The altitude of a triangle is increasing at a rate of 1 cm/ min while the area of the triangle is increasing at a rate of 2 cm2 / min. at what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2 ?

Ya, calculus and related rates, such fun!

everything is changing with respect to t

altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min

area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min

base=db/dt

alright

area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt

ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b

so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min

therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt

the base is decreasing at 1.6cm/min

A’b’c’ are the coordinates to the new location of the transformed triangle abc. find the distance between the coordinates points to use for your calculations.