In triangle ABC, BE and CF are two altitudes drawn from vertices B and C to the opposite sides. If AB = 6 cm, AC = 5 cm and altitude CF = 4 cm, then what is the length of altitude BE (in cm)?

Introduction / Context:
This problem uses properties of altitudes in a triangle and the fact that the area of a triangle can be expressed in multiple ways depending on which side is chosen as the base. By equating these different expressions for the same area, you can relate different altitudes to each other. Such questions check your understanding of basic triangle area formulas and your ability to connect them in a simple algebraic way, which is very common in quantitative aptitude and geometry sections of competitive exams.

Given Data / Assumptions:
• Triangle ABC is a non degenerate triangle.
• BE is an altitude from vertex B to side AC, so BE is perpendicular to AC.
• CF is an altitude from vertex C to side AB, so CF is perpendicular to AB.
• Side AB = 6 cm.
• Side AC = 5 cm.
• Altitude CF = 4 cm.
• We are required to find the length of altitude BE in centimetres.

Concept / Approach:
The area of triangle ABC can be written using any side as the base with the corresponding altitude. If we choose AB as the base, the area is (1 / 2) * AB * CF, because CF is perpendicular to AB. If we choose AC as the base, the area is (1 / 2) * AC * BE, because BE is perpendicular to AC. Both expressions represent the same geometric area, so they must be equal. Equating these two formulas allows us to solve directly for BE. This technique is a straightforward but powerful application of the area formula for triangles.

Step-by-Step Solution:
Step 1: Express the area of triangle ABC using base AB and altitude CF: Area = (1 / 2) * AB * CF. Step 2: Substitute AB = 6 cm and CF = 4 cm to get Area = (1 / 2) * 6 * 4. Step 3: Compute this value: Area = (1 / 2) * 24 = 12 square centimetres. Step 4: Now express the same area using base AC and altitude BE: Area = (1 / 2) * AC * BE. Step 5: Substitute AC = 5 cm and Area = 12 square centimetres into this expression: 12 = (1 / 2) * 5 * BE. Step 6: Solve for BE: (1 / 2) * 5 * BE = 12 implies 5 * BE = 24, so BE = 24 / 5 = 4.8 cm.

Verification / Alternative check:
You can quickly check the consistency by recomputing the area with the found value of BE. Using base AC = 5 cm and altitude BE = 4.8 cm, Area = (1 / 2) * 5 * 4.8 = (1 / 2) * 24 = 12 square centimetres, which matches the area calculated earlier using base AB and altitude CF. This confirms that BE = 4.8 cm is consistent and correct. The dimensions are also reasonable since altitudes in triangles with moderate side lengths often fall in similar numeric ranges.

Why Other Options Are Wrong:
Option 7.5 cm would produce an area of (1 / 2) * 5 * 7.5 = 18.75 square centimetres, which does not match 12 square centimetres. Option 3.33 cm is an approximate value and would give area about (1 / 2) * 5 * 3.33 which is close to 8.325 square centimetres, again not equal to 12. Option 5.5 cm leads to area (1 / 2) * 5 * 5.5 = 13.75 square centimetres, which is also inconsistent. Hence, only 4.8 cm preserves the equality of the two area expressions.

Common Pitfalls:
Learners sometimes confuse which altitude corresponds to which base, or they may mistakenly try to use Pythagoras theorem without having enough information. Another common error is forgetting to multiply by 1 / 2 or losing track of units. It is important to remember that for any triangle, the area formula (1 / 2) * base * height works for any side chosen as base, as long as the corresponding altitude is used. Equating two such expressions is a simple and reliable method for problems involving multiple altitudes.

Final Answer:
The length of altitude BE is 4.8 cm.