In triangle PQR, the altitudes from vertices P, Q and R meet the opposite sides at points X, Y and Z respectively, and all three altitudes intersect at point O. If PO = 6 cm, PX = 8 cm and QO = 4 cm, then what is the length, in centimetres, of altitude QY?

Introduction / Context:
This is a geometry question based on properties of altitudes and the orthocenter of a triangle. The orthocenter is the common point of intersection of the three altitudes. The problem uses a less known but powerful relation between segments into which the orthocenter divides the altitudes. Understanding such relations is important for advanced geometry problems that appear in higher level aptitude and Olympiad style exams.

Given Data / Assumptions:

• Triangle PQR has altitudes PX, QY and RZ drawn from vertices P, Q and R respectively.

• These altitudes meet at a common point O, which is the orthocenter.

• Along altitude PX, PO = 6 cm and full altitude PX = 8 cm.

• Along altitude QY, QO = 4 cm and we need to find full altitude QY.

• All altitudes are inside the triangle, so we consider an acute triangle.

Concept / Approach:
In any acute triangle, the orthocenter divides each altitude into two segments. If we denote one altitude with vertex segment length v and foot segment length f, there is a special relation: the product v * f is the same for all three altitudes. That is, PO * OX = QO * OY = RO * OZ, where X, Y and Z are the feet of the altitudes. We are given PO and full PX, so we can find OX, compute the product PO * OX, and then use this product to find OY and hence QY.

Step-by-Step Solution:
Given full altitude PX = 8 cm and PO = 6 cm. Therefore, OX = PX − PO = 8 − 6 = 2 cm. Compute product along altitude from P: PO * OX = 6 * 2 = 12. For altitude from Q, we know QO = 4 cm and need OY and QY. Use property: QO * OY = PO * OX (same product). So, 4 * OY = 12. OY = 12 / 4 = 3 cm. Full altitude QY = QO + OY = 4 + 3 = 7 cm.

Verification / Alternative check:
To verify the relation, one can construct a specific acute triangle and compute the altitudes and orthocenter coordinates using analytic geometry. For such a triangle, it can be checked that the product of the distance from the vertex to the orthocenter and the distance from the orthocenter to the foot of the altitude remains constant for each altitude. This geometric fact confirms that our use of PO * OX = QO * OY is valid. Substituting the numerical values again yields QY = 7 cm, which matches the earlier result.

Why Other Options Are Wrong:
The value 6 cm would correspond to OY = 2 cm and product 4 * 2 = 8, which does not match the product 12 from the first altitude. The values 5.8 cm and 6.3 cm are approximate values that might arise from guessing or incorrect algebra, but they do not satisfy the exact product relation. Only 7 cm yields OY = 3 cm and keeps the product equal to 12, satisfying the geometry property.

Common Pitfalls:
Many students are not familiar with this product property of altitudes and orthocenter and may try incorrect approaches, such as assuming the orthocenter divides each altitude in the same fixed ratio. Others may attempt to apply area formulas without enough information about side lengths or angles, leading to dead ends. It is important to know special theorems about the orthocenter in order to handle such questions efficiently.

Final Answer:
The length of altitude QY is 7 cm.