Question

How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?

• A. 16
• B. 17
• C. 24
• D. 2
• E. 23

Hint:

When dealing with triangle sides, always enforce the Triangle Inequality Theorem (\(a+b > c\), \(b+c > a\), \(c+a > b\)). For isosceles triangles, remember to split into two distinct cases: the sum involves the two identical sides, or the sum involves one identical side and the base.

Answer 1

The Correct Option is B

Step 1: Define the properties of the isosceles triangle.
Let the sides of the triangle be \(a, a, b\) (where \(a\) is the equal side).
The triangle inequality theorem states: \[ 2a > b \]

Step 2: Case 1 - The sum of the two equal sides is 12.
\(a + a = 12 \Rightarrow a = 6\)
The sides are \(6, 6, b\).
By triangle inequality: \[ 6 + 6 > b \Rightarrow b < 12 \]
So \(b = 1, 2, 3, \dots, 11\) → 11 triangles.

Step 3: Case 2 - The sum of an equal side and the unequal side is 12.
\(a + b = 12 \Rightarrow b = 12 - a\)
By triangle inequality: \[ 2a > 12 - a \Rightarrow 3a > 12 \Rightarrow a > 4 \]
Also: \[ 12 - a > 0 \Rightarrow a < 12 \]
So \(a = 5, 6, 7, 8, 9, 10, 11\).
Exclude \(a = 6\) (equilateral case).
Valid values: \(5, 7, 8, 9, 10, 11\) → 6 triangles.

Step 4: Total number of triangles:
\(11 + 6 = 17\)