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Triangle ABC is a right triangle with angle ACB as its right an

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Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 10 Nov 2018, 10:00
2
1
7
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

54% (02:58) correct 46% (03:03) wrong based on 35 sessions

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GMATbuster's Weekly Quant Quiz#8 Ques #1


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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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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 10 Nov 2018, 10:41
Answer after the typo correction will be A.
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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 10 Nov 2018, 10:53
1
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
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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 10 Nov 2018, 11:05
Triangle ABC is a right triangle with ∠∠ACB as its right 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 >52‾√52?


So in order to calculate the probability:

We need to find
Area of triangle APD(when BD=5root2)/Area of ABC

Area of ABC= USe 30 60 90:
Base:5root3
Height 5
Area: 12.5roo3

Area of APD:
Total area of abc- Area of BCD
BCD=USing pT theorem CD=5

Area=12.5

So probablilty:


1- Area BCD/AREa of ABC

1-12.5/12.5root3

Root 3-1/root3


Which after normalization is

3-root3/3
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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 10 Nov 2018, 11:36
BC=5
AC=5 root3
find pp to get dd :: bdd=5root2
cdd=5

now to get bd> 5root2, cd>5 is necessary

so p mst b in abdd

ac-cdd/ac=1-(cdd/ac)= 1-(5/5root3)=1-(root3/3)
=ans c
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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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New post 11 Nov 2018, 08:50
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Re: Triangle ABC is a right triangle with angle ACB as its right an  [#permalink]

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Re: Triangle ABC is a right triangle with angle ACB as its right an   [#permalink] 18 Nov 2019, 05:03
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