

Correct Answer: D

Solution :

Explanation Option (d) is correct. Let ABCD be trapezium and Al, BM the perpendiculars on DC. Let AL = BM = x. DL = \[\sqrt{169-{{x}^{2}}}\] \[\therefore \]      \[\sqrt{225-{{x}^{2}}}+\sqrt{169-{{x}^{2}}}+11=25\]             \[\sqrt{225-{{x}^{2}}}+\sqrt{169-{{x}^{2}}}=14\]                 (i) Also \[(225-{{x}^{2}})-(169-{{x}^{2}})=54\]                           (ii) Dividing (ii) and (i) \[\sqrt{225-{{x}^{2}}}-\sqrt{169-{{x}^{2}}=4}\]                                 (iii) Adding (i) and (iii) \[\sqrt[2]{225-{{x}^{2}}}=18\]             \[225-{{x}^{2}}=81\]             \[{{x}^{2}}=144\] \[\Rightarrow \]   \[x=12\] \[\therefore \] Area of trapezium = \[\frac{1}{2}\times 12\times (11+25)=216\,\,sq.\,\,m.\]