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06 May 2021, 02:45
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (01:40) correct
21%
(02:05)
wrong
based on 29
sessions
History
Date
Time
Result
Not Attempted Yet
In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle?
A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3
Attachment:
1.png [ 5.86 KiB | Viewed 1041 times ]
06 May 2021, 22:23
Kudos
Bookmarks
Bunuel
Let length of large triangle = l, => length of smaller triangles = \(\frac{l}{3}\)
Probability of hitting shaded region = Area of shaded region/Area of board
Area of shaded region = 3 * Area of smaller triangle = \(3 * \frac{\sqrt{3}}{4} * (\frac{l}{3})^2 = 3 * \frac{\sqrt{3}}{4} * \frac{l^2}{9} = \frac{\sqrt{3}}{4} * \frac{l^2}{3}\)
Area of board = \(\frac{\sqrt{3}}{4} * l^2\)
Probability =\(\frac{l^2}{3}/l^2 = \frac{1}{3}\) [\(\frac{\sqrt{3}}{4}\) got cancelled in both numerator and denominator]
So, correct answer is option C.
