SOLUTION: A boxing ring is in the shape of a square with 20 ft. on each side. How far apart are the fighters when they are in opposite corners of the ring?
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-> SOLUTION: A boxing ring is in the shape of a square with 20 ft. on each side. How far apart are the fighters when they are in opposite corners of the ring?
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You can put this solution on YOUR website! THIS PROBLEM IS A RIGHT TRIANGLE WITH 2 SIDES = TO 20 FEET AND THE HYPOTENUSE THE DISTANCE BETWEEN THE OPPOSITE CORNERS. THUS THE FORMULA A^2+B^2=C^2 THUS
20^2+20^2=C^2 OR 400+400=C^2 OR 800=C^2 OR C=28.28 FEET FOR THE DIAGONAL OF THE 20 FOOT SQUARE RING.
c^2 is the same as c*c or (c squared). 800 IS THE SAME AS 28.28*28.28 or 28.28 is the square root of 800 or 800=28.28*28.28 (or 28.28^2 or 28.28 squared=800)
another example: x^2=25 or x^2=x*x & 25=5*5 thus the sqare root of x^2=x & the square root of 25=5 then x=5.
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