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)

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

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 )

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