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