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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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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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Official Solution



Attachment:
Sol1.jpg
Sol1.jpg [ 74.99 KiB | Viewed 5744 times ]
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hi guys.
why can we simply say that BD should be (5, 10) ?

thank you very much
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Hello from the GMAT Club BumpBot!

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