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

Question

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

gmatphoru · Answer

Answer after the typo correction will be A.

sowmiyav · Answer

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

nitesh50 · Answer

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

mishugarg · Answer

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

GMATBusters · Answer

Official Solution https://gmatclub.com/forum/download/file.php?id=45041 Sol1.jpg

Albymana · Answer

hi guys. 
why can we simply say that BD should be (5, 10) ?

thank you very much

bumpbot · Answer

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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

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