Eteled said:
#47, so lets rewrite Pythagoras first, a^2 + b^2 = c^2 … b = (c^2 – a^2)^(1/2). Ok now chopping the triangle into two we get a = 3, c = 6 and plugging this into our equation for b we get b = (36 – 9)^(1/2) = sqrt(27) . You might have written sqrt(25) = 5. Just something you slipped on. They clean sqrt(27) up. 27 = 9*3, so they write sqrt(9*3) and then they pull out the 9. sqrt(27) = 3*sqrt(3). Now you have h. Put it into ½*b*h and you get ½*6*3*sqrt(3) = 9sqrt(3).
That is way too much work. I didn't use pythagoras during the ASTB. It is all triangle memorization. There are four basic triangles that need to be memorized. the 30-60-90, 45-45-90(isosceles right), 60-60-60 (equilateral) and 3-4-5.
30-60-90 triangle:
Properties: The three angles in this triangle are 30, 60 and 90 degrees. The shortest length, across from 30 degree angle is x, Angle across from 60 is x * sqrt(3). and hypoteneuse is 2*x.
Below was a nice triangle that was butchered by the forum.
/|30degrees
2x / |
/ | Xsqrt(3)
/__|
60 X 90
45-45-90 triangle (isosceles right)
Angles of 45 45 and 90 degrees. Short sides are both x and hypoteneuse is x * sqrt(2).
60-60-60(equilateral)
Angles of 60 all around. Each length is the same, hence the distance from each point is the same, so if 2 ships leave port at the same speed and are traveling 60 degrees difference in heading. The distance between the two ships is equal to their distance from port.
Also if you cut an equilateral in half (bisect) you will have two 30-60-90 triangles...hint hint hint the area will always be (0.5* X)^2 * sqrt(3)
3-4-5 (pythagorean Triple)
Right triangle with sides 3,4,5. Given a right triangle with sides any two of the 3-4-5 mix and the remaining side is obvious. This also works with multiples of 3-4-5, like 6-8-10 or 9-12-15 and so on.
a^2 + b^2 = c^2 Yeah but it take too much time. learn the above and you will finish with enough time to take a piss and get a drink of water.
John