MAKE SURE TO SCROLL DOWN AFTR HITTING "EDIT" AND DO NOT HIGHLIGHT EXISTING TEXT IT WILL DELETE IT.
We are all going to use the wiki to discuss problems assigned. Look below and you will find the problem that you are responsible to post the first answer for based on your Aquinas e-mail address. It is NOT o.k. to just write "I don't get it". Start by typing the problem (click on edit to start) word for word from the book. You Post your answer as if it were a test question ( You would not leave those blank you would take a stab at it!), HOW you got it (list specific page references and show work) and WHY you did that way. The rest of the class will make comments, on at least 3 other problems, corrections, and post their thoughts below your original response.
Mindscapes for Chapter 4.1 please post original response by Friday 9/24 (noon) and classmate responses by Sunday 9/26(noon):
4.1 (2) [mrh003] If a right triangle has legs of length 1 and 2, what is the length of the hypotenuse?
-You will first need to add together 1 squared and 2 squared which equals 5
- Then 5 is equal to c squared or the length of the hypotenuse, so we take the square root of 5 to find the length of the hypotenuse.
- So the answer to the question is the square root of 5.
If a triangle has one leg of length 1 and hypotenuse of length 3, what is the length of the other leg?
-We know that a squared + b squared = c squared, so we plug in the numbers.
-Now we have 1 squared + b squared = 3 squared, so 1+b squared = 9
-Then we subtract one from each side giving us b squared = 8
-Now we have our answer, the length of the other leg(b) is equal to the square root of 8.
Wow! Super job explaining this one...comments anyone? [SLewis](Comment): [SMD001] I'm not sure if this is how we are supposed to comment... But i was thinking for the first problem with the 2 legs of 1 and 2 that when you got your answer that you had to square root the 5 because it was in 5^2 form and same for the second problem when you have b^2=8 to get rid of the b^2 you would sq.root it making you sq.root the 8. Do we not go this far to solve them and leave them how they are or do we need to sq.root the end result if it is squared?
4.1 (#3) [SMD001]: If a right triangle has legs of length 1 and x, what is the length of the hypotenuse?
-I first pluged in the 1 and the x into the P. Theorem giving you= 1^2+x^2=C^2.
-I was thinking for this problem that you need at least one more side length being either the other leg(x) or the hypotenuse. But because you don't know the other lengths I believe all you can do is simplify the problem above.
-This gives you= 1+x^=C^2 Just like above what would "C" look like instead of C^2? [Slewis]
I think C = the square root of 1 + x^2
4.1 (8) klk002
Yes, there can be a triangle with ratio 1-2-3... or for example, 30 °,60°,90° degree. We are not talking about degrees of angles but rather side lengths...Can you build a triangle with side lengths a=1, b=2, and c=3? [SLewis]let me try again...with side lenghts... No, because the Pythagorean Theorem states that: In a right triangle, the square length of the hypotenuse is equal to the sum of the square of the lengths of the other two sides... SO 1 + 2 = 3 BUT 2+3 = is not equal to one...
This kind of triangle would have to be half an equilateral triangle, the side ratios are 1, square root of 3, and 2...
And you can have a triangle with natural numbers, 4, 5 and 6 ! How do you know that you can have a triangle with sides of lengths 4, 5, and 6? [SLewis]
1 + 4 = 5 and the square root of 5 does not = 3, so you can not build this triangle. Hello????Where are you folks responsible for those problems below? You will not receive participation points if you do not post your answers! [SLewis 9/25 @ 9:50pm!]
4.1 (4) [jtl001]
Suppose you know the base of a rectangle has a length of 4 inches and a diagonal has a length of 5 inches. Find the area of the rectangle.
-I need to find x in the P. Theorem as in x^2 + 4^2 = 5^2
-so x^2 = 25 - 16, x^2 = 9, x = 3
- going back to finding the area of the rectangle base = 4" height = 3" so 4 x 3 = 12 square inches of area
4.1 (13) [ljj001]
Oh not fun the second time to try this...
Okay deep breath if the formula is a^2+b^2=C^2 so I would think the solution should look like 1^2 +1^2 = c^2 next? 1+1=c/2, which would be c=1. Not sure if I have this right but I hope so. Part two would be square root of two but not sure how to show that on here. any feed back would be good.
(Comment)[trc002] This one is crazy, I still dont get it.
4.1 (15) [JEH003]
MAKE SURE TO SCROLL DOWN AFTR HITTING "EDIT" AND DO NOT HIGHLIGHT EXISTING TEXT IT WILL DELETE IT.
We are all going to use the wiki to discuss problems assigned. Look below and you will find the problem that you are responsible to post the first answer for based on your Aquinas e-mail address. It is NOT o.k. to just write "I don't get it". Start by typing the problem (click on edit to start) word for word from the book. You Post your answer as if it were a test question ( You would not leave those blank you would take a stab at it!), HOW you got it (list specific page references and show work) and WHY you did that way. The rest of the class will make comments, on at least 3 other problems, corrections, and post their thoughts below your original response.
Mindscapes for Chapter 4.1 please post original response by Friday 9/24 (noon) and classmate responses by Sunday 9/26(noon):
4.1 (2) [mrh003] If a right triangle has legs of length 1 and 2, what is the length of the hypotenuse?
-You will first need to add together 1 squared and 2 squared which equals 5
- Then 5 is equal to c squared or the length of the hypotenuse, so we take the square root of 5 to find the length of the hypotenuse.
- So the answer to the question is the square root of 5.
If a triangle has one leg of length 1 and hypotenuse of length 3, what is the length of the other leg?
-We know that a squared + b squared = c squared, so we plug in the numbers.
-Now we have 1 squared + b squared = 3 squared, so 1+b squared = 9
-Then we subtract one from each side giving us b squared = 8
-Now we have our answer, the length of the other leg(b) is equal to the square root of 8.
Wow! Super job explaining this one...comments anyone? [SLewis](Comment): [SMD001] I'm not sure if this is how we are supposed to comment... But i was thinking for the first problem with the 2 legs of 1 and 2 that when you got your answer that you had to square root the 5 because it was in 5^2 form and same for the second problem when you have b^2=8 to get rid of the b^2 you would sq.root it making you sq.root the 8. Do we not go this far to solve them and leave them how they are or do we need to sq.root the end result if it is squared?
4.1 (#3) [SMD001]: If a right triangle has legs of length 1 and x, what is the length of the hypotenuse?
-I first pluged in the 1 and the x into the P. Theorem giving you= 1^2+x^2=C^2.
-I was thinking for this problem that you need at least one more side length being either the other leg(x) or the hypotenuse. But because you don't know the other lengths I believe all you can do is simplify the problem above.
-This gives you= 1+x^=C^2
Just like above what would "C" look like instead of C^2? [Slewis]
I think C = the square root of 1 + x^2
4.1 (8) klk002
Yes, there can be a triangle with ratio 1-2-3... or for example, 30
°,60°,90° degree.
We are not talking about degrees of angles but rather side lengths...Can you build a triangle with side lengths a=1, b=2, and c=3? [SLewis]let me try again...with side lenghts... No, because the Pythagorean Theorem states that: In a right triangle, the square length of the hypotenuse is equal to the sum of the square of the lengths of the other two sides... SO 1 + 2 = 3 BUT 2+3 = is not equal to one...
This kind of triangle would have to be half an equilateral triangle, the side ratios are 1, square root of 3, and 2...
And you can have a triangle with natural numbers, 4, 5 and 6 !
How do you know that you can have a triangle with sides of lengths 4, 5, and 6? [SLewis]
1 + 4 = 5 and the square root of 5 does not = 3, so you can not build this triangle.
Hello????Where are you folks responsible for those problems below? You will not receive participation points if you do not post your answers! [SLewis 9/25 @ 9:50pm!]
4.1 (4) [jtl001]
Suppose you know the base of a rectangle has a length of 4 inches and a diagonal has a length of 5 inches. Find the area of the rectangle.
-I need to find x in the P. Theorem as in x^2 + 4^2 = 5^2
-so x^2 = 25 - 16, x^2 = 9, x = 3
- going back to finding the area of the rectangle base = 4" height = 3" so 4 x 3 = 12 square inches of area
4.1 (13) [ljj001]
Oh not fun the second time to try this...
Okay deep breath if the formula is a^2+b^2=C^2 so I would think the solution should look like 1^2 +1^2 = c^2 next? 1+1=c/2, which would be c=1. Not sure if I have this right but I hope so. Part two would be square root of two but not sure how to show that on here. any feed back would be good.
(Comment)[trc002] This one is crazy, I still dont get it.
4.1 (15) [JEH003]