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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?
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meters

An Olympic-size swimming pool is approximately 50 meters long by 25 meters wide. What-example-1
User ValerioMC
by
6.0k points

1 Answer

1 vote

Answer:

56m

Explanation:

Diagram:-


\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}

Length=50m

Breadth=25m

Diagonal be x

  • It will form right angle triangle Like below


\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}

So

➝x^2=50^2+25^2

➝x=√(50)^2+(25)^2

➝x=√2500+625

➝x=√3125

➝x=55.9m

➝x=56m(approx)

User Daedalus
by
6.4k points