Question

If A(6,1), B(8,2) and C(9,4) are the three vertices of a parallelogram ABCD. If E is the

midpoint of DC, find the area of ΔADE​

Answer 1

$\mathfrak{\bf{\underline{\underline{Given \ :}}}}$

$\mathsf{{A(6,1)}}$

$\mathsf{{B(8,2)}}$

$\mathsf{{C(9,4)}}$

Letѕ cσnѕíder the fσurth vєrtєх σf pαrαllєlσgrαm αѕ D$\mathsf{ (х,y)}$

We all knσw thαt, díαgσnαlѕ σf pαrαllelσgrαm ínterѕect eαch σther.

$\mathfrak{\bf{\underline{\underline{To \ Find} \ :-}}}$$\mathsf{Area \: of \:Triangle \: ADE = \: ? }$

$\mathfrak{\bf{\underline{\underline{ Solution \ :}}}}$$\mathsf {To \: find \: the \: coordinates \: of \: D}$

$\mathbf{Midpoint \: of \:BD} = \mathbf{Midpoint \: of \: AC}$

$\underline{\huge\mathfrak{Formula}} : - \\ ㅤㅤㅤㅤㅤ\mathbf{Midpoint = \frac{a + b}{2} }$

$\implies \: \bold{ \mathtt{(\frac{8 + x}{2} , \frac{2 + y}{2}) = (\frac{6 + 9}{2}, \frac{1 + 4}{2} } ) }$

$\implies \mathtt{ (\frac{8 + x}{2} , \frac{2 + y}{2} )= (\frac{15}{2} , \frac{5}{2}) }</p><p>$

$\implies \mathtt{ \frac{8 + x}{ \cancel2} = \frac{15}{ \cancel2} \: \: \: \: \: \: \: \: \frac{2 + y}{ \cancel2} = \frac{5}{ \cancel2} }$

$\implies{ \: \: \: \boxed{ \mathtt{x = 7}} \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \mathtt{y = 3}}}$

Therefσre, the cσ-σrdínαteѕ σf verteх D is ( 7,3 )

$\mathsf{Mipoints \: of \: DC \: i.e. \: coordinates \: of \: E \: is \: (\frac{9 + 7}{2}, \frac{3 + 4}{2} )}$

$\mathsf{coordinates \: of \: E = (8, \frac{7}{2} )}</p><p>$

Now,

$\mathsf{Area \: of \: Triangle \: ADE = </p><p>}$

$\implies \mathsf{ \frac{1}{2} \times 6(3 - \frac{7}{2} ) + 7( \frac{7}{2} - 1) + 8(1 - 3) }$

$\implies \mathsf{ | - \frac{ 3}{4} | }$

$\implies \: \mathsf{ \frac{3}{4} }$

$\boxed{\mathsf{Answer \ :- {\blue \ {Area \ of \ Triangle= \frac{3}{4}}}}}$