Find three different right-angled triangles

[aricho]

How do you do this?

Find three different right-angled triangles whose side lengths are all integers and whose area is 840 square units?

Is there a method that could be used?

 

[James R]

If the perpendicular sides are x and y, the area is

$$A = (1/2)xy = 840$$

So

$$xy = 1680$$

Choose three different integer values of x which divide 1680 to give an integer y. Then you have your answer.

 

[aricho]

Thanks, here's another

Thanks. Once again i was trying to make it too hard, looking at some pythagorean triad stuff, grrr.

How about this one?

The owner of a pet show bought a certain nu,ber of guinea pigs and half that many pairs of mice. He paid $2 for each guinea pic and $1 for each mouse. On every pet he placed a retail price that was 10 percent more than he paid for it. The mice are only sold in pairs.
After all but seven of the animals had been sold, the owner found that he had taken in an much money as he had paid for the animals

His potential profit was the retail value of the remaining 7 animals.

What is this value?

 

[honestrosewater]

If the perpendicular sides are x and y, the area is

$$A = (1/2)xy = 840$$

So

$$xy = 1680$$

Choose three different integer values of x which divide 1680 to give an integer y. Then you have your answer.

Or does the question mean the three triangles' combined area is 840? That would be
(1/2)ab + (1/2)cd + (1/2)ef = 840
You can simplify that to an easier-looking equation. You just want one solution, so you can let b = d = f = 1, and choose a, c, and e accordingly.

 

[aricho]

Guniea Pig Problem

Hey, anyone please read above to the guniea pig problem and have a crack?
Thanks

 

[honestrosewater]

Thanks. Once again i was trying to make it too hard, looking at some pythagorean triad stuff, grrr.

How about this one?

The owner of a pet show bought a certain nu,ber of guinea pigs and half that many pairs of mice. He paid $2 for each guinea pic and $1 for each mouse. On every pet he placed a retail price that was 10 percent more than he paid for it. The mice are only sold in pairs.
After all but seven of the animals had been sold, the owner found that he had taken in an much money as he had paid for the animals

His potential profit was the retail value of the remaining 7 animals.

What is this value?

What have you done so far? Can you figure out the retail prices?

 

[aricho]

i have:
X=guniea pigs
X/2= mice

x=2.20 each
x/2=1.10 each

I'm dead stuck

 

[whozum]

He paid $2 for each guinea pic and $1 for each mouse. On every pet he placed a retail price that was 10 percent more than he paid for it

You can set up a couple equations with that info.

 

[aricho]

so,
guniea pigs=2.20
mice=2.20 (pair)
is that what you mean?

 

[honestrosewater]

Notice it says he bought half as many pairs of mice, and paid $1 for each mouse. Did you copy the question correctly?

 

[aricho]

yep

i sure did, he paid $1 for each mouse.

 

[whozum]

1.1*Cost of guinea pigs and mice (in pairs) = Total profit + 7 cost animals.
Guinea pigs + Mice = Total animals
Guniea pigs * $2 + 2Mice * $2 = Total income.

something like that.

 

[aricho]

"Cost of guinea pigs and mice (in pairs) = Total profit + 7 cost animals"
How is this possible?

 

[whozum]

well his total income was 1.1*cost of guinea pigs and mice. in the end he had 7 animals left.


ugh I am going to bed. I am drunk

 

[HallsofIvy]

Let x be the number of guinea pigs. Then x/2 is the number of pairs of mice. We are told the he bought each guinea pig and each pair of mice for $2 and that he sold each for $2.20. The total cost to him of the animals is 2(x+ x/2)= 3x.
The hard part is that he had 7 animals left and we don't know how many were guinea pigs and how many were mice- so we don't know how many pairs of mice were left.

Okay, let's do it by "cases".
Since he bought an integer number of pairs of mice and only sold them in pairs, the number of mice left must be even and so the number of guinea pigs left must be odd.

1) He had one guinea pig left and 3 pairs of mice.
That would mean he sold x-1 guinea pigs and x/2- 3 pairs of mice: That means he brought in 2.2(x-1)+ 2.2(x/2-3)= 3.3x-8.8 dollars and that must be equal to the cost of the animals- 2(x+ x/2)= 3x: 3.3x- 8.8= 3x so .3x= 8.8 or x= 29 and 1/3. Not an integer.

2) He had three guinea pigs left and 2 pairs of mice.
That means he sold x-3 guinea pigs and x/2- 2 pairs of mice, bringing in 2.2(x-3)+ 2.2(x/2-2)= 3.3x- 11 equaling the cost of 3x: 3.3x- 11= 3x so .3x= 11 or x= 11/.3= 36 and 2/3. Not an integer.

3) He had 5 guinea pigs left and 1 pair of mice.
He sold x-5 guinea pigs and x/2- 1 pair of mice, bringing in 2.2(x-5)+ 2.2(x/2-1)= 3.3x-13.2= 3x so .3x= 13.2 and x= 44. An integer!

4) He had 7 guinea pigs left and no mice.
He sold x-7uinea pigs and x/2 pair of mice, bringing in 2.2(x-7)+ 2.2(x/2)= 3.3x- 15.4= 3x so .3x= 15.4 and x= 51 and 1/3. Not an integer.

He must have had 5 guinea pigs and one pair of mice left when he broke even. He will sell those for 2.2(6)= $13.20, his profit.

Although the problem doesn't ask it, he bought 44 guinea pigs and 22 pairs of mice.

 

[aricho]

"1) He had one guinea pig left and 3 pairs of mice.
That would mean he sold x-1 guinea pigs and x/2- 3 pairs of mice: That means he brought in 2.2(x-1)+ 2.2(x/2-3)= 3.3x-8.8 dollars and that must be equal to the cost of the animals- 2(x+ x/2)= 3x: 3.3x- 8.8= 3x so .3x= 8.8 or x= 29 and 1/3. Not an integer."

Why does this have to be an integer?

 

[honestrosewater]

"1) He had one guinea pig left and 3 pairs of mice.
That would mean he sold x-1 guinea pigs and x/2- 3 pairs of mice: That means he brought in 2.2(x-1)+ 2.2(x/2-3)= 3.3x-8.8 dollars and that must be equal to the cost of the animals- 2(x+ x/2)= 3x: 3.3x- 8.8= 3x so .3x= 8.8 or x= 29 and 1/3. Not an integer."

Why does this have to be an integer?

x needs to be an integer because x is the number of guinea pigs he bought. It's safe to assume he didn't buy a third of a guinea pig.

 

[aricho]

ofcourse

thanks, got it now.

Now, the Final one...

Bart simpson drove at a steady speed along the pacific highway with his mother beside him

"have you noticed," he said "that the McDonalds signs seem to be evenly spaced along the side of the road? i wonder how far apart they are"

Mrs Simpson glanced at her watch, then counted the number of mcdonalds signs they passed in one minute.

What a strange coincidence" exclaimed bart, "when you multiply the number by 10 it is exactly equal to the speed of our car in km/h (kilometers per hour)

Assuming that the car is traveling at a constant speed and the minute began and ended midway between the 2 signs, how far is it between one sign and the next?

I have the answer, it's 166.67 meters, found by:

Suppose they pass 6 signs, so the speed is 60km/h
therefore 1km/minute
1km/6signs=0.16(repepater) km

try with 8, 10,200, all equals the same in the end.

So, my question is, how do i write this algebraically, what is the formal method?