Questions on Geometry: Pythagorean theorem answered by real tutors!

From the problem, it is clear that t-1 is one of legs, since t-1 is
shorter than t+1.


But looking at (t+1) and (6-t), we can not say which of them is longer;
so, we can not determine in advance, which of these two is the hypotenuse.


Therefore, we should consider two cases separately (a) t+1 > 6-t  and  (b) t+1 < 6-t.


Case (a).  t+1 > 6-t.

           Then 6-t is the other leg, while t+1 is the hypotenuse.

           So, we write the Pythagorean equation and find t from it

               (t-1)^2 + (6-t)^2 = (t+1)^2

               t^2 - 2t + 1 + 36 - 12t + t^2 = t^2 + 2t + 1

               t^2 - 16t + 36 = 0

                =  =  = .

               So, one   root is t =  = 13.29  (approx.);
                   other root is t =  =  2.71  (approx.).

               The root t =  = 13.29 does not work for the problem,
                                                       since then 6-t is negative.

               The root t =  = 2.71 satisfies inequality t+1 > 6 -t.


               So, in this case the legs are  t-1 =   = 1.71;
                                         and  6-t =  = 3.29;

                           the hypotenuse is  t+1 =   = 3.71.


               So, in this case the hypotenuse is really longer than the legs,
               and the triangle inequalities are satisfied, which is easy to check.


               Thus this case (a) is one possible solution.



Case (b).  t+1 < 6-t.

           Then 6-t is the hypotenuse, while t+1 is the other leg.

           So, we write the Pythagorean equation and find t from it

               (t-1)^2 + (t+1)^2 = (6-t)^2

               t^2 - 2t + 1 + t^2 + 2t + 1 = 36 - 12t + t^2

               t^2 + 12t - 34 = 0

                =  =  = .


               So, one   root is t =  = 14.37  (approx.);
                   other root is t =  = -2.37  (approx.).


               These roots do not work for this problem since t = 14.37 will create negative 6-t,

                                                        while t = -2.37 creates both negative t+1 and t-1.


CONCLUSION.  In this problem, the only solution is  t =  = 2.71;

             the legs are  t-1 =   = 1.71  and  6-t =  = 3.29;

             the hypotenuse is  t+1 =   = 3.71.

The lessons to learn from my post

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

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

2x^2 + 2 = 2x

2x^2 - 2x + 2 = 0

x^2 - x + 1 = 0.


Use the quadratic formula and get two solutions in complex numbers

 =   and   = .    ANSWER


There is no real solutions.

This equation is equivalent to

     +  = 47x.


Write with the common denominator on the left side

     +  = 47x

     = 47x,

     = 47x,

    x^3 = ,

    x =  = 0.245384174  (rounded).    ANSWER

The shortest distance between two points is the straight line, connecting the points.

Surely, the hummingbird, being a disciplined rational creature, will choose the straight line to fly.


So, this problem is about a triangle.


You may even make one step forward and to see (or to guess/to recognize) a right-angled triangle here.


As soon as you identify a right angled triangle, apply the Pythagorean formula   


    d =  =  =  = 13 meters.    ANSWER


(5,12,13)  is a Pythagorean triple,  next after  (3,4,5).