CAT 2019

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Geometry – Right-Angled Triangles, Altitude on Hypotenuse, AM-GM Inequality
• Formula: For an altitude drawn from the right-angle vertex to the hypotenuse, if the altitude divides the hypotenuse into two segments of lengths (x) and (20 - x), then: $$AP^2 = x(20 - x)$$ The maximum value of (x(20 - x)) is achieved when (x = 20 - x), i.e., (x = 10), giving the maximum (AP = 10).

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2. Calculation

In triangle ABC, the right angle is at A, and BC is the hypotenuse of length 20 cm. AP is the altitude drawn from A perpendicular to BC, meeting BC at P. Let (BP = x), so (PC = 20 - x).

Using the geometric mean relation for the altitude on the hypotenuse (derived from the similarity of triangles BPA and APC):

$$AP^2 = BP \cdot PC = x(20 - x)$$

To maximize (AP), we maximize (f(x) = x(20 - x) = 20x - x^2).

By the AM-GM inequality:

$$x(20 - x) \leq \left(\frac{x + (20 - x)}{2}\right)^2 = \left(\frac{20}{2}\right)^2 = 100$$

Equality holds when (x = 20 - x), i.e., (x = 10).

So the maximum value of (AP^2 = 100), which gives:

$$AP_{\text{max}} = \sqrt{100} = 10 \text{ cm}$$

This occurs when P is the midpoint of BC, meaning the triangle ABC is an isosceles right-angled triangle with (AB = AC).

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3. Solution

Answer = Option A ✅

The maximum possible length of AP is 10 cm.