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प्रश्न
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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उत्तर
(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.
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