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**In ΔABC, Find the sides of the triangle, if:**

- AB = ( x - 3 ) cm, BC = ( x + 4 ) cm and AC = ( x + 6 ) cm
- AB = x cm, BC = ( 4x + 4 ) cm and AC = ( 4x + 5) cm

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#### Solution

(i) In right-angled ΔABC,

AC^{2} = AB^{2} + BC^{2}

⇒ ( x + 6 )^{2} = ( x - 3 )^{2} + ( x + 4 )^{2}

⇒ ( x^{2} + 12x + 36 ) = ( x^{2} - 6x + 9 ) + ( x^{2} + 8x + 16 )

⇒ x^{2} - 10x - 11 = 0

⇒ ( x - 11 )( x + 1 ) = 0

⇒ x = 11 or x = - 1

But length of the side of a triangle can not be negative.

⇒ x = 11 cm

∴ AB = ( x - 3 ) = ( 11 - 3 ) = 8 cm

BC = ( x + 4 ) = ( 11 + 4 ) = 15 cm

AC = ( x + 6 ) = ( 11 + 6 ) = 17 cm.

(ii) In right-angled ΔABC,

AC^{2} = AB^{2} + BC^{2}

⇒ ( 4x + 5 )^{2} = ( x )^{2} + ( 4x + 4 )^{2}

⇒ ( 16x^{2} + 40x + 25 ) = ( x^{2} ) + ( 16x^{2} + 32x + 16 )

⇒ x^{2} - 8x - 9 = 0

⇒ ( x - 9 )( x + 1 ) = 0

⇒ x = 9 or x = - 1

But length of the side of a triangle can not be negative.

⇒ x = 9 cm

∴ AB = x = 9 cm

BC = ( 4x + 4 ) = ( 36 + 4 ) = 40 cm

AC = ( 4x + 5 ) = ( 36 + 5 ) = 41 cm.

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