Right triangle?

Question

One leg of a right triangle is 1 ft longer than the other leg. If the triangle is inscribed in a circle and the hypotenuse of the triangle is the diameter of the circle, then what is the length of the radius for which the length of the radius is equal to the length of the shortest side?

Answer 1

Given that the radius is the size of the shorter leg, let this shorter side be r.
Given that the size of the longer leg is r+1.
Given that the size of the hypotenuse is 2r.
Using Pythagorean Theorem:

(2r)^2 = r^2 + (r+1)^2
4r^2 = r^2 + r^2 + 2r + 1
0 = 2r^2 - 2r -1

Using the quadratic formula:
r = { 2 +- sqrt[(-2)^2 - 4(2)(-1)]} / (2(2))
r = {2 +- sqrt[4 +8]} / 4
r = {2 +- sqrt[12]} / 4
r = {2 +- 2sqrt[3]} / 4
r = {1 +- sqrt[3]} / 2

Obviously the length cannot be negative, therefore the length of the radius is:
r = (1 + sqrt3) / 2

Answer 2

Okay. So, we have a right triangles with legs of length X and (X+1). By the Pythagorean theorem:

(Length of hypotenuse)^2 = x^2 + (x+1)^2

Now, we know the length of the hypotenuse is the diameter of a circle, and that circle's radius (half the diameter) is the length of the shorter leg (x). That means the length of the hypotenuse is 2x. Filling in to the above equation:

(2x)^2 = x^2 + (x+1)^2

Expanding the polynomial:

4x^2 = x^2 + (x^2 + 2x + 1)
4x^2 = 2x^2 + 2x +1
2x^2 - 2x + 1 = 0

Now just use the quadratic formula and your home free!

Answer 3

d = 2r; circle identity
h=d given; (hypotenuse equals diameter of circle)
r = L (the length of the radius is equal to the length of the shortest side)
L+1 is the other side.
h^2 = (L+1)^2 + L^2 by pythagora's theorem
h^2 = 2L^2 + 2L + 1 when we multiply out and combine like terms
Since r=L and h=d=2r
we have h=2L
(2L)^2 = 2L^2 + 2L + 1
4L^2 = 2L^2 + 2L + 1
0 = -2L^2 + 2L + 1 which is simply a quadratic to solve
L=(-2±sqrt(4-4(-2)(1)))/-4
L = 1/2 ± sqrt(3)/2
L^2 - L = 1/2
L^2-L+1/4=3/4
(L-1/2)=sqrt(3/4)
L=1/2±sqrt(3)/2
Since we have the answer as a length, and 1/2 - sqrt(3)/2 is negative, we have to take only the postive answer, which is 1/2 + sqrt(3)/2, which I think is 1.3-something.

Answer 4

r= length of the radious

s1 = r (length of the radius equal length of short side)
s2=s1+1 (one leg one foot longer than other) = r+1
h=diameter=2r

right triangle means s1^2 + s2^2 = h^2

r^2+(r+1)^2=(2r)^2

r^2+(r^2+2r+1)=4r^2
2r^2+2r+1=4r^2
2r+1=2r^2
0=2r^2-2r-1

If I have done the algebra correctly (and you should check it) then solving the quadratic above should give you the possible roots. If there is a postivie root then that would be the length "r".

Answer 5

since short leg =radius & hypotenuse=2 radiou, it is a 30-60-90 right triangle.
long leg =short leg *√3
s0 the long leg =x+1=x√3
1=x(√3-1)
x=1/(√3-1)=1.3660254037844386467637231707529

or the more general method:

The diameter can be found from
d^2-x^2+(1+x)^2
d^2=x^2+x^2+2x+1
d^2=2x^2+2x+1

since we want r=x & d=2r; d=2x
(2x)^2=2x^2+2x+1
4x^2=2x^2+2x+1 subtract RHS from each side
2x^2-2x-1=0
x=(2±√(4+4*2*1))/4
x=1/2±√(4+8) /4
x=1/2±1/2 √3
since x>0
x=(1+√3)/2=1.3660254037844386467637231707529

Answer 6

By Pythagoras r^2+(r+1)^2=(2r)^2
2r^2+2r+1=4r^2 So 2r^2-2r-1=0
r=((2+-sqrt(12))/4 (take the + sign) r=1/2(1 +sqrt3)

Answer 7

The triangle legs would the same on both side

Answer 8

fish