Question

Please could you find the answers to the questions in the attachment.

Answer 1

($\frac{x1+x2}{2} , \frac{y1+y2}{2}$)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