(\(\frac{x1+x2}{2} , \frac{y1+y2}{2}\))we need 3 equations
1. midpoint equation which is (\(\frac{x1+x2}{2} , \frac{y1+y2}{2}\)) when you have 2 points
2. distance formula which is D= \(\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}\)
3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height
so
x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=(\(\frac{11+19}{2} , \frac{10+6}{2}\))
midpoint=(\(\frac{30}{2} , \frac{16}{2}\))
midpoint= (15,8)
point x=(15,8)
y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=(\(\frac{5+21}{2} , \frac{8+0}{2}\))
midpoint=(\(\frac{26}{2} , \frac{8}{2}\))
midpoint=(13,4)
Y=(13,4)
legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= \(\sqrt{(19-11)^{2}+(6-10)^{2}}\)
D= \(\sqrt{(8)^{2}+(-4)^{2}}\)
D= \(\sqrt{64+16}\)
D= \(\sqrt{80}\)
D= \(4 \sqrt{5}\)
BC=\(4 \sqrt{5}\)
X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= \(\sqrt{(13-15)^{2}+(4-8)^{2}}\)
D= \(\sqrt{(-2)^{2}+(-4)^{2}}\)
D= \(\sqrt{4+16}\)
D= \(\sqrt{20}\)
D= \(2 \sqrt{5}\)
XY=\(2 \sqrt{5}\)
the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC=\(4 \sqrt{5}\) and XY=\(2 \sqrt{5}\)
AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= \(\sqrt{(21-5)^{2}+(0-8)^{2}}\)
D= \(\sqrt{(16)^{2}+(-8)^{2}}\)
D= \(\sqrt{256+64}\)
D= \(\sqrt{320}\)
D= \(4 \sqrt{2}\)
AD=\(4 \sqrt{2}\)
so we have
AD=\(4 \sqrt{2}\)
BC=\(4 \sqrt{5}\)
XY=\(2 \sqrt{5}\)
AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
(\(4 \sqrt{2}+4 \sqrt{5}\)) times 1/2 times \(2 \sqrt{5}\) =
(\(4 \sqrt{2}+4 \sqrt{5}\)) times \sqrt{5} [/tex] =
\(4 \sqrt{10}+4*5\)=\(4 \sqrt{10}+20\)=80 \(\sqrt{10}\)=252.982
X=(15,8)
Y=(13,4)
BC=\(4 \sqrt{5}\)
XY=\(2 \sqrt{5}\)
Area=80 \(\sqrt{10}\) square unit or 252.982 square units
