The sum of the two shorter sides of a rightangledtriangle is 18.Calculate the least possible length of hypoth?

Question

Who can help me with this maths problem?

Answer 1

x+y = 18.
We want to minimise x^2 + y^2.
Thats x^2 + (18-x)^2 = 2x^2 - 36x + 324.

If you know calculus, you can differentiate to get 4x - 36 = 0, so x = 9.
Or you could rewrite the equation as 2(x^2 - 18x + 162) = 2((x^2 - 18x + 81 + 81) = 2((x-9)^2 + 81) = 2(x-9)^2 + 162.
Then the minimum comes when x = 9.

Either way, the hypotenuse has length sqrt(9^2 + 9^2) = 9rt(2).

Answer 2

I think that the two shorter sides would both be nine. To find the length of the hypotenuse with the two shorter side lengths being nine you would have to do 9^2+9^2 which would be 81+81. That is equal to 162. Then you have to find the square root of 162. Since it isn't a perfect square you either have to reduce it down and keep it exact or you have to find the decimal. If you want to use keep it exact you have to first find the largest perfect square that goes into 162 which is 81. Then you have to find out what times 81 equals 162. The answer is 2 so you have to do the square root of 81 times the square root of 2. That would be 9 times the square root of 2. If you want to find the decimal you can just use the square root button on a calculator. I would find the decimal out for you but I don't have a calculator so I'll just stick with 9 times the square root of 2.

Answer 3

Okay well the sum of 2 of the sides has to be greater then the last one example side one is 4 side two is 5 so it would have to be like greater ten 4 less then 9

Answer 4

Given any 2 facets of a triangle: the sum of their lengths is greater than the 0.33 area and actual the linked fee of the difference of their lengths is below the dimensions of the 0.33 area. For this triangle, a probability measures, x, for the 0.33 area are |11 - 15 | < x < 11 + 15 4 < x < 26 answer: 4 isn't a a probability length for the 0.33 area

Answer 5

The shortest hypotenuse is when the two short sides are the same, ie 9 each. So the hypotenuse would be 9 x root 2, which I think would be around 12.6

Answer 6

A squared + B squared = Hypotenuse squared, so try different numbers that A+B = 18. 12.7 is the smallest number I got. 9 for each side. Keep trying and see.

Answer 7

let one of the legs be x
the other leg=18-x
minimise x^2+(18-x)^2
dH/dx=2x-2(18-x)
=4x-36
setting this to zero
4x=36
x=9
so the legs have got to be 9 each to minimise the length of the hypotenuse
in other words the right triangle should be isosceles

Answer 8

I think this is a calculus problem. You have to set up a primary equation, take the derivative, graph it, and then find the absolute minimum.