Question

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

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

Answer 1

(i) In right-angled ΔABC,

AC² = AB² + BC²

⇒ ( x + 6 )² = ( x - 3 )² + ( x + 4 )²

⇒ ( x² + 12x + 36 ) = ( x² - 6x + 9 ) + ( x² + 8x + 16 )

⇒ x² - 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,

AC² = AB² + BC²

⇒ ( 4x + 5 )² = ( x )² + ( 4x + 4 )²

⇒ ( 16x² + 40x + 25 ) = ( x² ) + ( 16x² + 32x + 16 )

⇒ x² - 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.