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

In a right-angled triangle ABC, the vertex A lies on a semicircle with BC as the diameter.

The perpendicular distance AP from vertex A to the hypotenuse BC represents the height of the triangle ABC when BC is considered the base.

The height AP is maximized when vertex A is at the highest point of the semicircle above diameter BC. This occurs when triangle ABC is an isosceles right-angled triangle.

In this case, the perpendicular AP is equal to the radius of the semicircle:
$AP=\frac{BC}{2}=\frac{20}{2}=10\text{cm}$

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