Curry
  • Curry
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Curry
  • Curry
i need help on 2/3/4
anonymous
  • anonymous
if you can, draw the figure to go along with it.
Curry
  • Curry
|dw:1328077516987:dw|

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More answers

anonymous
  • anonymous
this IS a tough one...
Curry
  • 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
  • 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
  • Curry
i got 2/3cm^2 for every 1 centimeter it moves
Curry
  • Curry
but i t
Curry
  • Curry
but i t
Curry
  • Curry
it is wrong cause there is no way that is possible
anonymous
  • anonymous
sorry, had to find some paper
anonymous
  • 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
  • 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
  • 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
  • Curry
hey here are some flaws with our thinking
anonymous
  • anonymous
ok
Curry
  • 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
  • Curry
same with number 2
anonymous
  • anonymous
good point
anonymous
  • anonymous
that's why part 1 is important. it tells you when you need to switch over.
Curry
  • Curry
at 3.6
anonymous
  • 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
  • Curry
im in the the us we just refer to it as trap in my class
Curry
  • Curry
since it uses the same formula i believe
anonymous
  • 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
  • anonymous
have you learned about piecewise functions yet?
anonymous
  • anonymous
it's been a long time since i had precalc
Curry
  • Curry
ye
Curry
  • Curry
number 4 doesnt work for 2/3
Curry
  • Curry
cause the rea has to be greater than any given side
Curry
  • Curry
area*
anonymous
  • anonymous
is this electronic submission? sounds like what 4 is asking for is an equation.
Curry
  • Curry
it is asking how much area is covered for every centimeter
anonymous
  • 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
  • Curry
yes
anonymous
  • anonymous
ok. didn't mean to be condescending or anything..
Curry
  • 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
  • Curry
for all 3/4/5 traingles
anonymous
  • anonymous
oh i see what you're saying now.
anonymous
  • 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
  • Curry
ye i see what ur saying
anonymous
  • anonymous
still, i am doing something wrong.
Curry
  • Curry
see thats my problem, but is that really possible
Curry
  • Curry
wait u r doing something wrong
Curry
  • Curry
wait nvmthats the same thing
anonymous
  • 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
  • anonymous
2/3* cm
Curry
  • Curry
then the area would be even smaller
anonymous
  • anonymous
huh.
anonymous
  • anonymous
lol
Curry
  • Curry
1 * 2/3 /2
Curry
  • Curry
is 2/6
anonymous
  • anonymous
that was a "huh, you're right" not a "huh?"
Curry
  • Curry
kk
Curry
  • Curry
such a carzy problem , my teacher solved it 10 min
Curry
  • Curry
but refused to chare this with us
Curry
  • Curry
we need sataleite
Curry
  • Curry
or zarkon
anonymous
  • anonymous
zarkon?
anonymous
  • anonymous
i'm dumb. dumb dumb dumb. i think it's going up by 4/3 not 2/3
Curry
  • Curry
still doesnt work
anonymous
  • anonymous
hmm
Curry
  • Curry
wow no one can answer this
Curry
  • Curry
wow no one can answer this
razor99
  • razor99
so u r really happy about it,
Curry
  • Curry
??
anonymous
  • 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
  • Curry
kk im a go to bed too.
anonymous
  • anonymous
currry do you have answers?
Curry
  • 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
  • Curry
4 and 2 are insanely unthinkable
Curry
  • Curry
ig u guys figure out the answer pleazse post
Curry
  • Curry
ig u guys figure out the answer pleazse post
Curry
  • Curry
check my answer for number 3
Curry
  • Curry
oh wait thats is true
Curry
  • Curry
so number 2 woulb be 500 secsonds
Curry
  • Curry
so number 2 woulb be 500 secsonds
Curry
  • 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
  • Curry
2. 8.3sec
Curry
  • Curry
3. 10.7241
Curry
  • Curry
i will check in the morning for answer. its 1am for me so im going to b ed
anonymous
  • anonymous
as far as i can tell, 6 minutes is correct.
anonymous
  • anonymous
8.3 seconds can't be right, though.
anonymous
  • 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
  • anonymous
taking the hypotenuse as the base.
anonymous
  • anonymous
|dw:1328083228792:dw|
anonymous
  • 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
  • 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
  • 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
  • anonymous
lol u r still here
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
4.8 meters.
anonymous
  • anonymous
so 3.6/4.8/6 triple
anonymous
  • anonymous
i THINK that's 1 and 2 conclusively solved now. @curry sorry for long-winded answer. 3 and 4 seem poorly worded.
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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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