Question

Let S be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in S with perimeter less than 30 is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is C

Let the side of right angled triangles are 
\(x, x + 1, x + 2\)
\(\therefore (x+2)^2 = (x+1 )^2 +(x)^2\)
\(⇒ x^2 + 4x + 4 = x^2 +2x +1 + x^2\)
\(⇒ x ^2 − 2x− 3 = 0\)
\(⇒ (x-3 )(x+ 1 )= 0\)
\(⇒ x = 3, x ≠ − 1\)
Perimeter of triangle \(= 3 x + 3\)
\(3x + 3 < 30\)
\(x + 1 < 10\)
\(x < 9\)
The possible triangle whose sides are 
\((3, 4, 5), (6, 8, 10)\)
\(∴\) Two triangles are possible
Therefore, the Correct Option is (C) 2

Answer 2

The correct option is (C): 2

Let the sides of the \(\triangle\) required be \(a, a+d, a+2d\) such that \(d>0\)

\(⇒3a+3d<30\)   ⋯(1)  &  \(a^2+(a+d)^2=(a+2d)^2\)

\(⇒a^2−2ad−3d^2=0\)

\(⇒a^2−3ad+ad−3d^2=0\)

\(⇒a(a−3d)+d(a−3d)=0\)

\(⇒a=3d \text{ }\text{ }\text{ }\text{ }12d<30 \text{ }\text{ }\text{ }(using (1))\)

\(⇒d=1,2\)

\(⇒a=3,6\)

Therefore, 2 triangles are possible.