document.write( "Question 1040514: The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place. \n" ); document.write( "
Algebra.Com's Answer #655323 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place. \n" ); document.write( "========== \n" ); document.write( "Area = b*h/2 = 24 \n" ); document.write( "Using 12 for the base, h = 4. \n" ); document.write( "The altitude from the right angle = 4. \n" ); document.write( "------- \n" ); document.write( "Label the 2 parts of the base c & d, and the 2 sides e & f \n" ); document.write( "c + d = 12 \n" ); document.write( "e^2 - c^2 = 16 \n" ); document.write( "f^2 - d^2 = 16 \n" ); document.write( "e^2 + f^2 = 12^2 \n" ); document.write( "------ \n" ); document.write( "Sub for c: c = 12 - d \n" ); document.write( "e^2 - d^2 + 24d - 144 = 16 \n" ); document.write( "e^2 = d^2 - 24d + 160 \n" ); document.write( "--- \n" ); document.write( "Sub for f: f^2 = d^2 + 16 \n" ); document.write( "e^2 + d^2 + 16 = 144 \n" ); document.write( "--> e^2 = 128 - d^2 \n" ); document.write( "---- \n" ); document.write( "128 - d^2 = d^2 - 24d + 160 \n" ); document.write( "2d^2 - 24d + 32 = 0 \n" ); document.write( "d^2 - 12d + 16 = 0 \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "================= \n" ); document.write( "The 2 solutions are c & d. Their sum is 12, the hypotenuse. \n" ); document.write( "6 + sqrt(20) and 6 - sqrt(20) \n" ); document.write( "--------- \n" ); document.write( "e^2 = c^2 + 16 \n" ); document.write( "= 36 + 20 + 12sqrt(20) \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |