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Question: 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?
- 5
- 5√2
- 6√2
- 10
- 8√2
Answer
Approach Solution (1)
Let p be the length of AP
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
p^2 = x (20 - x)
We have to maximize the value of p, which will be maximum when x = 20 - x
x = 10
Correct option: D
Approach Solution (2)
The length of AP would be maximum, when AB = AC (ABC is an Isosceles Right Triangle)
So, AB = AC = 10√2 cms
Applying Pythagoras theorem to APB, we get
AB^2 = AP^2 + PB^2
(10√2)^2 = 10^2 + x^2
x^2 = 100
x = 10 cms
Correct option: D
Approach Solution (3)
20^2 = P^2+B^2
For determining the maximum length 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
Correct option: D
“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?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
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