question_answer

Find the area of a trapezium ABCD in which AB || DC, AB = 77 cm, BC = 25 cm, CD = 60 cm and DA = 26 cm.

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 \[=\frac{1}{2}\times 137\times 24=1,644\,c{{m}^{2}}\]