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(20) A squre is inscribed in a circle with radius r. what is the probability that a randomly selected point within the circle will not be within the squre?

A. π-2/πr squre
B. π-2/π
C. (π-1/2)/π
D. 1-r/π
E. r/π


(18) In rectangle PQRS above, what is a+b om terms of x?

A. 90+X
B.90-X
C. 180+X
D. 270-X
E. 360-X

(look at the picture i attached)

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please explain me!! thanx

2007-07-11 16:14:52 · 5 answers · asked by ashylnn 1 in Science & Mathematics Mathematics

i can't attach picture. does anybody know how to attach picture not a website?

2007-07-11 16:18:30 · update #1

5 answers

first one is : B

Find the area of the circle first and teh square. Remeber the 45, 45, 90 angle rules. if radius is r then the length of the square is r rad2.. the probability in the end is the area of circle-area of square/ area of circle.. part/whole..

and i can't see the 2nd question..

2007-07-11 16:21:06 · answer #1 · answered by Aki 2 · 0 0

If the radius is r, then r/√2 is half the side of the square, so side is 2(r/√2) = r√2. Then area of square is (r√2)² = 2r². Since area of circle is πr², area of part of circle outside of square is πr² - 2r² = (π-2)r². Probability of hitting that is (π-2)r² / πr² = (π-2)/π.

2007-07-11 16:23:01 · answer #2 · answered by Philo 7 · 0 0

No pictures that I can see . . .

For the first problem, Find the area of the square, subtract from area of the circle. That is the area that is not within the square. Probability is that area, divided by the area of the whole thing (circle).
Answer is "B"

2007-07-11 16:23:58 · answer #3 · answered by bedbye 6 · 0 0

the first question i can answer, 2nd question i dont see a diagram


the diameter of the circle is 2R

the find the length of one of the sides of the square

A^2 + B^2=(2R)^2
A=B

so B=square root(2) R

area of square = 2R^2

area of circle=pieR^2

so the ratio of areas(circle -square/circle) is the probability (pie-2)/pie. the R^2 cancel out

the answer is B

2007-07-11 16:43:32 · answer #4 · answered by Anonymous · 0 0

That is a great picture.

2007-07-11 16:17:23 · answer #5 · answered by Anonymous · 0 0

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