An Olympic-size swimming pool is approximately 50 meters long by 25 meters wide. What distance will a swimmer travel if they swim from one corner to the
opposite?
type your answer...
meters

An Olympicsize swimming pool is approximately 50 meters long by 25 meters wide What distance will a swimmer travel if they swim from one corner to the opposite class=

Respuesta :

Answer:

56m

Step-by-step explanation:

Diagram:-

[tex]\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier (0,0)(0,0)(5,0)\qbezier (0,0)(0,0)(0,3)\qbezier (0,3)(0,3)(5,3)\qbezier (5,3)(5,3)(5,0)\qbezier (0,3)(5,0)(5,0)\qbezier (0,0)(0,0)(5,3)\put (-0.4,-0.2){\sf D}\put (-0.2,3.1){\sf A}\put (5.2,3.1){\sf B}\put (5.2,-0.2){\sf C}\put (2.4,1){\sf O}\put (2,-0.4){\bf 50\:m}\put (-1,1.5){\sf 25\:m}\end {picture}[/tex]

Length=50m

Breadth=25m

Diagonal be x

  • It will form right angle triangle Like below

[tex]\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 25m}\put(2.8,.3){\large\bf 50m}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf $\Theta$}\end{picture}[/tex]

So

➝x^2=50^2+25^2

➝x=√(50)^2+(25)^2

➝x=√2500+625

➝x=√3125

➝x=55.9m

➝x=56m(approx)