document.write( "Question 1044784: I need help i don't know how to solve Problem Solving like this :(
\n" ); document.write( "The base of an isosceles triangle is 8 m more than its altitude. If its area is 192 m^2, find the length of its sides.
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Algebra.Com's Answer #660169 by Cromlix(4381)\"\" \"About 
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Hi there,
\n" ); document.write( "Area of a triangle is = 1/2(base x height)
\n" ); document.write( "Make height (altitude) = x
\n" ); document.write( "Base = x + 8
\n" ); document.write( "Area = 1/2(base x altitude)
\n" ); document.write( "192 = 1/2(x(x + 8))
\n" ); document.write( "192 = 1/2(x^2 + 8x)
\n" ); document.write( "Multiply both sides by 2
\n" ); document.write( "384 = x^2 + 8x
\n" ); document.write( "Rearrange into ax^2 + bx + c form
\n" ); document.write( "x^2 + 8x - 384 = 0
\n" ); document.write( "Factorise
\n" ); document.write( "(x - 16)(x + 24) = 0
\n" ); document.write( "x + 24 = x = -24 (discount as -ve)
\n" ); document.write( "x - 16 = 0
\n" ); document.write( "x = 16
\n" ); document.write( "Altitude = 16 m
\n" ); document.write( "Base = 24 m
\n" ); document.write( "........
\n" ); document.write( "Now we can work out the size of the two
\n" ); document.write( "equal sides by dividing the base of 24 m
\n" ); document.write( "into 2 lots of 12 m.
\n" ); document.write( "We now have a right angled triangle
\n" ); document.write( "with base of 12 m and an upright of
\n" ); document.write( "16 m.
\n" ); document.write( "To find the hypotenuse we use Pythagoras' Theorem.
\n" ); document.write( "Namely:- Hypotenuse^2 = 12^2 + 16^2
\n" ); document.write( "Hypotenuse = √144 + 256
\n" ); document.write( "Hypotenuse = √400
\n" ); document.write( "Hypotenuse = 20 m
\n" ); document.write( ".............
\n" ); document.write( "Two equal sides = 20 m
\n" ); document.write( "Base = 24 m
\n" ); document.write( "Hope this helps :-)
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