a probability problem?

Question

A 10*10 cm square card is folded, a circle with radius 1 cm is placed so that it is entirely on the square. If the upper square is marked yellow, the lower part is marked red. What is the probability that the circle lies entirely in the red part?

Answer 1

the answer might vary since there's actually 2 ways to divide a square equally, the end results are either 2 rectangles (5X10) or 2 right-iso triangles. but from the word "upper" square, i assume you meant to divide into 2 rectangles since ppl usually places squares on 1 flat side and not on its vertex.

red part = yellow part = 5*10 = 50 cm^2

for circle to lie entirely on the square, the center should be bound to lie in the 8X8 square (1 cm margin on every side). for circle to lie entirely in red part, the center should be bound to lie in the rectangle with dimension 3X8.

P(entirely in red)
= (area of possible center) / (area of allowed center)
= 3*8 / (8*8)
= 3/8
= 37.5%

Answer 2

If the circle is to lie "entirely" in the card, then its center has to fall within a 9 * 9 cm square. However, for it to fall entirely in the red region, it would have to fall within a 9 * 4 cm rectangle. Hence, the probability is 4/9.

Addendum: Oh, RADIUS of 1 cm. Okay, 8 * 3 out of 8 * 8, or probability of 3/8. Hats off to manjyomesando1.

Answer 3

Area of square = 10*10 = 100
Area of circle = pi*r^2 = 3.142 *1*1 = 3.142

Probability of red part = 3.142 / 100 = 0.03142

checking;
range of prob : 0 < prob < 1

therefore, answer is 0.03142

Answer 4

10*10=100
3.14r^
3.14(1)^
3.14^
9.8596

100/9.8596 = about 10 (10.142 exactly) - I think that's the probability.