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, \[DL\,=\sqrt{225={{x}^{2}}}\] And \[\operatorname{MC}\,=\sqrt{169-{{x}^{2}}}\] \[\therefore \sqrt{225-{{x}^{2}}}+\sqrt{169-{{x}^{2}}}+11=25\] \[\sqrt{225-{{x}^{2}}}+\sqrt{169-{{x}^{2}}}=14\,\]                           (i) \[\operatorname{Also}\,\left( 225 -{{x}^{2}} \right)-\left( 169-{{x}^{2}} \right)=56\]         (ii) Dividing (ii) by (i) \[\sqrt{225-{{x}^{2}}}-\sqrt{169-{{x}^{2}}}=4\]            (iii) Adding (i) and (iii) \[^{2}\sqrt{225-{{x}^{2}}}=18\] \[225-{{x}^{2}}=81\] \[\therefore \, Area of trapezium =\frac{1}{2}\times 12\times \left( 11+25 \right)=216\,sq. m.\]