document.write( "Question 575111: If the area of an isosceles triangle is x sq. cm and each of the congruent sides measures y cm, find the measure of its base? \n" ); document.write( "
Algebra.Com's Answer #369436 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
It would be a much easier problem if it was an isosceles right triangle.
\n" ); document.write( "If you inadvertently left the word \"right\" out of your posting, I forgive you, but you will habv e to figure out the easier answer.
\n" ); document.write( "Other wise, read on.
\n" ); document.write( "I saw two \"simple\" ways to get to the answer. Either one is a nightmare for error prone people like me. (And I don't even want to think of how to work law of sines and/or law of cosines into a solution).
\n" ); document.write( "You could use Heron's formula relating the area of a triangle to the lengths of the sides.
\n" ); document.write( "You could also use the formula to calculate the area of a triangle having the lengths of two sides and the measure of the included angle.
\n" ); document.write( "I like the second option, because I can better visualize the meaning of my calculations.
\n" ); document.write( "If the length of the base is B and the measure of the vertex angle is B,
\n" ); document.write( "\"Area=+%281%2F2%29y%2Ay%2AsinB=%281%2F2%29y%5E2sinB\"
\n" ); document.write( "So \"x=%281%2F2%29y%5E2sinB\" and \"sinB=2x%2Fy%5E2\"
\n" ); document.write( "Then, either or \"cosB=-%281%2Fy%5E2%29sqrt%28y%5E4-4x%5E2%29\"
\n" ); document.write( "because usually there are two possible triangles for the same area and leg length.
\n" ); document.write( "The maximum area for isosceles triangles of leg length y will happen when sin(B)=1, meaning that it is an isosceles right triangle, and \"Area=y%5E2\".
\n" ); document.write( "Unless \"x=y%5E2\", there will be two possible triangles, an acute one and an obtuse one, that will have the same sin(B) for vertex angle B.
\n" ); document.write( "Because the altitude to the vertex splits the isosceles triangles into two right triangles with hypotenuse y, and a leg of length b/2 opposite an angle of measure B/2,
\n" ); document.write( "\"b%2F2=y%2Asin%28B%2F2%29\" and \"b=2ysin%28B%2F2%29\"
\n" ); document.write( "At this point, I consult a table of trigonometric identities and find that
\n" ); document.write( "\"sin%28B%2F2%29=sqrt%28%281-cosB%29%2F2%29\" (For a general angle B, there could be a minus sign, but not in this case).
\n" ); document.write( "So, substituting into the expression for b,
\n" ); document.write( "\"b=2y%2Asqrt%28%281-cosB%29%2F2%29\"
\n" ); document.write( "Next we substitute the expression for cosB found before:
\n" ); document.write( "\"b=2y%2Asqrt%28%281+%2B-+%281%2Fy%5E2%29sqrt%28y%5E4-4x%5E2%29%29%2F2%29\" = \"sqrt%284y%5E2%29%2Asqrt%28%281+%2B-+%281%2Fy%5E2%29sqrt%28y%5E4-4x%5E2%29%29%2F2%29\" = \"sqrt%284y%5E2%281+%2B-+%281%2Fy%5E2%29sqrt%28y%5E4-4x%5E2%29%29%2F2%29\" = \"sqrt%282y%5E2%281+%2B-+%281%2Fy%5E2%29sqrt%28y%5E4-4x%5E2%29%29%29\" = \"sqrt%282y%5E2+%2B-+2sqrt%28y%5E4-4x%5E2%29%29\"
\n" ); document.write( "So \"highlight%28b=sqrt%282y%5E2+%2B-+2sqrt%28y%5E4-4x%5E2%29%29%29\" \n" ); document.write( "
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