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13 May 2019, 23:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
30% (01:58) correct
70%
(02:13)
wrong
based on 64
sessions
History
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Not Attempted Yet
Numbers x and y are chosen independently and uniformly at random from the interval [0, 1]. Which of the following numbers is closest to the probability that x, y, and 1 are the side lengths of an obtuse triangle?
(A) 0.21
(B) 0.25
(C) 0.29
(D) 0.50
(E) 0.79
(A) 0.21
(B) 0.25
(C) 0.29
(D) 0.50
(E) 0.79
24 May 2019, 18:33
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We need the obtuse triangle whose sides are x,y and 1
As x and y lies in the interval of [0,1], the longest side of triangle must be 1
x^2 + y^2 < 1 {inequality holds for obtuse triangles)
Also we know that sum of two sides of triangle must be greater than the third side
x+y>1
We can use geometric probability to get our desired results.
Desired area is the area lie between two curves x^2 + y^2 < 1(circle of radius 1) and x+y>1(straight line)
Total area= square of side length 1
As we can see in the figure, Desired area(red portion)= {1/4(pi*1^2)}-(1/2*1*1)=pi/4-1/2
Total area= 1*1=1
The probability that x, y, and 1 are the side lengths of an obtuse triangle= (pi/4-1/2)/1=(pi/4-1/2)
As x and y lies in the interval of [0,1], the longest side of triangle must be 1
x^2 + y^2 < 1 {inequality holds for obtuse triangles)
Also we know that sum of two sides of triangle must be greater than the third side
x+y>1
We can use geometric probability to get our desired results.
Desired area is the area lie between two curves x^2 + y^2 < 1(circle of radius 1) and x+y>1(straight line)
Total area= square of side length 1
As we can see in the figure, Desired area(red portion)= {1/4(pi*1^2)}-(1/2*1*1)=pi/4-1/2
Total area= 1*1=1
The probability that x, y, and 1 are the side lengths of an obtuse triangle= (pi/4-1/2)/1=(pi/4-1/2)
Bunuel
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prob.png [ 8.16 KiB | Viewed 6011 times ]
30 May 2019, 03:37
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Can someone explain me the solution of this question. Not able to get the anser.
