in triangle abc,angle b=90 degree,find the sides of triangle if, ab=x-3 cm,bc=x+4cm and ac=x+6cm
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for right angle triangle
by Pythagoras theorem
ab^2 + bc^2 = ac^2
(x-3)^2 + (x+4)^2 = (x+6)^2
x^2 -6x +9 + x^2 + 8x + 16 = x^2 + 12x + 36
2x^2 +2x + 25 = x^2 + 12x + 36
x^2 -10x -11 = 0
x = (-b+-√(b^2 -4ac))/2a
x = {-(-10) +- √(100+44)}/2
=(10 +- √144)/2
= (10 +- 12)/2
x=(10+12)/2 or x = (10 -12)/2
x= 22/2 or x = -2/2
x =11 or x = -1
since side cannot be negative
so x = 11
there fore sides of triangle are
x-3 = 11-3=8
x+4=11+4=15
x+6=11+6=17
by Pythagoras theorem
ab^2 + bc^2 = ac^2
(x-3)^2 + (x+4)^2 = (x+6)^2
x^2 -6x +9 + x^2 + 8x + 16 = x^2 + 12x + 36
2x^2 +2x + 25 = x^2 + 12x + 36
x^2 -10x -11 = 0
x = (-b+-√(b^2 -4ac))/2a
x = {-(-10) +- √(100+44)}/2
=(10 +- √144)/2
= (10 +- 12)/2
x=(10+12)/2 or x = (10 -12)/2
x= 22/2 or x = -2/2
x =11 or x = -1
since side cannot be negative
so x = 11
there fore sides of triangle are
x-3 = 11-3=8
x+4=11+4=15
x+6=11+6=17
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