Total Number of Isosceles Triangles Formed – Special Conditions
June 24, 2024 2025-06-24 15:04Total Number of Isosceles Triangles Formed – Special Conditions

Total Number of Isosceles Triangles Formed – Special Conditions
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This post explores two classic types of questions often asked in competitive exams like CAT, XAT, SSC, etc., involving the number of isosceles triangles possible when:
- The sum of two sides is fixed.
- The product of two sides is fixed.
🔸 Case 1: Sum of Two Sides is Fixed
Question: If the sum of two sides of a triangle is 20, and all sides are natural numbers, then how many isosceles triangles are possible?
- \( a = b \): Then third side \( c = 20 - a \)
- We apply triangle inequality: \( a + b > c \), \( a + c > b \), \( b + c > a \)
The method shown uses pairwise values \( a + b = 20 \) and counts how many values of third side will maintain isosceles triangle condition.
Summary
- For values \( a = 1 \) to \( 6 \): only 1 isosceles triangle each.
- For values \( a = 7 \) to \( 9 \): 2 triangles each.
- For \( a = 10 \): (10, 10, x) → 19 values for x satisfying triangle inequality.
✅ Total Isosceles Triangles = 6 (1s) + 6 (2s) + 19 = 31
🔸 Case 2: Product of Two Sides is Fixed
Question: If the product of two sides is 100, and all sides are natural numbers, then how many isosceles triangles are possible?
We list all factor pairs \( a \times b = 100 \) such that \( a \leq b \) and for each, check triangle feasibility and isosceles nature.
Summary
- Pairs like (1,100), (2,50), (4,25), (5,20) → only 1 isosceles triangle each.
- For the pair (10,10): third side can be anything from 1 to 19 (from triangle inequality \( x < 20 \)).
✅ Total Isosceles Triangles = 4 (1s) + 19 (from 10,10) = 23
💡 Key Takeaways
- Use triangle inequality to filter valid triangles.
- Don’t forget to consider repeated or symmetric cases only once.
- When two sides are equal (isosceles), adjust third side range accordingly.
These techniques are frequently tested in aptitude exams. Mastering them will give you a sharp edge in geometry-based MCQs.























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