document.write( "Question 420202: A small commuter airline flies to three cities whose locations form the vertices of a right triangle. The total flight distance (from city A to city B to city C and back to city A) is 1400 miles. It is 600 miles between the two cities that are furthest apart (city A to city B). Find the other two distances between cities. \n" ); document.write( "
Algebra.Com's Answer #293704 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Farthest distance = 600 miles will be the hypotenuse\r \n" ); document.write( "\n" ); document.write( "the balance 800 is the sum of the two legs of the right triangle. \n" ); document.write( "if one side is x \n" ); document.write( "other side = 800-x \n" ); document.write( ".. \n" ); document.write( "x^2+(800-x)^2=600^2\r \n" ); document.write( "\n" ); document.write( "x^2+640000-1600x+x^2=360000 \n" ); document.write( "2x^2-1600x+280000=0 \n" ); document.write( "/2 \n" ); document.write( "x^2-800x+140000=0 \n" ); document.write( "solve using quadratic equation \n" ); document.write( "b^2-4ac= 120000 \n" ); document.write( "one side = 541.4 miles \n" ); document.write( "other side =258.6 miles \n" ); document.write( " |