# Area of a parallelogram.

## 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 ,  $Sin\theta = \cfrac{P_1}{AD}\;\Rightarrow AD=BC=\cfrac{P_1}{Sin\theta}$



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

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

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

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

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

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

Angle Bisector

Family of Straight lines.