AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm .Find AB.

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Solution

Given : A triangle right angled at B.
$A B = x - 3, B C = x + 4, A C = x + 6$
Thus, $A C^{2} = A B^{2} + B C^{2}$ (Pythagoras Theorem)
$(x + 6)^{2} = (x - 3)^{2} + (x + 4)^{2}$
$x^{2} + 12 x + 36 = x^{2} - 6 x + 9 + x^{2} + 8 x + 16$
$x^{2} - 10 x - 11 = 0$
$x^{2} - 11 x + x - 11 = 0$
$x (x - 11) + 1 (x - 11) = 0$
$(x + 1) (x - 11) = 0$
$x = 11, - 1$
Neglecting the negative value, $x = 11$
thus, $A B = 8$ cm, $A C = 17$ cm and $B C = 15$ cm

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AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm .Find AB.

Given : A triangle right angled at B.AB=x−3,BC=x+4,AC=x+6Thus, AC2=AB2+BC2 (Pythagoras Theorem)(x+6)2=(x−3)2+(x+4)2x2+12x+36=x2−6x+9+x2+8x+16x2−10x−11=0x2−11x+x−11=0x(x−11)+1(x−11)=0(x+1)(x−11)=0x=11,−1 Neglecting the negative value, x=11thus, AB=8 cm, AC=17 cm and BC=15 cm