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 - Mathematics

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Sum

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
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Solution


(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.

  Is there an error in this question or solution?
Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (A) [Page 159]

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Selina Concise Mathematics Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (A) | Q 12 | Page 159

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