Question 21265: Mr. Coleman our principal decided to redesign his rectangular rose bed into the shape of a right-angled triangle. The existing bed measured 24 by 35 feet. he discovered that he could make any one of three different right triangular beds, each equal in area to the existing bed and each having sides of an integral number of feet. As it was his custom to fence his beds, he naturally chose the bed with the smallest perimeter. What were the dimensions , in feet , of the new bed? What were the dimensions, in feet, of the other two alternatives?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! AREA OF RECTANGLE =24*35=840 SQ.FT.
AREA OF TRIANGLE = A*B/2,WHERE A AND B ARE THE 2 ADJACENT SIDES TO THE RIGHT ANGLE.
HENCE A*B=2*840=1680…FURTHER A,B,AND THE HYPOTENUSE SQ.RT.OF(A^2+B^2) SHOULD ALL BE INTEGERS
SO THE POSSIBILITIES ARE …..
A B C PERIMETER
15 112 113 240
24 70 74 168
40 42 58 140
SO THE BED WITH 40,42 AND 58 IS THE ONE WITH SMALLEST FENCING REQUIREMENT.
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