Please see attachment. He's stuck on how to figure out the last problem (the one on the right side of the attachment). I'm not sure how to help him. Can someone please help me?

Attachment 20159

Printable View

- Dec 19th 2010, 03:18 PMnikki712Geometric probability.
Please see attachment. He's stuck on how to figure out the last problem (the one on the right side of the attachment). I'm not sure how to help him. Can someone please help me?

Attachment 20159 - Dec 19th 2010, 03:23 PMdwsmith
The square is square so all sides are equal.

If we divide it horizontally, we have two smaller squares (left side) which are half shaded.

So we 50% shaded right and 1/2 of 50% shaded = 75%

Area of a circle = $\displaystyle \pi r^2$ and r = 7 so the area of our circle is $\displaystyle \pi 7^2=49\pi$

The diagonal of the square is 2*7=14 and the sides are equal. $\displaystyle c^2=a^2+b^2\Rightarrow c^2=a^2+a^2\Rightarrow c^2=2a^2$

c is the diagonal so c =14.

$\displaystyle 14^2=2a^2\Rightarrow 196=2a^2\Rightarrow \mbox{divide by 2} \ \ 98=a^2$

Area of a square is a*a so we have 98 is the area.

Area of shaded region is $\displaystyle 49\pi-98=49(\pi-2)$

Now divide by total area. $\displaystyle \displaystyle\frac{49(\pi-2)}{49\pi}=\frac{\pi-2}{\pi}$ - Dec 19th 2010, 03:38 PMArchie Meade
The probability of a point being in the shaded area is

$\displaystyle \displaystyle\frac{shaded\;region\;area}{circle\;a rea}$

Use Pythagoras' theorem to calculate half the length of the side of the square.

$\displaystyle \displaystyle\ x^2+x^2=7^2\Rightarrow\ 2x^2=7^2\Rightarrow\ x=\frac{7}{\sqrt{2}}$

$\displaystyle \displaystyle\ 2x=square\;side\;length=\frac{(2)7}{\sqrt{2}}=\fra c{7\sqrt{2}\sqrt{2}}{\sqrt{2}}=7\sqrt{2}$

Square area is $\displaystyle (2x)^2=(2)7^2$

The circle area is $\displaystyle {\pi}7^2$

Finally calculate the probability of a random point being in the shaded region.

$\displaystyle \displaystyle\ P=\frac{{\pi}7^2-(2)7^2}{{\pi}7^2}=\frac{{\pi}-2}{{\pi}}$ - Dec 19th 2010, 03:44 PMnikki712
Thanks!!!

Can anyone tell me if the problem on the left side of the picture is right? Just wanna make sure. :) - Dec 19th 2010, 03:45 PMdwsmith
Probability: is 75% for the first one and $\displaystyle \displaystyle \frac{\pi-2}{\pi}$ for the second one.

- Dec 19th 2010, 03:47 PMArchie Meade
Yes, the probability is the shaded region divided by the circle area.

The circle contains the total area.

The shaded region is the "difference" between the circle area and the square area.

Hence the square area is subtracted from the circle area to obtain the numerator of the probability fraction. - Dec 19th 2010, 03:49 PMdwsmith
- Dec 19th 2010, 04:40 PMnikki712
Well that sucks. I will tell him to solve the problem the same way you guys solved the other problem. Thanks for your help!!!

- Dec 20th 2010, 12:18 AMOpalg
- Dec 20th 2010, 06:36 AMSoroban
Hello, nikki712!

Another approach . . .

Quote:

A point is randomly chosen in the circle.

Find the probability that the point is in the shaded region.

Code:`* * *`

*:::::::::::*

o - - - - - - - o

*| |*

:| |:

*:| |:*

*:| * |:*

*:| * 7 |:*

:| * |:

*| * |*

o - - - - - - - o

*:::::::::::*

* * *

The area of the circle is: .$\displaystyle A_c \:=\:\pi r^2 \:=\:\pi(7^2) \:=\:49\pi$

The square is a rhombus.

The area of a rhombus is one-half the product of its diagonals.

. . $\displaystyle A_s \:=\:\frac{1}{2}(14)(14) \:=\:98$

Hence: .$\displaystyle P(\text{square}) \:=\:\dfrac{98}{49\pi} \:=\:\dfrac{2}{\pi}$

Therefore: .$\displaystyle P(\text{not-square}) \:=\:1 - \dfrac{2}{\pi} \:=\:\dfrac{\pi-2}{\pi} $

- Dec 20th 2010, 09:16 AMSoroban
Hello, nikki712!

For the first problem, draw a horizontal line through the center of the square.

Code:`*-------*-------*`

|:::::* |:*:::::|

|:::* |:::*:::|

|:* |:::::*:|

* - - - +:-:-:-:*

|:* |:::::*:|

|:::* |:::*:::|

|:::::* |:*:::::|

*-------*-------*

The square is divided into eight congruent right triangles.

Six of them are shaded.

Therefore . . .