Did the Math Department get this wrong?

[mileena]

I am going to be taking a geometry assessment test. I have a problem with four of the 20 questions in the sample practice assessment test they provide. Specifically with the answers the Math Department gives at the bottom of the test. Before I confront them, please tell me if I am crazy. How can a college math department be wrong in such a simple subject?? I hope I am wrong.

http://www.losmedanos.edu/Groups/Math/web/documents/MicrosoftWord-26challengeinfopacket.pdf


Problem 3)
" Find the measure of the interior angles of a regular 5-gon. "

Their answer is 180 degrees

Is not the answer 540 degrees?? [(n-2) * 180] ?


Problem 7)
"Find the area of the equilateral triangle with side length of 10 yds. "

Their answer is 25 radical 3.

My answer is (25 radical 3) / 2 since the area of a triangle is (bh)/2


Problem 9b)
Can you look at the diagram from the link above. I might be wrong, but I think there is no way to know which leg of the larger triangle the 12" side of the smaller triangle corresponds to. Therefore, I can't find the ratio of lengths and then apply the Pythagorean Theorem to the smaller triangle.


Problem 17)
" Find the area of the circular garden, where the length of the fence around the garden is 62.8m. Round your answer to 1 decimal place. "

Their answer is 313.8 meters squared.

My answer is 314.0 meters squared (approximately equals 100pi)


[Also, problem 5 is wrong in that they neglected to provide the measurement of the other angle (90 degrees) in the answer.]

 

[D H]

I hope I am wrong.

Your wishes have been granted!

Problem 3)
" Find the measure of the interior angles of a regular 5-gon. "

Their answer is 180 degrees

Is not the answer 540 degrees?? [(n-2) * 180] ?

Read their answer again. It's 108 degrees (540/5), not 180. The question is admittedly a bit vague, but I would take it as asking about the angle subtended by each of the interior angles rather than the sum of those angles.

Problem 7)
"Find the area of the equilateral triangle with side length of 10 yds. "

Their answer is 25 radical 3.

My answer is (25 radical 3) / 2 since the area of a triangle is (bh)/2

Their answer is correct. What is the base of the triangle? The height?

Problem 9b)
Can you look at the diagram from the link above. I might be wrong, but I think there is no way to know which leg of the larger triangle the 12" side of the smaller triangle corresponds to. Therefore, I can't find the ratio of lengths and then apply the Pythagorean Theorem to the smaller triangle.

Look at the pictures. Each triangle has a long leg and a short one. They give the length of the long leg for each. You don't need to know how long the shorter leg is to answer the question.

Problem 17)
" Find the area of the circular garden, where the length of the fence around the garden is 62.8m. Round your answer to 1 decimal place. "

Their answer is 313.8 meters squared.

My answer is 314.0 meters squared (approximately equals 100pi)

A=\pi\left(\frac{62.8}{2\pi}\right)^2 = 313.84\cdots
Round this to 1 place and you get 313.8. You apparently made the mistake of rounding in your intermediate calculations.

 

[HallsofIvy]

I am going to be taking a geometry assessment test. I have a problem with four of the 20 questions in the sample practice assessment test they provide. Specifically with the answers the Math Department gives at the bottom of the test. Before I confront them, please tell me if I am crazy. How can a college math department be wrong in such a simple subject?? I hope I am wrong.

http://www.losmedanos.edu/Groups/Math/web/documents/MicrosoftWord-26challengeinfopacket.pdfProblem 3)
" Find the measure of the interior angles of a regular 5-gon. "

Their answer is 180 degrees

Is not the answer 540 degrees?? [(n-2) * 180] ?

Unfortunately both of your links show only blank so I cannot be certain what the exact question is or what the answer is supposed to be. Yes, the total interior angles in any 5-gon is 540 degrees. Each angle in a regular pentagon is 540/5= 108 degrees. (NOT 180.)

Problem 7)
"Find the area of the equilateral triangle with side length of 10 yds. "

Their answer is 25 radical 3.

My answer is (25 radical 3) / 2 since the area of a triangle is (bh)/2

Dropping a perpendicular to one side divides the equilateral triangle into two right angles having hypotenuse 10 and one leg 5. The other leg (the altitude of the equilateral triangle) has length sqrt(100- 25)= sqrt(75)= 5sqrt(3). So bh/2= (10)(5sqrt(3))/2= 25sqrt(3). (The area of each right triangle is 25sqrt(3)/2. That may be what you calculated.)

Problem 9b)
Can you look at the diagram from the link above. I might be wrong, but I think there is no way to know which leg of the larger triangle the 12" side of the smaller triangle corresponds to. Therefore, I can't find the ratio of lengths and then apply the Pythagorean Theorem to the smaller triangle.

Again, I can't read the problem.

Problem 17)
" Find the area of the circular garden, where the length of the fence around the garden is 62.8m. Round your answer to 1 decimal place. "

Their answer is 313.8 meters squared.

My answer is 314.0 meters squared (approximately equals 100pi)

That "62.8" is misleading (I can't help but wonder if intentionally!) It is exactly 20 times 3.14 so it seems perfectly reasonable to use that for pi, doesn't it? But if you use the more accurate 3.141592, the diameter is 62.8/3.141592= 19.98987 so the radius is 9.99493 m and the area is (3.141592)(9.33493)= 31.4 square meters.

[Also, problem 5 is wrong in that they neglected to provide the measurement of the other angle (90 degrees) in the answer.]

Can't see the problem!

 

[mileena]

Your wishes have been granted!

Haha. Funny.

Read their answer again. It's 108 degrees (540/5), not 180. The question is admittedly a bit vague, but I would take it as asking about the angle subtended by each of the interior angles rather than the sum of those angles.

Ok, I just read the question too quickly. If I had read more closely, I would have also divided by 5. Ugh. My bad.

Their answer is correct. What is the base of the triangle? The height?

Ugh. You're right. I just did the area of the smaller 30-60-90 triangle which is part of the equilateral triangle. I tried doing the test too fast!

Look at the pictures. Each triangle has a long leg and a short one. They give the length of the long leg for each. You don't need to know how long the shorter leg is to answer the question.

But the triangles they give may not be to scale. The shorter leg might actually be longer than what appears to be the longer leg.

A=\pi\left(\frac{62.8}{2\pi}\right)^2 = 313.84\cdots
Round this to 1 place and you get 313.8. You apparently made the mistake of rounding in your intermediate calculations.

Ok, I wish I knew how to do those mathematical symbols on a post like you!

This is what I did:

2(pi)(r) = 62.8
r = 31.4/pi

A = (pi)(31.4/pi)squared =
(pi)(31.4/pi)(31.4/pi) =
[(31.4)(31.4)]pi =
985.96/pi =
985.96/3.14 =
314.0

Did I do something wrong? Isn't 3.14 a good enough approximation of pi?

 

[mileena]

Unfortunately both of your links show only blank so I cannot be certain what the exact question is or what the answer is supposed to be.

HallsofIvy, I do appreciate your help, despite the link not working!

What browser are you using? It worked for me on both Firefox and IE.

Here is the link in full, with a space placed after each of its three periods (.) so it will appear on this website (maybe that will help?):

http://www. losmedanos. edu/Groups/Math/web/documents/MicrosoftWord-26challengeinfopacket. pdf

[remove the 3 spaces above after the periods, and copy]

Yes, the total interior angles in any 5-gon is 540 degrees. Each angle in a regular pentagon is 540/5= 108 degrees. (NOT 180.)

Thank you. As D H also pointed out, I even gave their answer wrong! "108" is correct.

Dropping a perpendicular to one side divides the equilateral triangle into two right angles having hypotenuse 10 and one leg 5. The other leg (the altitude of the equilateral triangle) has length sqrt(100- 25)= sqrt(75)= 5sqrt(3). So bh/2= (10)(5sqrt(3))/2= 25sqrt(3). (The area of each right triangle is 25sqrt(3)/2. That may be what you calculated.)

Yep, you are correct. I calculated the wrong triangle. Trying to do the work too fast!

Again, I can't read the problem.

Can't see the problem!

I wish I could paste the similar triangle drawings here, but I don't know how!

That "62.8" is misleading (I can't help but wonder if intentionally!) It is exactly 20 times 3.14 so it seems perfectly reasonable to use that for pi, doesn't it? But if you use the more accurate 3.141592, the diameter is 62.8/3.141592= 19.98987 so the radius is 9.99493 m and the area is (3.141592)(9.33493)= 31.4 square meters.

I agree. They should have told you what value of pi to use. I am used to expressing answers in just terms of pi, so the whole problem is not really that valuable.

 

[Ray Vickson]

Haha. Funny.



Ok, I just read the question too quickly. If I had read more closely, I would have also divided by 5. Ugh. My bad.



Ugh. You're right. I just did the area of the smaller 30-60-90 triangle which is part of the equilateral triangle. I tried doing the test too fast!



But the triangles they give may not be to scale. The shorter leg might actually be longer than what appears to be the longer leg.




Ok, I wish I knew how to do those mathematical symbols on a post like you!

This is what I did:

2(pi)(r) = 62.8
r = 31.4/pi

A = (pi)(31.4/pi)squared =
(pi)(31.4/pi)(31.4/pi) =
[(31.4)(31.4)]pi =
985.96/pi =
985.96/3.14 =
314.0

Did I do something wrong? Isn't 3.14 a good enough approximation of pi?

Whether or not 3.14 is an adequate approximation to π (symbol found on the "quick symbols panel at the top or side of the input panel!) depends on how much accuracy you want in the final result. Do the people who are marking this work regard 314.0 as a good-enough approximation to 313.84≈3i3.8? When in doubt, just keep some more significant figures---if you are using a calculator it won't kill you to increase the accuracy.

Note: in some problems (not this one), rounding off too soon can be disastrous, especially if/when the final answer involves subtraction two moderate-to-large nearly-equal terms to get the final answer. Premature rounding can produce order-of-magnitude errors in such cases, and can even give answers having the wrong sign, etc.