hard question please help me
ABC is a triangle with sides AB = 6m, BC =8m, and AC = 10m. A line k in the plane of the triangle ABC moves along the segment AC at the rate of 1cm per sec. The line starts at A and ends at C and is always perpendicular to AC.
1. How long does it take the line to reach the point B?
2. How long does it take the line to bisect the area of the triangle?
3. What is the area of the region that the line sweeps in its movement after 6min 45 sec?
4.What is the area of the region that the line sweeps at any give moment from the start of its movement?

- Curry

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- Curry

i need help on 2/3/4

- anonymous

if you can, draw the figure to go along with it.

- Curry

|dw:1328077516987:dw|

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## More answers

- anonymous

this IS a tough one...

- Curry

yep it was a precalc problem. i solved 1,3,4 and i need 2 now but im not sure if my answer for 3 and 4 are right but i am sure of number 1

- anonymous

the key is in part 4. if you can figure that out, you should be able to do the rest very easily. give me a moment.

- Curry

i got 2/3cm^2 for every 1 centimeter it moves

- Curry

but i t

- Curry

but i t

- Curry

it is wrong cause there is no way that is possible

- anonymous

sorry, had to find some paper

- anonymous

i think you have it right, actually. you have a basic 3/4/5 triangle here. when the line k is moving across, it is creating a smaller 3/4/5 triangle because it is perpendicular to AC.

- anonymous

so the area of the small triangle will be the base 1 cm/s times the height 4/3 cm/s divided by 2; multiply by the amount of time passed.

- anonymous

and to get 2), you figure out how big the total triangle is, and then divide by 2 divide by 1 cm/s, i think.

- Curry

hey here are some flaws with our thinking

- anonymous

ok

- Curry

for number 4 that is true only as long as u reach the altitude after that each one cm is going to create a trapezoid

- Curry

same with number 2

- anonymous

good point

- anonymous

that's why part 1 is important. it tells you when you need to switch over.

- Curry

at 3.6

- anonymous

if you're in the US, technically i think it's a trapezium, not a trapezoid; since it won't have parallel sides lol

- Curry

im in the the us we just refer to it as trap in my class

- Curry

since it uses the same formula i believe

- anonymous

i learned (b1 + b2)*h/2. i am very curious as to what b2 will be. however, there's an easier way, and one that's already in place; split it into 2 triangles. the second triangle will have a new area/s equation.

- anonymous

have you learned about piecewise functions yet?

- anonymous

it's been a long time since i had precalc

- Curry

ye

- Curry

number 4 doesnt work for 2/3

- Curry

cause the rea has to be greater than any given side

- Curry

area*

- anonymous

is this electronic submission? sounds like what 4 is asking for is an equation.

- Curry

it is asking how much area is covered for every centimeter

- anonymous

i may be misunderstanding you, but the area swept out changes with every centimeter. this is time dependent, not distance dependent... also remember that k is moving at one CENTImeter per second, and the triangle has sides in meters.

- Curry

yes

- anonymous

ok. didn't mean to be condescending or anything..

- Curry

but if it moves 1 centimeter then thats 1 sec after 1 centimeter the 2 legs are 1cent imeter and 4/3 cm which creates an area of 2/3
but the area has to be greater than the sides

- Curry

for all 3/4/5 traingles

- anonymous

oh i see what you're saying now.

- anonymous

that somehow doesn't seem right, though. if you take 3/4/5 and then divide all sides by two to get 1.5/2/2.5, if you multiply 1.5 by 2 and then divide by 2, you get 1.5; am i thinking about this wrong?

- Curry

ye i see what ur saying

- anonymous

still, i am doing something wrong.

- Curry

see thats my problem, but is that really possible

- Curry

wait u r doing something wrong

- Curry

wait nvmthats the same thing

- anonymous

ok, here's the problem: when you increase the distance along AC by 1, you're not increasing the area by 2/3cm^2, you're actually increasing the *length of k* by 2/2 cm

- anonymous

2/3* cm

- Curry

then the area would be even smaller

- anonymous

huh.

- anonymous

lol

- Curry

1 * 2/3 /2

- Curry

is 2/6

- anonymous

that was a "huh, you're right" not a "huh?"

- Curry

kk

- Curry

such a carzy problem , my teacher solved it 10 min

- Curry

but refused to chare this with us

- Curry

we need sataleite

- Curry

or zarkon

- anonymous

zarkon?

- anonymous

i'm dumb. dumb dumb dumb. i think it's going up by 4/3 not 2/3

- Curry

still doesnt work

- anonymous

hmm

- Curry

wow no one can answer this

- Curry

wow no one can answer this

- razor99

so u r really happy about it,

- Curry

??

- anonymous

OK, I'm commenting so that I have this saved for tomorrow. Curry, post if you don't get help while I'm asleep. I'll bring it to the higher-math helpers' attention tomorrow.

- Curry

kk
im a go to bed too.

- anonymous

currry do you have answers?

- Curry

im sure this is one of those prob lems that only zarkon or satelite can solve
i have the answer for 1 which im sure baout for 3 i got 10.7241

- Curry

4 and 2 are insanely unthinkable

- Curry

ig u guys figure out the answer pleazse post

- Curry

ig u guys figure out the answer pleazse post

- Curry

check my answer for number 3

- Curry

oh wait thats is true

- Curry

so number 2 woulb be 500 secsonds

- Curry

so number 2 woulb be 500 secsonds

- Curry

hey fool if u happen to find all the answers plz post it with the WORK. I got these answers plz do check them
1.6min
2.8.3sec
3.10.7241 as the area
4. ?

- Curry

2. 8.3sec

- Curry

3. 10.7241

- Curry

i will check in the morning for answer. its 1am for me so im going to b ed

- anonymous

as far as i can tell, 6 minutes is correct.

- anonymous

8.3 seconds can't be right, though.

- anonymous

yeah, but that's not question 1 haha. question 1 is just asking how long it takes for k to equal the height. *edited for spelling/grammar*

- anonymous

taking the hypotenuse as the base.

- anonymous

|dw:1328083228792:dw|

- anonymous

ok, now let's get back to part 2. first, we have to note what we're looking for; we want the time when the area swept out by k is half of the total area. the total area is 6*8/2 = 24, so we want 24/2 = 12.
easy part done.
we know just by looking that it's going to be past point B. so we can include that area as already accounted for, and just look for the remainder.

- anonymous

hopefully this isn't too confusing, but we have area 1 as follows: |dw:1328085095628:dw| and we want to figure out how big area 2 needs to be: |dw:1328085234534:dw|

- anonymous

As you already figured out, Curry, and I obtusely (oh, haha! a math joke...sorry i'll stop now) didn't understand, we have a trapezoid as the second area.
Area 1 is, of course, 4.8*3.6/2 = 8.64 m^2
Area 2, therefore, needs to be 12 - Area1 = 3.36 m^2
we'll set this equal to the equation for the trapezoid, which is (k+B2)*x/2 = 3.36

- anonymous

lol u r still here

- anonymous

we know that k = 4.8 m, so we need to know what B2 is so that we can solve for x.
|dw:1328085662851:dw| this is a little harder than figuring out the growth of the triangle segment.

- anonymous

one way to think about it is to take the imaginary little triangle at the top and figure out it's growth instead |dw:1328085960597:dw| for each increment of 1 cm to the right, the small triangle will grow 3/4 cm in height. since we want B2, we'll take \[k - 3/4 \times x\]

- anonymous

The area of the trapezoid will therefore be (k + (k-(3/4)*x)/2
i don't like their equation builder so i'm going to stop using it until i learn it better.

- anonymous

set that equal to 3.36 m, and we have a quadratic: .375x^2 - 4.8 x + 3.36 = 0. solve this for x to get 12.06 m and .743 m
we can throw out 12.06 because that would put us beyond the end of the original triangle, but .743 is viable.
finally, to get the answer for 2, we add .743 to 3.6, and multiply by 100 to get the number of seconds: 434.3 seconds., or 7 minutes 14ish seconds.

- anonymous

4.8 meters.

- anonymous

so 3.6/4.8/6 triple

- anonymous

i THINK that's 1 and 2 conclusively solved now. @curry sorry for long-winded answer. 3 and 4 seem poorly worded.

- anonymous

if 3 is asking: "what is the total area swept out by k after 6:45", the answer is Area1 = 8.64, Area2 = 2.084
Area1 + Area2 = 10.7241 meters. so looks like he got that part.

- anonymous

So now, we have to ask again what 4) is really looking for... cuz the way i read it, sounds like it wants an equation. the area swept out per centimeter is constantly changing.

- anonymous

If it wants an equation, it's a bit challenging but doable:
By sneaky algebraic methods, the area can be described as:
Area = (2/3)*x^2 for x <3.6
Area = 8.64 + 4.8*(x-3.6)+.375*(x-3.6)^2 for x>3.6
The -3.6 in there should shift the equation correctly so that everything is accounted correctly.
I think there's an even sneakier way to combine the two equations into one, but i honestly can't remember/figure out how to do it. it's after 4 am where i live. I encourage you to plug in various values for x between 0 and 10 to see if this works correctly!
If it wants a flat number, I'm afraid I don't know how to help you... I hope that the rest of this proves useful, and good luck with your assignment!

- anonymous

for reference: x = 0 should give 0, x = 3.6 should give 8.64, x = 4.343 should give about 12, and x = 10 should give 24.

- dumbcow

just to add to the conversation: I found an equation for area thats a bit less complicated
x = t/100, where t is in sec
|dw:1328091690260:dw|
From the 2 similar triangles we can make the following relationship
\[\frac{10-x}{y} = \frac{8}{10} \rightarrow y = \frac{5(10-x)}{4}\]
The 3rd side of the triangle then would be
\[\frac{3(10-x)}{4}\]
Area of the triangle
\[\rightarrow \frac{1}{2}(10-x)*\frac{3(10-x)}{4} = \frac{3}{8}(10-x)^{2}\]
Now total area of large triangle is 24, so area of part the line has passed is the difference
for 0

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