Let $ABC$ be a right-angled triangle with hypotenuse $BC$ of length $20$ cm. If $AP$ is perpendicular on $BC$, then the maximum possible length of $AP$, in cm, is

We have a right-angled triangle ABC where the right angle is at vertex A, BC is the hypotenuse with length 20 cm, and AP is the altitude (height) dropped from A perpendicular to BC.

Key insight: We need to find when this altitude AP is maximum.

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The triangle ABC has the same area whether we calculate it using the two legs as base and height or the hypotenuse as base and altitude AP as height.

Let's say the two legs are AB = x and AC = y.

Area = $\dfrac{1}{2} \times x \times y$

Area = $\dfrac{1}{2} \times 20 \times AP$

Since both expressions equal the same area:

$\dfrac{1}{2} \times x \times y = \dfrac{1}{2} \times 20 \times AP$

$xy = 20 \times AP$

Therefore: $AP = \dfrac{xy}{20}$

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Now we need to maximize AP, which means maximizing the product xy.

Since ABC is a right triangle with hypotenuse 20:

$x^2 + y^2 = 20^2 = 400$

What values of x and y give us the maximum product xy?

This happens when x = y (when the triangle is isosceles)!

For any two numbers with a fixed sum of squares, their product is maximum when they are equal.

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When x = y, our constraint becomes:

$x^2 + x^2 = 400$

$2x^2 = 400$

$x^2 = 200$

$x = \sqrt{200} = 10\sqrt{2}$

So when x = y = $10\sqrt{2}$:

$AP = \dfrac{xy}{20} = \dfrac{(10\sqrt{2})(10\sqrt{2})}{20} = \dfrac{100 \times 2}{20} = \dfrac{200}{20} = 10$

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The maximum altitude occurs when the right triangle is isosceles (both legs equal). This creates the tallest possible triangle for a given hypotenuse length.

Therefore, the maximum possible length of AP is 10 cm.