A square that its side length is 2 cm. What is the larger number of points we can put inside it (the distance between each tow points must be larger than 1.5 cm) ?
Is it 2 or 4 or 5 or 9 or infinity?
Do not forget to explain your answer.
2007-01-12 02:12:33 · 11 answers · asked by T.S.M 2 in Science & Mathematics ➔ Mathematics
drowing tools is not allowed.
2007-01-12 02:26:05 · update #1
Its trickey...
and the answer is 4
2007-01-20 01:13:13 · answer #1 · answered by Sandeep K 3 · 0⤊ 0⤋
Let the square be named ABCD, starting from the top left and going clock-wise. Imagine that line AC has been joined and that there is a point E on AC so that AE is say 0.1 cm. Imagine a point F between A and E on the line AE (except A and E). Since the diagonal AC would be of length sq.root of 8=approx. 2.8 cm, it is obvious that there would be an infinite no. of points between F and C on the line AC whose distance from F would be more than 1.5 cm. Likewise, there would be an infinite no. of points on AE (like F) which lie between A and E, and each of these points would similarly have other points on AC at a distance of more than 1.5 cm. This argument can be extended further.
This shows that the answer to the question is infinity.
2007-01-14 03:26:11 · answer #2 · answered by greenhorn 7 · 0⤊ 0⤋
On each of the two perpendicular sides (each 2cm long) of the square, you can mark only 2 points at a distance of 1.5 cm from each other. Through these points draw lines parallel to the perpendicular sides. These lines will intersect in 4 points which will give the desired points.
Answer = 4
2007-01-20 00:39:37 · answer #3 · answered by Anonymous · 0⤊ 0⤋
My answer is also 4.
Just visualize a square with circles which should not overlap. Only 4 of 1.5 cm can move inside the square.
2007-01-19 03:51:41 · answer #4 · answered by Tess 2 · 0⤊ 0⤋
Obviously,the answer would not be 4. All the other squares that i can draw within would fall between 1.5 cm to 1.9999999...cm respectively. Theoretically, the answer may be infinity but i dont think it we can do it in practice.
2007-01-19 19:03:56 · answer #5 · answered by ben2938 1 · 0⤊ 0⤋
4
A square has for equal sides, To maximize the number of points assume starting point on one of the edges. Then, eliminate the overlapping area with radius = 1.5 cm ( The point on which less area covered will give off the maximum number of points that can be put).
2007-01-18 19:57:03 · answer #6 · answered by 2cute4u 2 · 0⤊ 0⤋
4
Draw a square with side 2cm. Put one point in its center. Draw a circle with radius 1.5 around that point. The square is completely contained in the circle. Therefore there are no other space to put a point.
The other way is to draw 4 points, one at each corner. Then draw 4 circles -> No space left.
2007-01-12 02:20:04 · answer #7 · answered by catarthur 6 · 0⤊ 0⤋
all of us understand the board is a sq., so each and each and every fringe of the board contains n*2cm squares, and the diagonal by the board is 20*sqrt(2) Pythagoras -> x^2 + y^2 = z^2 (2n)^2 + (2n)^2 = (20*sqrt(2))^2 8n^2 = 800 n^2 = 100 n = sqrt(100) = 10 so each and each and every side has 10 of the 2cm squares in length i.e. a 10 x 10 board, or 10*10 = 100 squares in complete
2016-12-02 04:07:52 · answer #8 · answered by Anonymous · 0⤊ 0⤋
ok. you have a 2x2 cm square. that = 4 cm^2/1.5cm you find out the rest and let me know. what the heck is a tow point. Do not forget to explain your question! infinity---no way.
2007-01-19 20:35:10 · answer #9 · answered by cowboybabeeup 4 · 0⤊ 0⤋
2*2=4/1.5=2.66666or maybe infinity? if you can over lap
2007-01-19 23:04:32 · answer #10 · answered by JUST ME 3 · 0⤊ 0⤋