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

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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

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unraveled · Accepted Answer

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).

madzaka · Answer

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

yashikaaggarwal · Answer

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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freedom128 · Answer

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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monikakumar · Answer

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

eakabuah · Answer

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.

y605ash · Answer

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

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ndawar · Answer

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

bumpbot · Answer

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

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

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