triangles??!!?

Question

i am doing some coursework for maths and i need 3 right angle triangles with a perimeter of 1000. it may sound easy but u cant just put 3 lengths cause it may not work, if that makes sense.

if anyone can give me 3 right angle triangles with a perimeter of 1000 it would be such a great help.

i will award the highest star amount

thnx

i need 3 right angle triangles with the perimeter of 1000. but u cant just put 3 lengths that add to 1000 because it may not work in the construction of the triangle, like the hypotinuse may be 600 and the others 100 and 300, i dont think that would work!?

Answer 1

You can use the following right angled triangles
1) 3,4,5
3+4+5= 12
1000/12=250/3
So, now we multiply 250/3 by all the sides
The first side will be 3*250/3= 250
The second side will be 4*250/3= 333 1/3
The third side will be 5*250/3= 416 2/3
The total length is 1000

2) 5,12,13
5+12+13=30
1000/30= 100/3
So, now we multiply 100/3 by all the sides
The first side will be 5*100/3= 166 2/3
The second side will be 12*100/3= 400
The third side will be 13*100/3= 433 1/3
The total length is 1000



3) 8,15,17
8+15+17=40
1000/40=25
So, now we multiply 25 by all the sides
The first side will be 8*25= 200
The second side will be 15*25= 375
The third side will be 17*25= 425
The total length is 1000

Answer 2

There is only one set of integers that satisfies the conditions -
a = 200, b = 375, c = 425
where a^2 + b^2 = c^2 and a + b + c = 1000.

You could say a = 375 and b = 200, but it's essentially the
same triangle.

Some other relatively easy solutions for a, b and c are:

(1) a = 23+7/16, b = 488, c = 488+9/16

(2) a = 40, b = 479+1/6, c = 480+5/6

(3) a = 100, b = 444+4/9, c = 455+5/9

(4) a = 131+17/18, b = 424, c = 444+1/18

(5) a = 166+2/3, b = 400, c = 433+1/3

(6) a = 218+3/4, b = 360, c = 421+1/4

(7) a = 280, b = 305+5/9, c = 414+4/9

(8) a = 250, b = 333+1/3, c = 416+2/3

There are quite a few others, but the fractions
get more difficult.

Answer 3

The Right Triangle Theory is 3 4 5
To Get a triangle with 1000 use the following

3+4+5=12

Side 1 = 3*(1000/12) = 250
Side 2 = 4*(1000/12)=333.3333333
Side 3 = 5*(1000/12)=416.6666667

The Sum Of These Sides = 1000

Answer 4

To be right triangles, the three sides must work in the

Pythagorean Theorem A^2 + B^2 = c^2

Just pick three numbers to which the Pythagorean Theorem applies. Ex. 3,4,5

Then let x represent a commom ratio number such that

3x + 4x + 5x = 1000 or 12x = 1000 x = 83 1/3

The sides would be 250, 333 1/3, and 416 2/3


Other ex. 5,12,13 and 7,24,25

Answer 5

You may start with most popular 3, 4 and 5 sides which has perimeter 12 linear units

What you need is a perimeter 1000

So Multiply 3,4 and 5 by 1000/12 and you have

3000/12, 4000/12 and 5000/12 is one triangle!

You can workout many such groups that basically links to Pythagoras triplets!

I can give you clue for "where to search groups of Pythagoras triplets"

of course 3,4 and 5 which relate hypotenuse 5

Next try 25, (24, 7) ---> another pair (20,15)

Next try 65, (63, 16) ---> more (60,25), (56, 33), (52,39)

Next try 125 hypotenuse (try for sets yourself)

Next try 625 hypotenuse (try for sets yourself)


Regards!

Answer 6

for a right triangle: x^2 + y^2 = z^2
also X + Y + Z = K = 1000
Z= K - X - Y
(K - X - Y)^2 = X^2 + Y^2
=> K^2 - 2KX -2KY-2XY = 0
X=K/2 * (K-2Y)/(K-Y)

Pick a Y, Solve for X and Then Solve for Z,

Let Y = 200 => X = 375 => z= 425

Another approach is to take the standard triangles like the 3,4,5 triangel whose perimeter is 12 and scale to 1000.

i.e side a: 1000/12*3 = 250
b: 1000/12 * 4 = 333.333333...
c: 1000/12 * 5 = 416.666666..
a+b+c = 1000

Answer 7

aa+bb=cc

a+b+c=1000

You can use 3:4:5 triangle times 1000/12

=> 250; 333.333; 426.666

You add sides of triangle e.g. 3+4+5 = 12 to get divisor
1: 1: root 2 triangle 1+1+1.41 = 3.41
approx 293 : 292 : 415

You can make up the last from your own triangle

Answer 8

The equations that need to be satisfied are:
a²+b²=c²
a+b+c=1000
c-hypothenuse
As you see there are 2 equations for 3 unknowns; a solution to reduce the system to a 2 equation system would be to fix one angle and have a relationship between the catheti. In all situations you will have tg(alpha)=b/a e.g. for a 45deg angle you will have a=b and the equations will become:
2a²=c²
2a+c=1000
solve these for a and c
The method is valid for all angles for which you know the value of the tangent.
I am afraid though that the value of 1000 was not very well chosen. Please do the routine work yourself. Good luck!

Answer 9

am not good at math, however what do you mean by 3 right angle triangles ????
do you mean you want :
- (3) triangles
- each triangle have a Right angle
- the sum of all the sides of the (3) triangles together equal 1000
or
- each triangle have the perimeter of 1000 ??

any way good luck in your homework

Answer 10

I do not understand, are the sum of the perimeters supposed to equal 1000 or do each triangle's perimeter need to be 1000?