in a right angled triangle if perpendicular is( x-5) and base is 1 and hypotenuse is( x+1-3+x-2). Find x.
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using Pythagoras theorem, we have
(2x-4)²=(x-5)²+1²,
4x²+16-16x=x²+25-10x+1,
then
4x²-x²-16x+10x+16-25-1=0,
3x²-6x-10=0,
therefore
x=[6+-√{(-6)²-4×3×-10}]/2×3,
x=[6+-√(36+120)]/6,
x=[6+-√156]/6
x=[6+-2√39]/6,
(2x-4)²=(x-5)²+1²,
4x²+16-16x=x²+25-10x+1,
then
4x²-x²-16x+10x+16-25-1=0,
3x²-6x-10=0,
therefore
x=[6+-√{(-6)²-4×3×-10}]/2×3,
x=[6+-√(36+120)]/6,
x=[6+-√156]/6
x=[6+-2√39]/6,
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