51. The figure below illustrates an isosceles right
triangle with legs of length 1, along with onefourth of a circle centered at the right-angle
vertex of the triangle. Using the result that
the shortest path between two points is a line
segment, explain why this figure shows that
2(squareroot of 2) <2
I see this is the second time you posted this.
Of course you realize that we can't see any diagram, but I think I can draw it from your description
First of all, you must have a typo.
Did you mean 2√2 < π ??
Since your statement of2√2 < 2 is false, (2√2 = appr. 2.8)
In the triangle let the hypotenuse be h
h^2 = 1^2 + 1^2 = 2
h = √2
the arc is (1/4)(2π) = π/2
then because of the straight between two points is always less than a curve between the same two points,
√2 < π/2
multiply by 2
2√2 < π
To explain why the figure shows that 2√2 < 2, let's consider the two paths: one path along the hypotenuse of the triangle and one path along the curved boundary of the circle.
The hypotenuse of the triangle is the shortest path between the two points, which have a distance of 2. Since it is an isosceles right triangle, the hypotenuse has a length of √2 times the length of each leg (1). Therefore, the length of the hypotenuse is √2.
On the other hand, we have one-fourth of a circle centered at the right-angle vertex. If we were to follow the curved boundary of the circle, the length of the path would be more than going straight along the hypotenuse.
In the given figure, the curved boundary of the circle is divided into four segments, and one of the segments is shown. The length of the curved boundary in this segment is actually greater than the hypotenuse of the triangle.
Since the shortest path between two points is a line segment, the hypotenuse of the triangle is shorter than following the curved boundary of the circle. Therefore, we can conclude that 2√2 < 2.