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Posted on 3/11/21 at 9:25 am to moontigr
For the area of the star I get 314.76 square cm.
Right or wrong? If it is correct I’ll post the steps, if you need them.
Right or wrong? If it is correct I’ll post the steps, if you need them.
Posted on 3/11/21 at 10:35 am to moontigr
For the star, can't you split the octagon into two trapezoids (one on each side, the one of the left is already defined for you) and a rectangle between the two trapezoids?
Area for each trapezoid would be: .5(b1+b2)*h = .5(4+9.7)*2.8=19.18
Area of the rectangle is = l*w=9.7*4=38.8
Area of each triangle = .5*b*h=.5*4*7=14
Then add them all together
Edit: as someone else noted, the octagon as labeled is physically impossible in plane geometry. The formula for the area of a regular octagon is (2s^2)*(1+sqrt(2)) which gives you 77.25 (meaning that OPs original answer would be correct). However if you calculate the area via the method I suggested you get 77.16. Obviously both cannot be true, so the figure must be mislabeled. Based on the way the figure is labeled, I'm guessing the method they want the student to use is dividing it into 2 trapezoids & one rectangle then summing the areas of the 3 shapes.
Area for each trapezoid would be: .5(b1+b2)*h = .5(4+9.7)*2.8=19.18
Area of the rectangle is = l*w=9.7*4=38.8
Area of each triangle = .5*b*h=.5*4*7=14
Then add them all together
Edit: as someone else noted, the octagon as labeled is physically impossible in plane geometry. The formula for the area of a regular octagon is (2s^2)*(1+sqrt(2)) which gives you 77.25 (meaning that OPs original answer would be correct). However if you calculate the area via the method I suggested you get 77.16. Obviously both cannot be true, so the figure must be mislabeled. Based on the way the figure is labeled, I'm guessing the method they want the student to use is dividing it into 2 trapezoids & one rectangle then summing the areas of the 3 shapes.
This post was edited on 3/11/21 at 10:50 am
Posted on 3/11/21 at 11:12 am to nctiger71
Yeah, that’s wayyyyy off.
Posted on 3/11/21 at 11:42 am to moontigr
Area of the Star:
Points:
(b*h)/2
(7*2)/2 = 7
7*8 = 56
Body:
(b*h)/2
(2.8*2.8)/2 = 3.92
3.92*4 = 15.68
(cross in middle)
b*h
4*9.7 = 38.8
(38.8*2)-(4*4) = 61.6 (subtract because you're double counting the center square)
Total
56+15.68+61.6 = 133.28
You don't have to use the Pythagorean theorem on either of your problems.
If Teach says that's wrong then tell her I'd like to have a word.
ETA: Tell her the 9.7 dimension is actually 9.6 given the other dimensions, which would make the area 132.48
ETA2: Tell her the answer is impossible because of an error. The 4, 2.8, and 9.7 cm dimensions can not be what they are listed as, probably due to significant digits.
Points:
(b*h)/2
(7*2)/2 = 7
7*8 = 56
Body:
(b*h)/2
(2.8*2.8)/2 = 3.92
3.92*4 = 15.68
(cross in middle)
b*h
4*9.7 = 38.8
(38.8*2)-(4*4) = 61.6 (subtract because you're double counting the center square)
Total
56+15.68+61.6 = 133.28
You don't have to use the Pythagorean theorem on either of your problems.
If Teach says that's wrong then tell her I'd like to have a word.
ETA: Tell her the 9.7 dimension is actually 9.6 given the other dimensions, which would make the area 132.48
ETA2: Tell her the answer is impossible because of an error. The 4, 2.8, and 9.7 cm dimensions can not be what they are listed as, probably due to significant digits.
This post was edited on 3/11/21 at 12:25 pm
Posted on 3/11/21 at 12:00 pm to moontigr
Do you know the correct area answers?
Is the area for the star 174.76?
Is the area for the star 174.76?
This post was edited on 3/11/21 at 12:10 pm
Posted on 3/11/21 at 12:10 pm to nctiger71
No... that’s what we’ve been trying to figure out! We got 133.25/130 but those answers are wrong. They’re both somewhere between 129-135
Posted on 3/11/21 at 12:13 pm to moontigr
Everything in both problems can be simplified down into squares and rectangles.
Posted on 3/11/21 at 12:42 pm to moontigr
quote:
They’re both somewhere between 129-135
How do you know the answers are between 129-135?
Posted on 3/11/21 at 12:53 pm to nctiger71
Well, if you read this thread pretty much everyone who has solved it has gotten within that range. Also, others who I work with have also landed in that range.
Posted on 3/11/21 at 1:12 pm to moontigr
On the 3rd try I get 134.40 for the star.
I had two error that I fixed. Think that's correct?
I had two error that I fixed. Think that's correct?
Posted on 3/11/21 at 1:17 pm to moontigr
1. A = 155.49 cm^2
P= 74.24 cm
2. A = 129 cm^2
P = 67.2 cm
3. Star design cost: $23.32
Rocket design cost: $19.35
4. Star. Greater area.
P= 74.24 cm
2. A = 129 cm^2
P = 67.2 cm
3. Star design cost: $23.32
Rocket design cost: $19.35
4. Star. Greater area.
Posted on 3/11/21 at 1:21 pm to nctiger71
quote:It's unworkable because the dimensions don't jive. You'd have to change each one, in turn, to correspond with the others then present a bank of possible answers. The teacher can pick the correct one out of the bunch.
Think that's correct?
Posted on 3/11/21 at 2:11 pm to WildManGoose
quote:
(38.8*2)-(4*4) = 61.6 (subtract because you're double counting the center square)
This was my mistake.
Posted on 3/11/21 at 2:19 pm to moontigr
I didn't frick your wife; Im not doing your Kid's homework.
Posted on 3/11/21 at 3:05 pm to WildManGoose
quote:
It's unworkable because the dimensions don't jive.
I see what you mean. There is a .1 cm error in either the longest dimension of the center polygon or in the base of the star points. I guess this is some of that 2 + 2 = 5 math.
Posted on 3/11/21 at 3:47 pm to WildManGoose
quote:
It's unworkable because the dimensions don't jive.
Correct. If we are to believe the statement that all sides of the octagon are equal then the line segment labeled 9.7 should actually be 9.657 (length of medium diagonal in a regular octagon = side * (1+sqrt2).
Since this problem apparently expects the correct answer to two significant digits after the decimal, the rounding to 9.7 creates a problem. OP should let the teacher know that it is not appropriate to expect an answer with two significant digits after the decimal when some of the input variables have only one significant digit after the decimal.
This question is problematic for a number of reasons and has really upset me. Maybe a trigger warning in the OP would be wise as well.
Posted on 3/11/21 at 4:13 pm to tigercross
These are the correct answers as provided by the teacher....
Area of star: 133.16 (Look at the method described by tigercross above)
Area of rocket: 129
Area of star: 133.16 (Look at the method described by tigercross above)
Area of rocket: 129
This post was edited on 3/11/21 at 4:17 pm
Posted on 3/11/21 at 4:17 pm to tigercross
quote:
Since this problem apparently expects the correct answer to two significant digits after the decimal, the rounding to 9.7 creates a problem. OP should let the teacher know that it is not appropriate to expect an answer with two significant digits after the decimal when some of the input variables have only one significant digit after the decimal.
Tigercross has it correct. SigFigs matter always. Mathematically, your answer can never be more precise than your input, and practically, your measurements can only be as precise as your tools.
Posted on 3/11/21 at 4:39 pm to moontigr
quote:
These are the correct answers as provided by the teacher....
Area of star: 133.16 (Look at the method described by tigercross above)
Tell her to show her work. Outside of whatever lesson she is teaching, the area of a regular (all sides equal) octogon=2a^2(1+v2), with "a" being the length of the side.
Area of an octogon with a side length of 4 = 77.25
Area of each triangle point = 14 (x4)
Total area = 133.25 if she wants 2 decimals.
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