The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is
Explanation:
Let the sides of the triangle be x, x + 1 and x + 2.
Now, x + x + 1 + x + 2 ≤ 100 ⇒ 3x ≤ 97 ⇒ x ≤ 32.33 ...(1)
Also, the given triangle is an acute triangle, hence (x + 2)2 < (x + 1)2 + x2 ⇒ x2 + 4 + 4x < x2 + 2x + 1 + x2 ⇒ x2 - 2x - 3 > 0 ⇒ (x - 3)(x + 1) > 0 ⇒ x < -1 or x > 3 ...(2)
From (1) and (2), we get x can take any value out of 4, 5, 6, ..., 31 and 32 i.e., 29 values.
Hence, option (b).
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