A) \[~204\,c{{m}^{2}}\]
B) \[~1644\text{ }c{{m}^{2}}\]
C) \[~1645\,c{{m}^{2}}\]
D) \[~1600\,c{{m}^{2}}\]
Correct Answer: B
Solution :
We have \[AB||DC,\] Let \[AP=x\,cm,\]\[QB=y\,cm\]and \[DP=CQ=h\,cm\] In\[\Delta \Alpha PD,\]by Pythagoras theorem \[{{h}^{2}}={{26}^{2}}-{{x}^{2}}\] ?(i) In \[\Delta QBC,\]by Pythagoras theorem \[{{h}^{2}}={{25}^{2}}-{{y}^{2}}\] ?(ii) From (i) & (ii), we get \[{{26}^{2}}-{{x}^{2}}={{25}^{2}}-{{y}^{2}}\] \[676-{{x}^{2}}=625-{{y}^{2}}\] \[{{x}^{2}}-{{y}^{2}}=51\] ?(iii) Also, \[AB=x+y+60=77\Rightarrow x+y=17\] \[\Rightarrow \]\[x=17-y\] ?(iv) Putting the value of \[x\]in (iii), we get \[(17-{{y}^{2}})-{{y}^{2}}=51\] \[\Rightarrow \]\[289+{{y}^{2}}-34y-{{y}^{2}}=51\] \[\Rightarrow \]\[29-51=34y\Rightarrow 238=34y\] \[\therefore \]\[y=7\] Now, from (ii) \[{{h}^{2}}=625-49=576\] \[\therefore \] \[h=24\,cm\] Now, Area of trapezium ABCD = \[=\frac{1}{2}\][sum of parallel sides] \[\times \]height \[=\frac{1}{2}\times 137\times 24=1,644\,c{{m}^{2}}\]You need to login to perform this action.
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