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Let the shortest side of the triangle be x m. Then, Hypotenuse = (2x + 6) m and third side = (2x + 6 – 2) m = (2x + 4) m According to Pythagoras theorem, we have (2x + 6) 2 = x 2 + (2x + 4) 2 ⇒ 4x 2 + 24x + 36 = x 2 + 4x 2 + 16x + 16 ⇒ 24x - 16x + 36 - 16 = x 2 ⇒ 8x + 20 = x 2 ⇒ x 2 – 8x – 20 = 0 ⇒ x 2 –10x + 2x - 20 = 0 ⇒ x(x– 10) + 2(x – 10) = 0 ⇒ (x – 10) (x + 2) = 0 ⇒ x – 10 = 0 or x + 2 = 0 ⇒ x = 10 or x = –2 As length of the side cannot be negative, it is rejected. ∴ 2x + 4 = 20 + 4 = 24 and 2x + 6 = 20 + 6 = 26 Thus, the side of the triangle are 10 m, 24 m and 26 m.

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Quadratic Equations

Long Answer Type

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Let the shortest side of the triangle be x m. Then,
Hypotenuse = (2x + 6) m
and third side = (2x + 6 – 2) m
= (2x + 4) m
According to Pythagoras theorem, we have
(2x + 6)² = x² + (2x + 4)²⇒ 4x² + 24x + 36 = x² + 4x² + 16x + 16
⇒ 24x - 16x + 36 - 16 = x²⇒ 8x + 20 = x²⇒ x² – 8x – 20 = 0
⇒ x²–10x + 2x - 20 = 0
⇒ x(x– 10) + 2(x – 10) = 0
⇒ (x – 10) (x + 2) = 0
⇒ x – 10 = 0 or x + 2 = 0
⇒ x = 10 or x = –2
As length of the side cannot be negative, it is rejected.
∴ 2x + 4 = 20 + 4 = 24 and 2x + 6 = 20 + 6 = 26
Thus, the side of the triangle are 10 m, 24 m and 26 m.

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