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 #842526 by MathLover1(20850) $\"\"$ $\"About$

You can put this solution on YOUR website!
point $\"X\"$ is the intersection of the two diagonals $\"TW\"$ and $\"UV\"$ of the cubical box illustrated. \r
\n" ); document.write( "\n" ); document.write( "The shortest distance from point $\"T\"$ to point $\"Z\"$ is $\"4sqrt%283+%29cm\"$ . and it is
\n" ); document.write( " $\"TZ+\"$ = solid diagonal\r
\n" ); document.write( "\n" ); document.write( "Again, by the Pythagorean theorem we know that\r
\n" ); document.write( "\n" ); document.write( " $\"%28TZ%29%5E2+=+%28WZ%29%5E2+%2B+%28WT%29%5E2\"$
\n" ); document.write( " $\"%284sqrt%283+%29cm%29%5E2+=+%28WZ%29%5E2+%2B+%28WT%29%5E2\"$ ........since cube, all sides are equal, so $\"WV=WU=WZ\"$ and $\"%28WT%29%5E2=%28WV%29%5E2%2B%28WU%29%5E2=%28WZ%29%5E2%2B%28WZ%29%5E2=2%28WZ%29%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"16%2A3cm%5E2+=+%28WZ%29%5E2+%2B+2%28WZ%29%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"16%2A3cm%5E2+=+3%28WZ%29%5E2\"$
\n" ); document.write( " $\"%2816%2A3cm%5E2%29%2F3+=+%28WZ%29%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"%28WZ%29%5E2=16cm%5E2\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"WZ=4cm\"$ ........since cube, $\"ZY=4cm\"$ \r
\n" ); document.write( "\n" ); document.write( "point $\"X\"$ is the intersection of the two diagonals, $\"+XZ=XY\"$ \r
\n" ); document.write( "\n" ); document.write( "triangle is isosceles, and altitude from $\"X\"$ to $\"ZY\"$ will bisect $\"+ZY\"$ , let say at point $\"M\"$ and $\"XM\"$ divides triangle $\"XYZ\"$ into two right triangles\r
\n" ); document.write( "\n" ); document.write( " altitude from $\"X+\"$ to $\"WV\"$ will bisect $\"WV\"$ , let say at point $\"N\"$ , and the length of $\"XN+=2cm\"$ \r
\n" ); document.write( "\n" ); document.write( "now, connect points $\"N\"$ and $\"M\"$ and you got right triangle $\"XNM\"$ whose side $\"XM\"$ is altitude from $\"X\"$ to $\"ZY\"$ \r
\n" ); document.write( "\n" ); document.write( "sides of this right triangle are:\r
\n" ); document.write( "\n" ); document.write( " $\"XN+=2cm\"$
\n" ); document.write( " $\"NM=WZ=4cm\"$ \r
\n" ); document.write( "\n" ); document.write( "find altitude $\"XM\"$ :\r
\n" ); document.write( "\n" ); document.write( " $\"%28XM%29%5E2=%282cm%29%5E2%2B%284cm%29%5E2\"$
\n" ); document.write( " $\"%28XM%29%5E2=4cm%5E2%2B16cm%5E2\"$
\n" ); document.write( " $\"%28XM%29%5E2=20cm%5E2\"$
\n" ); document.write( " $\"XM=sqrt%2820cm%5E2%29\"$
\n" ); document.write( " $\"XM=sqrt%284%2A5%29cm\"$
\n" ); document.write( " $\"XM=2sqrt%285%29cm\"$ \r
\n" ); document.write( "\n" ); document.write( "then, the area of triangle $\"XYZ\"$ will be:\r
\n" ); document.write( "\n" ); document.write( " $\"A=%281%2F2%29ZY%2AXM\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"A=%281%2F2%294cm%2A2sqrt%285%29cm\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"A=%281%2Fcross%282%29%294cm%2Across%282%29sqrt%285%29cm\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"A=4sqrt%285%29cm%5E2\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "as you can see, my solution is confirmed by two other tutors
\n" ); document.write( "wondering what do you have against me?\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );