Step-by-step explanation:
Since PT = 2PM and AR = 2AD,
The midsegment MD must be the average of PA and TR:
Therefore we have
2(20x + 4) = 10 + (34x + 16)
=> 40x + 8 = 34x + 26
=> 6x = 18
=> x = 3
Hence length of MD
= 40(3) + 8 = 128.
I’m not sure if the 2 is counted in the equation or not
Since x=4, substitute it for x in the other equation
4+3y=29
3y=25
Y=8.333333333
2(4)+3y=29
8+3y=29
3y=37
y=12.33333333
It has to be the last one, D. I did it over 3 times to make sure.
