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The length is 3*sqrt(5) = 6.7082, approx.
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What is the length of a right triangle if the base is 8 and hypotenuse is 10?
Use the Pythagorean theorem. a^2 + b^2 = c^2 a = base b = length c = hypotenuse 8^2 + b^2 = 10^2 b^2 = 10^2 - 8^2 b = sqrt(10^2 - 8^2) b = sqrt(100 - 64) b = sqrt(36 b = 6 ------------------the length
How can you express in radical form the length of the diagonal of a square whose sides are each 2?
Let 'a' and 'b' be the length of one side and diagonal of a square. Pythagorus's theorem as applied to a square: a^2 + a^2 = b^2. Substituting a = 2 into the equation, we have b^2 = 2^2 + 2^2 = 8. b = sqrt(8) = 2 * sqrt(2). Q.E.D. ===========================
Can you solve for b in the equation 12 equals negative 4 over 2 time negative 2 plus b?
12 = -4/2 * -2 + b 12 = -2 * -2 + b 12 = 4 + b 8 = b
If A (-2 -4) and B (-8 4) what is the length of the line?
Since you know that it is a parallelogram (not parallellogram!) you already know that the opposite sides are mutually parallel. All you need to do is to establish that any pair of adjacent sides are of equal length, or equivalently, the squares of these lengths are equal. The squared length of the line AB where A = (p,q) and B = (r,s) is (p - r)^2 + (q - s)^2.
Why does the area of a triangle quadruple when the sides are doubled?
Since the area is (length of base)*(height)/2 {call these dimensions B & H) A1 = B*H/2 With dimensions doubled, A2 = (2*B)*(2*H)/2 = 4*B*H/2 = 4*A1. By not simplifying to 2*B*H, it's easier to see that it is four times the original area. It is 4 times because the two length dimensions are multiplied, and 2 * 2 = 4.
