question_answer

The values of t, if the area of the pentagon ABODE be \[\frac{45}{2}\] sq. units where \[A=(1,3),\] \[B=(-2,5),\] \[C=(-3,-1),\] \[D=(0,-2),\]\[D=(0,-2)\] and \[E=(2,t)\] are _____.                   

A)  \[-1.17\]                     

B)  \[-1,106\]       

C)  \[1,17\]           

D)         \[1,18\]           

Correct Answer: A

Solution :

Area of pentagon ABCDE = Area of \[\Delta ABC+\] Area of \[\Delta ACD+\]Area of \[\Delta ADE\]             Now, area of \[\Delta ABC=\] \[=\frac{1}{2}\left| 1(5+1)-2(-1-3)-3(3-5) \right|=10\,sq.\]units Area of \[\Delta ACD=\frac{1}{2}\left| [1(-1+2)-3(-2-3)+0] \right|=8sq.\]units. Area of \[\Delta AED=\frac{1}{2}\left| [1(t+2)+2(-2-3)+0] \right|\]             \[=\frac{1}{2}|(t+2-10)=\frac{1}{2}|t-8|\] \[\Rightarrow \] Area of pentagon             \[ABCDE=\frac{45}{2}\Rightarrow \,(t-8)=\pm 9\] \[\Rightarrow \] \[t-8=9\]or \[t-8=-9\]\[\Rightarrow \]\[t=17\] or \[t=-1\]