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
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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
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. $\displaystyle \displaystyle\frac{49(\pi-2)}{49\pi}=\frac{\pi-2}{\pi}$
The probability of a point being in the shaded area isQuote:
Originally Posted by nikki712
$\displaystyle \displaystyle\frac{shaded\;region\;area}{circle\;a rea}$
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.
$\displaystyle \displaystyle\ P=\frac{{\pi}7^2-(2)7^2}{{\pi}7^2}=\frac{{\pi}-2}{{\pi}}$
Thanks!!!
Can anyone tell me if the problem on the left side of the picture is right? Just wanna make sure. :)
Probability: is 75% for the first one and $\displaystyle \displaystyle \frac{\pi-2}{\pi}$ for the second one.
Yes, the probability is the shaded region divided by the circle area.Quote:
Originally Posted by nikki712
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.
5/6 isn't correct for either question.Quote:
Originally Posted by nikki712
Well that sucks. I will tell him to solve the problem the same way you guys solved the other problem. Thanks for your help!!!
The reason that 5/6 is wrong is that the six triangles are not all the same size. The two triangles in the central diamond are larger than the four triangles around the outside, in fact their area is twice as large.Quote:
Originally Posted by nikki712
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
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*:| * 7 |:*
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*| * |*
o - - - - - - - o
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* * *
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} $
Hello, nikki712!
For the first problem, draw a horizontal line through the center of the square.
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|:::::* |:*:::::|
|:::* |:::*:::|
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* - - - +:-:-:-:*
|:* |:::::*:|
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The square is divided into eight congruent right triangles.
Six of them are shaded.
Therefore . . .
