The perimeter is composed of two straight parts and two semicircles. We can use this to break down the problem.
We can find the straight parts easily. They are given in the problem.
straight parts: 82 x 2 = 164 m
The two semicircles make a circle. We just have to find the circumference of a circle with a diameter of 66 cm to get the length of the semicircles.
semicircles: 2(π)(66/2) = 66(3.14) = 207.24 m
answer: 164+207.24 = 371.24 m
1) I'm not sure what an open operation is.
2) D
3) No
4) Yes
5) D
6) hmm
7) there doesn't seem to be a 6 or 7
8) C
ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Start circle: πd = (3.14)(19) = 59.7
Move diagonally to the circle with the radius of 6.2.
Second circle: 2πr = 2(3.14)(6.2) = 39
Move upwards to the circle with the radius of 10.5
third circle: 2πr = 2(3.14)(10.5) = 66
Move right to the circle with the diameter of 16.6
Fourth circle: πd = (3.14)(16.6) = 52.2
Move down to the circle with the diameter of 7.7
fifth circle: πd = (3.14)(7.7) = 24.2
Move down to the circle with the diameter of 50
Sixth circle: πd = (3.14)(50) = 157.1
Move left to the circle with the radius of 11.8
Seventh circle: 2πr = 2(3.14)(11.8) = 74.1
Move down to the circle with the radius of 38
Eight circle: 2πr = 2(3.14)(38) = 238.8
Move right to the circle with the diameter of 1.1
ninth circle: πd = (3.14)(1.1) = 3.5
Move right to the circle with the radius of 14.8
10th circle = 2πr = 2(3.14)(14.8) = 93
Move up to the end.
Hope this helps :)
