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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 the maximum possible length of AP, in cm, is
  • a)
    10
  • b)
    5
  • c)
    8√2
  • d)
    6√2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm....

Let p be the length of AP.
It is given that 
∠BAC = 90 and ∠APC = 90
Let ∠ABC = θ, then ∠BAP = 90 - θ and ∠BCA = 90 - θ
So ∠PAC = θ
Triangles BPA and APC are similar
p2 = x(20 - x)
We have to maximize the value of p, which will be maximum when x = 20 - x
x = 10
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Most Upvoted Answer
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm....
Let's call the length of AP as x.
Since triangle ABC is right-angled, we can use the Pythagorean Theorem to find the length of AB.
AB^2 = BC^2 - AC^2
AB^2 = 20^2 - x^2
AB^2 = 400 - x^2
Now, we want to find the maximum possible length of AP, so we need to find the maximum possible length of AB. To do this, we need to minimize the value of x^2. Since x is a positive value, the minimum possible value of x^2 is 0 (when x = 0). Therefore, the maximum possible length of AP is when AB is at its maximum, which occurs when x = 0.
So, the maximum possible length of AP is 0 cm.
Therefore, the correct option is:
a) 10
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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 the maximum possible length of AP, in cm, isa)10b)5c)8√2d)6√2Correct answer is option 'A'. Can you explain this answer?
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