In a right isosceles triangle with a hypotenuse of length 1, the lengths of the legs can be found using the Pythagorean theorem. Setting the length of each leg as "x," the equation becomes x² + x² = 1², simplifying to 2x² = 1. Solving for x gives x² = 1/2, leading to x = √(1/2). Rationalizing the denominator results in x = √2/2. Thus, the length of each leg in the triangle is √2/2.
#1
urekmazino
3
0
For a right isosceles triangle (45-45-90) of hypotenuse 1, solve for the length of the unknown legs. Give an exact answer and rationalize the denominator in the final answer.
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
A power has two parts. Base and Exponent.
A number 423 in base 10 can be written in other bases as well:
1. 4* 10^2 + 2*10^1 + 3*10^0 = 423
2. 1*7^3 + 1*7^2 + 4*7^1 + 3*7^0 = 1143
3. 7*60^1 + 3*60^0 = 73
All three expressions are equal in quantity. But I have written the multiplier of powers to form numbers in different bases. Is this what place value system is in essence ?