3)
First we find the circumference of the spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=1
C=2π*1
C=2π
C=6.283185307179586476925286766559
so the first wrap eats up 6.283185307179586476925286766559cm of the 100 centimeters (1 meter) of twine, then increases the diameter to 3 cm. we have 100-6.283185307179586476925286766559 centimeters of twine left. (93.71cm)
So we find the circumference of the 2nd wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=1.5
C=2π*1.5
C=3π
C=9.4247779607693797153879301498385
The second wrap eats up 9.4247779607693797153879301498385cm of the remaining 93.71. . .cm of twine, then increases the diameter to 4 cm. we have 84.292036732051033807686783083602 centimeters of twine left.
So we find the circumference of the 3rd wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=2
C=2π*2
C=4π
C=12.566370614359172953850573533118
The third wrap eats up 12.566370614359172953850573533118cm of the remaining 84.29. . .cm of twine, then increases the diameter to 5 cm. we have 71.725666117691860853836209550484 centimeters of twine left.
So we find the circumference of the 4th wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=2.5
C=2π*2.5
C=5π
C=15.707963267948966192313216916398
The fourth wrap eats up 15.707963267948966192313216916398cm of the remaining 71.72. . .cm of twine, then increases the diameter to 6 cm. we have 56.017702849742894661522992634087 centimeters of twine left.
So we find the circumference of the 5th wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=3
C=2π*3
C=6π
C=18.849555921538759430775860299677
The fifth wrap eats up 18.849555921538759430775860299677cm of the remaining 56.01. . .cm of twine, then increases the diameter to 7 cm. we have 37.16814692820413523074713233441 centimeters of twine left.
So we find the circumference of the 6th wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=3.5
C=2π*3.5
C=7π
C=21.991148575128552669238503682957
The sixth wrap eats up 21.991148575128552669238503682957cm of the remaining 37.16. . .cm of twine, then increases the diameter to 8 cm. we have 15.176998353075582561508628651453 centimeters of twine left.
So we find the circumference of the 7th wrap spool:
C=2πr
We need the radius, but have the diameter, so divide by 2
r=4
C=2π*4
C=8π
C=25.132741228718345907701147066236
The 7th wrap wrap eats up the final 15.176998353075582561508628651cm of twine.
He will wrap 6 complete wraps and one incomplete wrap. The answer to "a" is 6.
the 7th wrap's 15.176998353075582561508628651453cm is 60.387357729738339422209408431284% of the remaining 25.132741228718345907701147066236 circumference.
(15.1769.../25.1327...)*100
The answer to "b" is 60.4%.
I hope you had time to do this and it wasn't a timed test, if so they should have stopped at half a meter of twine.
4)
a. 360 degrees divided up evenly 5 times = 360/5 = 72 degrees.
b. we know 72 degrees is the central angle of the triangle, we also know two of the sides are 5 in long (circle diameter divided by 2). Using "The Law of Cosines": (you should be allowed to google this, memorization is for birdbrains.)
side c² = side a² + side b² − 2(side a* side b * cos(central angle)
Putting in the values we know: c²= 5² + 5² − 2 × 5 × 5 × cos(72º)
So: c²=
Take the square root: c = √34.55 = 5.88 to 2 decimal places
"b" is 5.88 in
"c" is 5.88in * 5 = 29.39 in.
For D, I imagine they're looking for you to find the area of a triangle then multiply by 5, rather than memorizing the formula for the area of a pentagon. We have enough information to use the area for a pentagon formula already with the length of a side, but to do it by triangles, we need to know the "apothem" of the pentagon, which is the same as the height of the triangle which invokes Uncle Pythagoras. "a² +b² = c²"
2.94²+b²=5² so:
8.6436+b²=25
b²=16.3564
b=4.044
and that just got us the height to use in the formula for the area of a triangle:
At = (height * base) / 2
At = (4.044*5.88)/2
At = 11.89
The area of the pentagon is 5 times the area of those triangles, so
Ap = 5(At) = 5*11.89 = 59.45 square inches.
"d" is 59.45 square inches.
Checking with the formula for the area of a regular pentagon with a known side:59.48 square inches. Real close.
e)
This question is total feces. The answer is A=r[tex] \sqrt{2-2cos(360/n)} [/tex] just a bunch of algebra on the law of cosines, inserting "360/n" for the variable in the cosine operation. Don't feel bad for one second about this question, your teacher should have culled it, and if they coined it, they should be shot. Using trig is much more useful to learn than deriving it. You don't need to know how to assemble a car to drive one. This is information you GOOGLE, we won't ever find ourselves rubbing two sticks together to make fire, and needing to re-derive all our geometry and trigonometry again.
