Question

In the given figure, PQRS is a parallelogram with PS = 7 cm, PT = 4 cm and PV = 5 cm. What is the length of RS in cm? (The diagram is representative.)

• A. $\dfrac{20}{7}$
• B. $\dfrac{28}{5}$
• C. $\dfrac{9}{2}$
• D. $\dfrac{35}{4}$

Hint:

In a parallelogram, if two perpendicular distances to two different sides are given, equate the two area formulas: \[ \text{(known side)}\times(\text{its altitude})=\text{(unknown side)}\times(\text{its altitude}). \] This avoids trigonometry or coordinates.

Answer 1

The Correct Option is B

Key facts about a parallelogram.
Opposite sides are parallel: $PS \parallel QR$ and $PQ \parallel RS$.
The area of a parallelogram is ``base $\times$ corresponding altitude’’ and is independent of which side you pick as the base.
The altitude to a base is the perpendicular distance between the two lines containing a pair of opposite sides that are parallel to that base.
What the given perpendiculars represent.
$PT \perp QR$ with $PT=4$. Since $PS \parallel QR$, the perpendicular distance from any point on $PS$ to the line $QR$ equals the altitude corresponding to the base $PS$. Thus, $PT$ is the altitude to base $PS$.
$PV \perp RS$ with $PV=5$. Since $PQ \parallel RS$, the perpendicular distance from any point on $PQ$ to the line $RS$ equals the altitude corresponding to the base $RS$. Thus, $PV$ is the altitude to base $RS$.
Compute the area two ways. \[ \text{Area} = (\text{base } PS)\times(\text{altitude to } PS) = 7\times 4 = 28 \ \text{cm}^2. \] Using $RS$ as the base: \[ \text{Area} = (\text{base } RS)\times(\text{altitude to } RS) = RS \times 5. \] Equate the two area expressions: \[ RS \times 5 = 28 ⇒ RS = \frac{28}{5}\ \text{cm}. \] (Alternative coordinate/vector check.)
Place $QR$ on the $x$–axis with $Q=(0,0)$ and $R=(7,0)$ (since $QR\equiv PS$ and $PS=7$). Let $P=(x_0,4)$ so that $PT=4$. Then $S=P+(R-Q)=(x_0+7,4)$; the side $RS$ has direction vector $S-R=(x_0,4)$ and length \[ |RS|=\sqrt{x_0^2+4^2}. \] The area of the parallelogram is also $|QR|\times$ (vertical separation between the parallels $QR$ and $PS$) $=7\times 4=28$.
Using base $RS$, the perpendicular distance to its opposite side is $PV=5$, hence area $=|RS|. 5$. Therefore $|RS|=28/5$, independent of $x_0$, confirming the earlier result. \[ \boxed{\displaystyle RS=\frac{28}{5}\ \text{cm}} \]