IPMAT 2020 - Quantitative Aptitude Question 12
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IPMAT 2020 - Quantitative Aptitude
Perimeter ≤ 100
sides = a – 1 + a + a + 1 ≤ 100
= 3a ≤ 100
a ≤ 100/3
a = 33
Sides = 32,33,34
Smallest possible triangle = 4, 5, 6 because given triangle is & 3, 4, 5 gives a right triangle
a2 + b2 > c2
42 + 52 > b2
162 + 252 > 36
41 > 36
Hence 4, 5, 6 is acute triangle
Total No. of Triangles = 32 – 4 + 1
= 29
Set of possible = {(4, 5, 6), (5, 6, 7), (6, 7, 8) ….(32, 33, 34)}