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 = \[\frac{1}{2}\times 12\times (11+25)=216\,\,sq.\,\,m.\]You need to login to perform this action.
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