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If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is (SSC CHSL 2014)
- a)3 units
- b)2 units
- c)4 units
- d)5 units
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If the sides of a right angled triangle are three consecutive integers...
Let the three sides are x, (x + 1) and (x + 2)
In a right angle triangle, sum of square of base and
height is equal to the square of hypotenuse
∴ (x + 2)2 = x2 + (x + 1)2
4x + 4 = x2 + 2x + 1
x2 – 2x – 3 = 0
(x – 3) (x + 1) = 0 ⇒ x = 3
Thus, three sides are x = 3 unit
x + 1 = 3 + 1 = 4 unit
x + 2 = 3 + 2 = 5 unit
Smallest side = x = 3 unit.
In a right angle triangle, sum of square of base and
height is equal to the square of hypotenuse
∴ (x + 2)2 = x2 + (x + 1)2
4x + 4 = x2 + 2x + 1
x2 – 2x – 3 = 0
(x – 3) (x + 1) = 0 ⇒ x = 3
Thus, three sides are x = 3 unit
x + 1 = 3 + 1 = 4 unit
x + 2 = 3 + 2 = 5 unit
Smallest side = x = 3 unit.
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Community Answer
If the sides of a right angled triangle are three consecutive integers...
Let the three sides are x, (x + 1) and (x + 2)
In a right angle triangle, sum of square of base and
height is equal to the square of hypotenuse
∴ (x + 2)2 = x2 + (x + 1)2
4x + 4 = x2 + 2x + 1
x2 – 2x – 3 = 0
(x – 3) (x + 1) = 0 ⇒ x = 3
Thus, three sides are x = 3 unit
x + 1 = 3 + 1 = 4 unit
x + 2 = 3 + 2 = 5 unit
Smallest side = x = 3 unit.
In a right angle triangle, sum of square of base and
height is equal to the square of hypotenuse
∴ (x + 2)2 = x2 + (x + 1)2
4x + 4 = x2 + 2x + 1
x2 – 2x – 3 = 0
(x – 3) (x + 1) = 0 ⇒ x = 3
Thus, three sides are x = 3 unit
x + 1 = 3 + 1 = 4 unit
x + 2 = 3 + 2 = 5 unit
Smallest side = x = 3 unit.
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If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is (SSC CHSL 2014)a)3 unitsb)2 unitsc)4 unitsd)5 unitsCorrect answer is option 'A'. Can you explain this answer?
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