Six boys stood equally spaced on a circle of radius 40 feet. each boy walked to all of the other non-adjacent persons on the circle, shook their hands and then returned to his original spot on the circle before the next boy started his trip to shake hands with all of the other non-adjacent boys on the circle. after all six boys had done this, what is the least distance in feet that could have been traveled

answer: plz put a pic in so i can answer it

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

answer is c

step-by-step explanation:

81/12=324/x

x=48