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# Geometry/Probability Problem

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In triangle $$ABC$$, we have $$\angle B = 60^o$$$$\angle C = 90^o$$, and $$AB = 2$$.

Let $$P$$ be a point chosen uniformly at random inside $$ABC$$. Extend ray $$BP$$ to hit side $$AC$$ at $$D$$. What is the probability that $$BD \lt \sqrt2$$?

Thank you!!!

Apr 1, 2020

#1
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Triangle ABC is 30-60-90

AB = 2        BC = AB/2 = 1     AC =sqrt [(AB)² - (BC)²] = 1.732050808

CD = sqrt[(BD)² - (BC)²] = 1

Probability that BD < sqrt(2) = (CD) / (AC) = 0.577350269  or  57.7350269 %

I've corrected my error with  red "ink"

Nobody's perfect Apr 1, 2020
edited by Guest  Apr 1, 2020
edited by Guest  Apr 1, 2020
edited by Guest  Apr 1, 2020
edited by Guest  Apr 1, 2020
#3
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Previous comment deleted. Guest has done the duty of showing work.

edited by AnExtremelyLongName  Apr 1, 2020
#2
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The restriction $$BD<\sqrt{2}$$ creates a 45 - 45 - 90 triangle WITHIN the 30 - 60 - 90 triangle.

We easily calculate the area of the 30 - 60 - 90 triangle.

Small leg = 1

Long leg = $$\sqrt{3}$$

Area = $$\frac{\sqrt{3}}{2}$$

We easily calculate the area of the 45 - 45 - 90 triangle.

Leg = 1

Leg = 1

Area = $$\frac{1}{2}$$

Now the probability of BD being less than root 2 is the area of the 45 - 45 - 90 triangle divided by 30 - 60 - 90 triangle.

Now you do it!

Apr 1, 2020
#4
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Thank you AnExtremelyLongName!

MathsAreFun  Apr 1, 2020
#5
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I think I know how to do it now!

MathsAreFun  Apr 1, 2020
#6
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Yes please don't just give an answer to me, If you are, at least give an explanation too!

Apr 1, 2020
#7
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I'm glad you want to learn #8
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A  =  30°

D                2

1

C          1                 B  =  60°

BC  =  1

AC  =   √3

If   BD  = √2

Then  DC  =   √[ (2)  - 1 ] =  1

The  area of  triangle  ABC  = (1/2) BC)(AC)  =  (1/2)(1)(√3) =  √3/2

Triangle  DCB  will  have  a max  area  when BD  = √2

And  its area  =  (1/2)(CD)(CB)  = (1/2)(1)(1)  = 1/2

So......the probability  that  BD < √2   will occur  when P  falls inside triangle  DCB

And this probability is     (1/2)  / (√3) /2  =   1 / √3   Apr 1, 2020