In trianglePQR ; X,Y and Z are respectively the mid points of sides PQ , QR and PR. If ar (triangle XPZ) = 12 cm² , find ar( triangle ZYR)

Answer :

We have our diagram , As :

Here X , Y and Z are respectively the mid points of sides PQ , QR and PR.
And
Area of $∆$ XPZ  =  12 cm²
As given X  and Z are respectively the mid points of sides PQ and PR. So , from conserve of mid point theorem , we get

XZ  | |  QR 
And
$$\begin{gathered} \frac{\mathrm{PX}}{\mathrm{XZ}} = \frac{\mathrm{PQ}}{\mathrm{QR}} \\ \Rightarrow \frac{\mathrm{PX}}{\mathrm{XZ}} = \frac{2 \mathrm{PX}}{\mathrm{QR}} ( \mathrm{As} \mathrm{given} \mathrm{X} \mathrm{is} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{PQ} ) \\ \Rightarrow \frac{1}{\mathrm{XZ}} = \frac{2}{\mathrm{QR}} \\ \Rightarrow \mathrm{XZ}\hspace{0.17em} = \frac{\mathrm{QR}}{2} \\ \mathrm{And} \\ \mathrm{QY} = \mathrm{YR} = \frac{\mathrm{QR}}{2} ( \mathrm{As} \mathrm{given} \mathrm{Y} \mathrm{is} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{QR} ) \\ \mathrm{So}, \\ \mathbf{XZ}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{QY}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{YR}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{(}\mathbf{\hspace{0.17em}}\mathbf{1}\mathbf{ }\mathbf{)} \end{gathered}$$

In $∆$ PXZ and $∆$ ZYR 

PZ  =  ZR                                      (  As given Z is mid point of PR )

$\angle$ PZX  =  $\angle$ ZRY             (  Corresponding angles as we know XZ  | | QR and take PR as transversal line )

XZ  =  YR                                   ( From equation 1 )

So,
$∆$ PXZ $\cong$$∆$ ZYR        ( By SAS rule )

So,

Area of $∆$  PXZ  =  Area of $∆$  ZYR       , As we know congruent triangle have same area .
So,
Area of $∆$  ZYR  =  12 cm²                                      ( Ans )