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)?

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)

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Is the answer to this 0.18?

If it is, i’ll post my approach.

milindSeptember 11, 2008 at 5:05 pm

No, the answer is much larger than that

outtimedSeptember 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)

milindSeptember 11, 2008 at 5:25 pm

The answer is 1- root(3)/3

outtimedSeptember 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

AmitSeptember 11, 2008 at 7:18 pm

yeah right @ Amit, but we will find it easier if we compare areas and not sides

outtimedSeptember 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🙂

AmitSeptember 11, 2008 at 7:40 pm