Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm.

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

A. 5
B. 5√2
C. 6√2
D. 10
E. 8√2

For BC to be a hypotenuse, triangle ABC is right angled at A. For AP to be maximum triangle ABC has to be an isosceles triangle.
Thus,
Hypotenuse BC = √2*a where a = AB = AC
20 = √2 * a
a = 10√2

Area of triangle = \(\frac{1}{2}\)*AB * AC = \(\frac{1}{2}\) * AP * BC
10√2 * 10√2 = AP * 20
AP = 10

Answer D.
Note: For a better understanding you need to consider that the right angled triangle ABC is inside a circle where BC is diameter of the circle. So, in the half the circle with triangle ABC, the largest perpendicular distance from the circumference to side BC is from center of the circle(here P being the center of the circle).
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AP Max is when we have an isosceles triangle

Thus we have 10/10 for the base and 10*sqrt(2) for the sides of the triangle

We use AB^2+BP^2=AP^2

(10*sqrt(2))^2=10^2+AP^2

AP=10

Answer D

Answer is supposedly D , i.e.10
H^2=P^2+B^2
20^2=P^2+B^2
For determining the maximum lenth of AP perpendicular to hypotenuse.
The sides of the triangle should be maximum as well. That is √20&√20
Now in ∆ABP and ∆APC
AB^2=AP^2+BP^2 = AP^2 = AB^2-BP^2-----(1)
And,
AC^2= AP^2+CP^2 = AP^2 = AC^2-CP^2-----(2)
From equation 1 & 2
AB^2-BP^2 = AC^2-CP^2
20-BP^2 = 20-CP^2
BP=CP=10
Putting values we get AP as 10.
So OA is 10

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Let BC = 20 (hypotenuse) and AP = height of the ABC with base BC.

AP is maximum when AB=AC (with the side angle of 45 degree each). So, AB=AC = 20/√2 = 10√2.

Area of ABC:
0.5*AB*AC = 0.5*AP*BC
AB*AC = AP*BC
(10√2)*(10√2) = AP*20
---> AP = 200/20 = 10

FINAL ANSWER IS (D)

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AP would be the median and the max value would be if the traingle is iscoeles right triangle, where it bisects both the angle at vertex and the hypotenuse, the half of hypotenuse is equal to the median AP.
Hence 10
D

We are given that \(ABC\) is a right-angled triangle with the hypotenuse \(BC=20cm\). We are given that \(AP\) is perpendicular to \(BC\). We are to determine the maximum possible length of \(AP\) in cm.

It is worth noting that a maximum perpendicular to hypotenuse \(BC\) is only possible if triangle \(ABC\) is a right isosceles triangle, implying that angle \(B\)=angle \(C=45\) or the length of side \(AB=\) the length of side \(AC\). This will enable point \(P\) to be the midpoint of hypotenuse \(BC\).
Triangle \(ABP\) is a right isosceles triangle with length \(BP =\) length \(AP\). But since \(P\) is the midpoint of \(BC\), then \(AP=10cm\).

The answer is therefore D.

D) 10 ; Perpendicular from the hypotenuse to the vertex is ½ the length of hypotenuse

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AB/AC =AP/PC
AB*AC=20AP
LET PC = x
20=(AP^2 +x^2)/x
x(20-x)=AP^2
by putting x=10
so,10*10=AP^2
AP=10
D is ANS

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