all of these are applications of the distance formula
remember a^2+b^2=c^2
so the distance (c) is [tex] \sqrt{a^{2}+b^{2}} [/tex]
a and b are legnths of legs so we find the legnths
the distance between points (x1,y1) and (x2,y2) is
D=[tex] \sqrt{(x2-x1)^{2}+(y2-y1)^{2}} [/tex]
input points and find distance
1. (3,5) ot (7,3)
D=[tex] \sqrt{(7-3)^{2}+(3-5)^{2}} [/tex]=√(16+4)=√20=4.47
answer is A
2. (-3,-2) to (4,6)
D=[tex] \sqrt{(4-(-3))^{2}+(6-(-2))^{2}} [/tex]=√(49+64)=√113=10.63
answer is 10.6 units
3. from (1,7) to (9,-2)
D=[tex] \sqrt{(9-1)^{2}+(-2-7)^{2}} [/tex]=√(64+81)=√145=12.04
answer is B
4. (-1,-2) to (6,2)
D=[tex] \sqrt{(6-(-1))^{2}+(2-(-1))^{2}} [/tex]=√(49+16)=√65=8.06
round
8.1 units
5. apply multipule times
A=(2,2)
B=(12,2)
C=(9,8)
D=(2,8)
find the distance between
A to B
B to C
C to D
D to A
we can see the distances between AB, CD and AD easily
just since 1 variable doesn't change, we just find the change in the other variable
AB=10 units
CD=7 units
DA=6 units
now
CB is from (12,2) to (9,8)
D=[tex] \sqrt{(9-12)^{2}+(8-2)^{2}} [/tex]=√(9+36)=√45=6.7
add everybody
AB+BC+CD+DA=10+6.7+7+6=29.7
we see that they put a trick and see option D, the not careful people will chose that but the quesiton askes for the perimiter
answer is B
