# [SOLVED] 3 problems

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• Mar 25th 2009, 07:32 AM
GreenMile
[SOLVED] 3 problems
1. What is shadowed surface of cube if its side is $\displaystyle a=2\sqrt{2}$
(Image is on attachment).

2.On the box, we have 5 red balls, 4 dark balls, and 6 white balls. If we catch one ball, what is probability to catch a white one?

3. If we have a segment [AB], A(5,-1), B(x,2) and it's middle point is O(0,0), what is value of x?

Thank you.
• Mar 25th 2009, 07:50 AM
thisisme
1. Where is the attachment, can't see it.

2. 5 + 4 + 6 = 15 balls in total. 6 are white so its 6/15=2/5, ie the probability is 2/5.

3. Given a line with two points (A,B) and (C,D), the midpoint , (E,F), is found as E=(A+C)/2 and F=(B+D)/2.

So, in this example we have
A=5, B=-1, C=x, D=2, E=F=0
Using basic algebra we find A=-C (subbing in E=0).
A=5 thus C=-5
Wait... If you apply the same reasoning we find that B=-D, which is false so I think you may have given some incorrect numbers. If your point A and point B co-ordinates are correct then it is your midpoint that is incorrect.
• Mar 25th 2009, 07:54 AM
GreenMile
Ups, sorry. There is attachment.
• Mar 25th 2009, 08:03 AM
thisisme
That is a rectangle. Area of a rectangle=base times height

Base=a
Height= sqrt(2(a^2))=a*sqrt2 (by pythagoras formula)
if a=2*sqrt2 then,
Base=2*sqrt2, and
Height=2*(sqrt2)*(sqrt2)=2*2=4
Base*Height=4*2*sqrt2=8sqrt2

Area=8sqrt2
• Mar 26th 2009, 12:42 AM
GreenMile
I have just another problem so I don't see very reasonable to open new thread just for that.

At this square $\displaystyle |DE|=2|EC|$. Find $\displaystyle cosx$.
• Mar 26th 2009, 05:59 AM
Grandad
Cosine of angle
Hello GreenMile
Quote:

Originally Posted by GreenMile
I have just another problem so I don't see very reasonable to open new thread just for that.

At this square $\displaystyle |DE|=2|EC|$. Find $\displaystyle cosx$.

It's generally better to open a new thread, but here's the answer to this one.

If the length of the side of the square is $\displaystyle a$, then $\displaystyle |DE| = \tfrac{2}{3}a$

$\displaystyle \Rightarrow |AE|^2 = a^2 + (\tfrac{2}{3}a)^2 = a^2 + \tfrac{4}{9}a^2 = \frac{13a^2}{9}$, by Pythagoras.

$\displaystyle \Rightarrow |AE| = \frac{\sqrt{13}}{3}a$

$\displaystyle \Rightarrow \cos x = \frac{|DE|}{|AE|} = \frac{2}{3}a \div \frac{\sqrt{13}}{3}a = \frac{2}{\sqrt{13}}$

Grandad