Question

In the given figure, PQRS is a rhombus. X, Y and Z are the mid-points of PQ, PS and SR respectively. If PR = 6 cm and PQ = 5 cm, then the value of (XY + YZ) equals

• A. 14 cm
• B. 7 cm
• C. 11 cm
• D. 9 cm

Answer 1

The correct option is B 7 cm


The diagonals of a rhombus bisect each other at right angles. Let PR and SQ intersect at O.
$\therefore \angle POQ=90°$
In $\Delta POQ$, by Pythagoras theorem,
$(PQ{)}^2=(PO{)}^2+(OQ{)}^2$
$\Rightarrow (5{)}^2=(3{)}^2+(OQ{)}^2$
$\therefore PO=\frac{PR}{2}=\frac{6}{2}=3cm$ and $PQ=5cm$
PR = 6 cm and PQ = 5 cm
$\Rightarrow OQ=\sqrt{25-9}=\sqrt{16}$
$\Rightarrow OQ=4cm$
Thus, QS = 2OQ = 8 cm
In $\Delta PSQ,XY||SQ$ and Y, X are mid-points of PS and PQ respectively.
$\therefore XY=\frac{1}{2}QS$ (Mid-point theorem)
$\Rightarrow XY=\frac{8}{2}=4cm$ ......(i)
Similarly, in $\Delta PSR,YZ||PR$ and Y, Z are mid-points of PS and SR respectively.
$\therefore YZ=\frac{1}{2}PR$ (Mid-point theorem)
$\Rightarrow YZ=\frac{1}{2}\times 6=3cm$ ......(ii)
$\therefore XY+YZ=4+3=7cm$ [From (i) and (ii)]
Hence, the correct answer is option (b).