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10 Nov 2018, 10:00
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58% (03:10) correct
42%
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Triangle ABC is a right triangle with \(\angle\)ACB as its right angle, \(\angle\) ABC = 60deg , and AB = 10 cm. Let P be randomly chosen inside triangle ABC, and extend BP to meet AC at D. What is the probability that BD >\(5\sqrt{2}\)?
A) (2-\(\sqrt{2}\))/2
B) 1/3
C) (3-\(\sqrt{3}\))/3
D) 1/2
E) (5-\(\sqrt{5}\))/5
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10 Nov 2018, 10:53
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AB=10cm (given)
30:60:90=1:sqrt(3):2
so, BC=5 cm
AC= 5sqrt(3) cms
Maximum possible value for CD= 5sqrt(3)
For BD>5 sqrt(2)
CD>5
Possible range of values for CD= 5 sqrt(3)-5
Total probability = possible range of values/ maximum possible value
=(5sqrt(3)-5)/(5 sqrt 3)
=1-(1/sqrt(3))
so, option C
30:60:90=1:sqrt(3):2
so, BC=5 cm
AC= 5sqrt(3) cms
Maximum possible value for CD= 5sqrt(3)
For BD>5 sqrt(2)
CD>5
Possible range of values for CD= 5 sqrt(3)-5
Total probability = possible range of values/ maximum possible value
=(5sqrt(3)-5)/(5 sqrt 3)
=1-(1/sqrt(3))
so, option C
