Area of a parallelogram.

Parallelogram area


Area of a parallelogram.

The area of a parallelogram ABCD can be found by calculating area of two triangles ABD and triangle BCD. 

In $\Delta$ADB ,  $Sin\theta = \cfrac{P_1}{AD}\;\Rightarrow AD=BC=\cfrac{P_1}{Sin\theta}$  

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In $\Delta$BCD , $Sin\theta =\cfrac{P_2}{DC}\;\Rightarrow DC=AB=\cfrac{P_2}{Sin\theta}$

Area of parallogram represent perpendicular from D on AB and BC

                   $\therefore$ Area  ABCD = $2.(\cfrac{1}{2}.AB.AD.Sin\theta)$


Let equation of AB :  $L_1=a_1x+b_1y+c_1=0$

                            BC :  $L_2=a_2x+b_2y+c_2=0$

                            DC :  $L_3=a_1x+b_1y+c_3=0$

              and       AD :  $L_4=a_2x+b_2y+c_4=0$

                                   $\Rightarrow P_1=\cfrac{\vert c_1 – c_3 \vert}{\sqrt{a_1^2+b_1^2}}\;and\;P_2=\cfrac{\vert c_2 – c_4 \vert}{\sqrt{a_2^2+b_2^2}}$

                             $\therefore$ Area =$\cfrac{P_1P_2}{Sin\theta}$

                                                                $=\cfrac{\vert c_1 – c_3\vert \vert c_2 – c_4 \vert}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}Sin\theta}$

Angle between AB and AD is   $tan\theta=\vert \cfrac{m_{AB}-m_{AD}}{1+m_{AB}m_{AD}} \vert$

                                                       $tan\theta=\vert \cfrac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2} \vert$

                                               $Cosec^2\theta = 1+Cot^2\theta = 1+\Bigl(\cfrac{a_1a_2+b_1b_2}{a_1b_2-a_2b_1}\Bigr)^2$

                                             $Cosec^2\theta = \cfrac{(a_1b_2-a_2c_1)^2+(a_1a_2+b_1b_2)^2}{(a_1b_2-a_2b_1)^2}$

                                      $\Rightarrow Sin\theta =\cfrac{\vert a_1b_2 – a_2b_1 \vert}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$

                                     $\therefore$ Area of parallelogram $= \cfrac{\vert c_1 – c_3 \vert \vert c_2 – c_4 \vert}{\vert a_1b_2 – a_2b_1\vert }$ 

Note:-  (1)   If $P_1=P_2\;\therefore$ parallelogram ABCD is a rohmbus

                     $\therefore$ Area of rohmbus $=\cfrac{1}{2}d_1d_2$

                    Where $d_1\,,\,d_2$ are length of diagonals AC and BD

              (2)   Area of rohmbus 

                     Area of rohmbus = $4\vert \cfrac{1}{2}.(OB).(OC) \vert$

                                                    =$\vert 2.\cfrac{-c}{m_2}.c \vert$

                                                    =$2.\cfrac{c^2}{\vert m_2 \vert}$ 

             Area of rohambus

Angle Bisector

Family of Straight lines.

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