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 \[t=-1\]You need to login to perform this action.
You will be redirected in
3 sec