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?
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Source: NCTM Mathematics Teacher, November 2006

SOLUTION
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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
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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

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About mvtrinh

Retired high school math teacher.
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