|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
13 May 2019, 23:30
Kudos
Bookmarks
C
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
Date
Time
Result
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
Kudos
Bookmarks
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
Attachments
prob.png [ 8.16 KiB | Viewed 6056 times ]
30 May 2019, 03:37
Kudos
Bookmarks
Can someone explain me the solution of this question. Not able to get the anser.
