The numerical-initial equation "64 S on a CB" is a cryptic crossword clue that translates to "64 squares on a chessboard." It references the standard 8x8 grid found on a chessboard.
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150 square yards are 1350 square feet.
Texas has an area of 268600 square miles.
8,718.77 square feet.
Comins Mansfield has written: 'Chessboard delights' -- subject(s): Chess problems 'The modern two-move chess problem' -- subject(s): Chess problems
The word "square" has four phonemes: /s/ /k/ /w/ /ɛər/.
The answer is 3.Proof:According to Pythagoras, for any right-angled triangle (one that has one angle equal to 90 degrees), the square of the longest side equals the sum the squares of the two shorter sides.So, if the sides of the equilateral triangle are of length s, a normal from any apex will divide the opposite side into equal lengths that are each equal to a half of s (= s x 0.5)The normal is part of a right-angled triangle which has its longest side of length s, the normal as its next-shorter side and its shortest side is of length (s x 0.5)We have been told that the square of the normal = 0.75 (A)We can calculate the square of the length of a half side as:(s x 0.5) x (s x 0.5) = s^2 x 0.25 (s squared times a quarter) (B)So the sum of these squares (A) + (B) = s^2 x 0.25 + 0.75According to Pythagoras, (A) + (B) = the square of the longest side s = s^2So s^2 = [(s^2 x 0.25) + 0.75]Using algebra, we can reduce this to: [s^2 - (s^2 x 0.25)] = 0.75So (s^2 x 0.75) = 0.75 so s^2 = 1 so s =1So the answer to the question " what is the sum of its three sides" is 1 + 1 + 1 = 3 ---- ; Above Correct Answer and Proof by Martinel. : Joined: Thu Feb 07, 2008 12:00 am
Tennessee is the 36th largest state by area with 42,143.27 square miles.
240 Calories in a normal sized Gregg's sausage roll, however the jumbo sausage roll is probably half again if not double!
The formula for the area of a square is: s * s where s = length of a side The formula for the perimeter of a square is: 4 * s where s = length of a side
For a square with side S. There will be the 4 small triangles each with sides {S/sqrt(2),S/sqrt(2),S}. Then also you can 'see' 4 larger triangles, each with sides {S,S,S*sqrt(2)}