For All Your Quant Queries
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Triangle ABC is right angled at C. m<ABC=60 and AB=10. Let P be a point randomly chosen inside triangle ABC, such that BP extended meets CA at D. What is the probability that BD>5root(2)?
Written by Implex
September 11, 2008 at 3:46 pm
Posted in Geometry, Probability, Problem of the week
Tagged with Geometry, Probability, Problem of the week
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Is the answer to this 0.18?
If it is, i’ll post my approach.
September 11, 2008 at 5:05 pm
No, the answer is much larger than that
September 11, 2008 at 5:15 pm
oops.. misunderstood the problem. BD>5root2.
We get the triangle with sides 5root(3) and 5
Now, for BD=5root(2), we get an isosceles triangle with sides 5 each. For this the angle DBC=45deg.
So for the small 15 deg part, we get BD>5root(2)
so, prob =15/60=.25 (much larger to nahi hai par)
September 11, 2008 at 5:25 pm
The answer is 1- root(3)/3
September 11, 2008 at 5:42 pm
I think we are comparing sides here. For BD to be 5sqrt2 CD = 5.
For all values of 5<CD<5root3 the condition is satisfied .Hence probability = (root3-1)/root3
September 11, 2008 at 7:18 pm
yeah right @ Amit, but we will find it easier if we compare areas and not sides
September 11, 2008 at 7:24 pm
Yes, thats the correct way, So if we fix CD=5 then for the condition to be satisfied P should lie in tri(ABD) out of possible tri(ABC). Nice method🙂
September 11, 2008 at 7:40 pm
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