## Parallelogram

$ABCD$ is a parallelogram with $A(0,0),B(20,10)$, and $D(10,y)$. If the area of the parallelogram is $600$, what is the value of $y$?

Source: NCTM Mathematics Teacher, November 2006

SOLUTION

Draw a perpendicular $\overline{DE}$ to $\overline{AB}$. If the x-coordinate of $E$ equals $2e$, its y-coordinate equals $e$ because $\overline{AB}$ has a slope of $1/2$.
Use distance formula to calculate $AB$
$AB=\sqrt{(20-0)^2+(10-0)^2}$
$=\sqrt{400+100}$
$=\sqrt{500}$
$=10\sqrt{5}$
Area of $ABCD=AB\times DE$
$600=10\sqrt{5}\times DE$
$DE=600/(10\sqrt{5})$
$=12\sqrt{5}$
Since $\overline{DE}$ is perpendicular to $\overline{AB}$, its slope equals $\textrm{-}2$
$\textrm{-}2=(e-y)/(2e-10)$
$\textrm{-}2(2e-10)=e-y$
$\textrm{-}4e+20=e-y$
$y=5e-20\qquad\qquad (1)$
Use distance formula to calculate $DE$
$DE=\sqrt{(2e-10)^2+(e-y)^2}$
$12\sqrt{5}=\sqrt{(2e-10)^2+(e-y)^2}$
Squaring both sides
$720=(2e-10)^2+(e-y)^2$
Substitute the value of $y$ from Eq. $(1)$
$720=(2e-10)^2+(e-5e+20)^2$
$720=(2e-10)^2+(20-4e)^2$
$720=(2e-10)^2+[\textrm{-2}(2e-10)]^2$
$720=(2e-10)^2+4(2e-10)^2$
$720=5(2e-10)^2$
$144=(2e-10)^2$
$12=2e-10$
$e=11$
Substitute the value of $e$ into Eq. $(1)$
$y=5e-20$
$=55-20$
$=35$

Answer: $35$

Alternative solution

We can find the coordinates of $C$ by adding the x-coordinate of $B$ and $D$ and the y-coordinates of $B$ and $D$
$C=(30,y+10)$
Area of $ABCD$=area of rectangle $AFCG$-area of the four triangles $AFD,DFC,BCG$, and $ABG$
$600=30(y+10)-1/2[(y+10)10+30(10)+(y+10)10+30(10)]$
$600=30y+300-1/2(10y+100+300+10y+100+300)$
$600=30y+300-1/2(20y+800)$
$600=30y+300-10y-400$
$700=20y$
$y=35$