document.write( "Question 969785: Each side of the squares equals to 7units.\r
\n" ); document.write( "\n" ); document.write( "How do I calculate the area of an equilateral triangle inside the square with sides 7units?\r
\n" ); document.write( "\n" ); document.write( "How do I work out the area of a circle inside a square with sides equal to 7 units?\r
\n" ); document.write( "\n" ); document.write( "What would the ratio be of area of triangle:area of circle?
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Algebra.Com's Answer #592681 by KMST(5328)\"\" \"About 
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EQUILATERAL TRIANGLE:
\n" ); document.write( "The area of a triangle with sides of lengths \"a\" , \"b\" , and \"c\" ,
\n" ); document.write( "opposite angles of measures \"A\" , \"B\" , and \"C\" ,
\n" ); document.write( "can be calculated as \"area=b%2Ac%2Asin%28A%29\" :
\n" ); document.write( "
\n" ); document.write( "In the case of an equilateral triangle,
\n" ); document.write( "all sides have the same length, \"b\" ,
\n" ); document.write( "and all angles measure \"60%5Eo\" ,
\n" ); document.write( "so \"area=b%5E2%2Asin%2860%5Eo%29=b%5E2sqrt%283%29%2F2=approximately\"\"0.866b%5E2\" .
\n" ); document.write( "
\n" ); document.write( "If your equilateral triangle inside a square looks like this ,
\n" ); document.write( "then \"b=s=7\" and \"area=s%5E2%2Asin%2860%5Eo%29=%28sqrt%283%29%2F2%29%2As%5E2=approximately\"\"0.866%2As%5E2\" .
\n" ); document.write( "\"area=7%5E2%2Asqrt%283%29%2F2=49sqrt%283%29%2F2=approximately\"\"0.866%2A49=42.434\" .
\n" ); document.write( "However, it may not be the expected answer,
\n" ); document.write( "but you may be able to fit a slightly larger triangle if you \"tilt\" it,
\n" ); document.write( "like this .
\n" ); document.write( "
\n" ); document.write( "CIRCLE:
\n" ); document.write( "The larges circle that you would be able to fit inside a square has a diameter as long as the side of the square:
\n" ); document.write( " If the square side length is \"s\", the circle radius is \"r=s%3F2\"
\n" ); document.write( "The area of a circle of radius \"r%7D%7D+is+%7B%7B%7Bpi%2Ar%5E2\" ,
\n" ); document.write( "so a circle of radius \"s%2F2%7D%7D+would+have+an+area+of%0D%0A%7B%7B%7Barea=pi%2A%28s%2F2%29%5E2=pi%2As%5E2%2F4\" .
\n" ); document.write( "In particular, for \"s=7\"
\n" ); document.write( "the area would be \"area=pi%2A7%5E2%2F4=approximately\"\"38.485\" .
\n" ); document.write( "
\n" ); document.write( "For the triangle and circle described above,
\n" ); document.write( "the ratio of their areas would be
\n" ); document.write( "\"1.103\" .
\n" ); document.write( "(The side of the square \"s=7\" did not matter, as long as we use the same square size to fit the triangle and the circle).
\n" ); document.write( "
\n" ); document.write( "LARGER TRIANGLE:
\n" ); document.write( " In the right triangle corners, according to the Pythagorean theorem,
\n" ); document.write( "\"b%5E2=x%5E2%2Bx%5E2\" and \"b%5E2=%28s-x%29%5E2%2Bs%5E2\" , so
\n" ); document.write( "\"x%5E2%2Bx%5E2=%28s-x%29%5E2%2Bs%5E2\"
\n" ); document.write( "\"x%5E2%2Bx%5E2=s%5E2-2sx%2Bx%5E2%2Bs%5E2\"
\n" ); document.write( "\"x%5E2=-2sx%2B2s%5E2\"
\n" ); document.write( "\"x%5E2%2B2sx=2s%5E2\"
\n" ); document.write( "\"x%5E2%2B2sx%2Bs%5E2=2s%5E2%2Bs%5E2\"
\n" ); document.write( "\"%28x%2Bs%29%5E2=3s%5E2\" , and since \"x%2Bs%3E0\" and \"s%3E0\"
\n" ); document.write( "\"x%2Bs=sqrt%283s%5E2%29\"
\n" ); document.write( "\"x%2Bs=s%2Asqrt%283%29\"
\n" ); document.write( "\"x=s%281-sqrt%283%29%29\" , and \"x%5E2=s%5E2%281-sqrt%283%29%29%5E2\" .
\n" ); document.write( "Now, we knew that for a triangle, \"area=%28sqrt%283%29%2F2%29b%5E2\" ,
\n" ); document.write( "and that this tilted triangle has \"b%5E2=x%5E2%2Bx%5E2=2x%5E2\" ,
\n" ); document.write( "so \"b%5E2=2%281-sqrt%283%29%29%5E2%2As%5E2\" and \"0.928s%5E2\" .
\n" ); document.write( "That makes the area of thte tilted triangle
\n" ); document.write( "approximately \"0.928%2A7%5E2=45.472\" ,
\n" ); document.write( "and the ratio of areas for the tilted triangle and circle would be
\n" ); document.write( " .
\n" ); document.write( "That ratio is approximately \"1.182\" .
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