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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
(2019)
- a)8√2
- b)6√2
- c)5
- d)10
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm....
In ΔABC, let AC = a and AB = b
Then by pythagoras theorem for ΔABC,
a2 + b2 = 202 = 400 ...(i)
In ΔABC and ΔPAC,
∠BAC = ∠APC (= 90°)
∠C = ∠C (common)
∴ ΔABC ~ ΔPAC (by AA)
[Ratio of corresponding sides of two similar triangles are equal]
...(ii)For maximum value of AP, we have to maximize the product ab.
Applying AM ≥ GM inequality, we get
⇒ ab ≤ 200
Hence, the maximum value of ab = 200
From eq. (ii), it is clear that value of AP will be maximum when value of ab will be maximum
∴ maximum value of AP = 200 / 20 = 10
Most Upvoted Answer
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm....
In ΔABC, let AC = a and AB = b
Then by pythagoras theorem for ΔABC,
a2 + b2 = 202 = 400 ...(i)
In ΔABC and ΔPAC,
∠BAC = ∠APC (= 90°)
∠C = ∠C (common)
∴ ΔABC ~ ΔPAC (by AA)
[Ratio of corresponding sides of two similar triangles are equal]
...(ii)For maximum value of AP, we have to maximize the product ab.
Applying AM ≥ GM inequality, we get
⇒ ab ≤ 200
Hence, the maximum value of ab = 200
From eq. (ii), it is clear that value of AP will be maximum when value of ab will be maximum
∴ maximum value of AP = 200 / 20 = 10
Community Answer
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm....
We can use the Pythagorean theorem to find the relationship between the lengths of the sides of the right-angled triangle:
AB^2 + BC^2 = AC^2
Since AC is the hypotenuse and BC is given to be 20 cm, we have:
AB^2 + 20^2 = AC^2
Simplifying, we get:
AB^2 + 400 = AC^2
Since AP is perpendicular to BC, we can consider the right-angled triangle APB. Using the Pythagorean theorem again, we have:
AP^2 + AB^2 = BP^2
Since the maximum possible length of AP is asked for, we want to maximize AP^2. To do this, we want to minimize BP^2. The smallest possible value for BP^2 is 0 (when BP = 0), so we can write:
AP^2 + AB^2 ≥ 0
AP^2 ≥ -AB^2
AP^2 ≥ -400
AP ≥ √(-400)
Since we're dealing with real numbers, the square root of a negative number is not defined. Therefore, there is no maximum possible length for AP. The answer is (2019) d) No maximum length.
AB^2 + BC^2 = AC^2
Since AC is the hypotenuse and BC is given to be 20 cm, we have:
AB^2 + 20^2 = AC^2
Simplifying, we get:
AB^2 + 400 = AC^2
Since AP is perpendicular to BC, we can consider the right-angled triangle APB. Using the Pythagorean theorem again, we have:
AP^2 + AB^2 = BP^2
Since the maximum possible length of AP is asked for, we want to maximize AP^2. To do this, we want to minimize BP^2. The smallest possible value for BP^2 is 0 (when BP = 0), so we can write:
AP^2 + AB^2 ≥ 0
AP^2 ≥ -AB^2
AP^2 ≥ -400
AP ≥ √(-400)
Since we're dealing with real numbers, the square root of a negative number is not defined. Therefore, there is no maximum possible length for AP. The answer is (2019) d) No maximum length.
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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(2019)a)8√2b)6√2c)5d)10Correct answer is option 'D'. Can you explain this answer?
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