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Question 20 : 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

- 10
- 8\\sqrt{2})
- 6\\sqrt{2})
- 5

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Now, the length of AP would be maximum, when AB = AC (ABC is an Isosceles Right Triangle)

So, AB = AC = 10cms

Applying Pythagoras theorem to APB, we get

AB^{2} = AP^{2} + PB^{2}

(10^{2} = 10^{2} + x^{2}

x^{2} = 100

x = 10 cms

The question is **"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" **

**Choice A** is the correct answer.

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