Question

If the diagonal of the parallelogram are congruent, then prove that it is a rectangle. XYZW is a rectangle. If XY+YZ=7 and XZ+YW=10, then find XY.

Answer 1

Hint: In this question it is given that if the diagonal of parallelogram are equal, i.e, $XZ = YW$, then we have to show that it is a rectangle. So, for this we need to draw the diagram first.

Complete step-by-step answer:
We have,
$\
  XY + YZ = 7........(1) \\
  XZ + YW = 10.......(2) \\
\ $
We know that diagonal of the rectangle are equal , then
$XZ = YW$
Now, from equation (2)
We get
$\
  XZ + YW = 10 \\
   \Rightarrow XZ + XZ = 10 \\
   \Rightarrow XZ = 5 \\
\ $
Similarly, $YW = 5$
Now, in $\
  \vartriangle XYZ \\
    \\
\ $Using Pythagoras theorem,
$\
  X{Z^2} = X{Y^2} + Y{Z^2} \\
   \Rightarrow {5^2} = {(XY + YZ)^2} - XY.YZ \\
   \Rightarrow 25 = 49 - 2XY.YZ \\
   \Rightarrow XY.YZ = 12 \\
    \\
\ $
NOW,${(XY - YZ)^2} = X{Y^2} + Y{Z^2} - 2XY.YZ$
$\
   \Rightarrow {\left( {XY - YZ} \right)^2} = 25 - 24 \\
   \Rightarrow {\left( {XY - YZ} \right)^2} = 1 \\
   \Rightarrow XY - YZ = 1..........(3) \\
\ $
Adding (1) and (2), we get
$\
  2XY = 8 \\
   \Rightarrow XY = 4 \\
\ $
Hence, $XY = 4$

Note: To show that the parallelogram is a rectangle, we need to show that every angle is $90^\circ $because if one angle is $90^\circ $ then it indirectly implies that every is $90^\circ $.
So,as we know that in a parallelogram the opposite angles are always equal.