AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm. Find AC.

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Solution

Given a triangle, right angled at B.
Thus, $A C^{2} = A B^{2} + B C^{2}$ (Pythagoras theorem)
$(4 x + 5)^{2} = (4 x + 4)^{2} + x^{2}$
$16 x^{2} + 25 + 40 x = 16 x^{2} + 16 + 32 x + x^{2}$
$x^{2} - 8 x - 9 = 0$
$x^{2} - 9 x + x - 9 = 0$
$(x - 9) (x + 1) = 0$
$x = - 1, 9$
Neglect the negative value, $x = 9$
Thus, $A C = 41$ cm, $A B = 9$ cm and $B C = 40$ cm.

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AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm. Find AC.

Given a triangle, right angled at B.Thus, AC2=AB2+BC2 (Pythagoras theorem)(4x+5)2=(4x+4)2+x216x2+25+40x=16x2+16+32x+x2x2−8x−9=0x2−9x+x−9=0(x−9)(x+1)=0x=−1,9Neglect the negative value, x=9Thus, AC=41 cm, AB=9 cm and BC=40 cm.