document.write( "Question 1205618: \r
\n" ); document.write( "\n" ); document.write( "Point X is the intersection of the two diagonals TW and UV of the cubical box illustrated. The shortest distance from point T to point Z is $\"4sqrt%283%29\"$ cm. Find the area in square centimetres of triangle XYZ. \n" ); document.write( "

Algebra.Com's Answer #842531 by ikleyn(52817) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Point X is the intersection of the two diagonals TW and UV of the cubical box illustrated.
\n" ); document.write( "The shortest distance from point T to point Z is $\"4sqrt%283%29\"$ cm.
\n" ); document.write( "Find the area in square centimetres of triangle XYZ.
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document.write( "It is well known fact that if \"a\" is the length of the edge of a cube,\r\n" );
document.write( "then the longest 3D-diagonal of the cube, connecting the opposite corners (vertices)\r\n" );
document.write( "is  .\r\n" );
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document.write( "In this problem, you are given that the longest 3D-diagonal of the cube is  cm.\r\n" );
document.write( "It means that the edge of the cube is 4 cm long.\r\n" );
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document.write( "OK.  To find the area of triangle XYZ, we need to know its height (its altitude).\r\n" );
document.write( "This altitude is the hypotenuse of the right angled triangle with the legs a/2 = 2 cm\r\n" );
document.write( "and a = 4 cm, so the altitude of the triangle XYZ is\r\n" );
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document.write( "    h =  =  =  cm.\r\n" );
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document.write( "Now the area of the triangle XYZ is half the product of its base a = ZY = 4 cm by the altitude h\r\n" );
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document.write( "     =  =  = 8.944 cm^2  (approximately).    ANSWER\r\n" );
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