(i) Find a |
(ii) Find b |
(iii) Area of the parallelogram. |
A)
(i) (ii) (iii) 1 \[-1\] 10sq. units
B)
(i) (ii) (iii) 1 \[-2\] 6 sq. units
C)
(i) (ii) (iii) 1 1 6 sq. units
D)
(i) (ii) (iii) \[-1\] \[-1\] 6 sq. units
Correct Answer: C
Solution :
Let \[A(-2,1),\,B(a,0),C(4,b)\]and \[D(1,2)\]be the vertices of a parallelogram. Since, diagonals of parallelogram bisect each other. \[\therefore \] Let \[E(x,y)\] be the point where diagonals bisect each other. Hence, coordinates of E from AC \[x=\left( \frac{-2+4}{2} \right)=1\] ?..(i) and \[y=\frac{1+b}{2}\] ?..(2) and coordinates of E from BD \[x=\frac{a+1}{2}\] and \[y=\frac{2+0}{2}=1\] [from (3)] \[\Rightarrow \] \[\frac{a+1}{2}=1\] [from (1)] \[\Rightarrow \] \[a=2-1\] \[\Rightarrow \]\[a=1\] and \[\frac{1+b}{2}=1\,\,\,\Rightarrow b=2-1\,\,\,\Rightarrow \,\,b=1\] (i) Value of \[a=1\] (ii) Value of \[b=1\] (iii) Value of ABCD = Area of \[\Delta ABC+\] Area of \[\Delta ACD\] Now, area of \[\Delta ABC=3sq.\] units Area of \[\Delta ACD=3sq.\] units Area of parallelogram \[=3+3=6\,sq.\] units.You need to login to perform this action.
You will be redirected in
3 sec