Question

# If $ABCD$ be parallelogram and $P$ be the midpoint of the intersection of the diagonals. if $O$ is any point, then $OA+OB+OC+OD$is?

A

$3OP$

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B

$4OP$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

$2OP$

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D

$\frac{1}{2}OP$

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B $4OP$Explanation for the correct options:Finding the value for :Given, $ABCD$ be parallelogram and $M$ be the midpoint of the intersection of the diagonals.Step2. Finding the value of :Now $\begin{array}{rcl}OA+OB+OC+OD& =& \left(OP+PA\right)+\left(OP+PB\right)+\left(OP+PC\right)+\left(OP+PD\right)\\ & =& 4OP+\left(PA+PB+PC+PD\right)\\ & =& 4OP+\left(PA-PA+PB-PB\right)\\ & =& 4OP\end{array}$For parallelogram $P$ bisects the two diagonal. Hence, correct option is $\mathbf{\left(}\mathbit{B}\mathbf{\right)}$

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Related Videos
Applications of Cross Product
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
Select...