Question

In a parallelogram PQRS, A is the point on RS such that AR = AS. Diagonals of PQRS intersect each other at point ‘0’. If a point B lies on PR such that $PB=\dfrac{3PR}{4}$, then $\left( OB+AB \right)$ is equal to?

Answer 1

Hint: For solving this type of question we should know about what a parallelogram is and how the points intersect each other in it. So, the parallelogram is a four-sided plane rectilinear figure in which the opposite sides are parallel to each other. A parallelogram consists of 4 vertices and it also consists of 4 edges. The parallelogram is a quadrilateral type. It has no lines of symmetry.

Complete step by step answer:
According to the question, our parallelogram is PQRS. And here A is a point on RS. So, according to our question, the figure would be,

As it is given in the question that AR = AS, so here A will divide RS from the middle. Now B point is like as on PR as it divides PR in $\dfrac{3}{4}$, which means,
$PB=\dfrac{3}{4}PR$
So, now, the figure according to the question becomes as follows,

Therefore, $AR=AS=\dfrac{RS}{2}$
And, $OP=OB=\dfrac{PR}{2}$
So, we have,
$RB=OB=\dfrac{OR}{2}=\dfrac{1}{4}PR$
So, AB is equal to the height of $\Delta ARB$, so, we have,
$\begin{align}
  & AB=\sqrt{{{\left( \dfrac{RS}{2} \right)}^{2}}-{{\left( OB \right)}^{2}}} \\
 & \Rightarrow AB=\sqrt{{{\left( \dfrac{RS}{2} \right)}^{2}}-{{\left( \dfrac{PR}{4} \right)}^{2}}} \\
\end{align}$
Therefore,
$AB+OB=\dfrac{PR}{4}+\sqrt{{{\left( \dfrac{RS}{2} \right)}^{2}}-{{\left( \dfrac{PR}{4} \right)}^{2}}}$
So, the value of AB + OB is $\dfrac{PR}{4}+\sqrt{{{\left( \dfrac{RS}{2} \right)}^{2}}-{{\left( \dfrac{PR}{4} \right)}^{2}}}$.

Note: When we solve this question then we have to make triangles which help us to solve the question in an easy way and they give the correct answer possibilities much. These are also of help to calculate the complete area of a parallelogram and with the help of triangles we can easily calculate that.