Philip Calvert
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Textbook 6B, Practice A, page 40,# 6
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Yes, thank you! 
volume of prisms and cylinders Textbook 6B, Practice A, page 40,# 7b
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Thank you! 
Philip Calvert started following Textbook 6B, Practice A, page 39, 4a and b, Textbook 6B, Review 5, page 42, #6a, Textbook 6B, Review 5, page 41, #3 and and 2 others

The answer for 6a is 4r + (pi x r). I don't understand how pi is used here. Two curves surround two sides of the square, but I don't understand how this fact is identified as "pi x r".

Just a quick one here: the Answer Key I have gives only the area, but not the perimeter, and the book asks for both. We calculated the perimeter as 82.82 cm. Is that correct?

For this one, I'm confused about how to explain it. 7 is the radius I'm pretty sure, but the tricky part is that the quarter circle is larger than the semicircle. Therefore, multiplying 22/7 by 49 and by 1/2 produces 77. Multiplying 22/7 by 49 and by 1/4 produces 38.5, so it would seem the ratio of A:B is 2:1. I know the opposite is true, but I'm stuck on how to explain this. Could you please assist?

Textbook 6B, Practice A, page 39, 4a and b
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Ok, thanks Jenny. 
volume of prisms and cylinders Textbook 6B, Practice A, page 40,# 7b
Philip Calvert posted a topic in Grade 6  Standards Edition
We correctly calculated the volume of the syrup as 715.4 inches cubed in 7a, but were unsuccessful with 7b: 1/2 x 20 (base) x 9 (height) x 15 (Height) = 1 350. The difference between the volume of the container and of the syrup is 1 350  715.4 = 634.4, but the book's answer is 8! What have we done incorrectly? 
I think the volume of the rectangle is 7 x 14 x 42 = 4 116, and the volume of the semicircle is 3.14 x 49 x 42 = 6 462.1. Four rectangles in the float means 4 116 x 4 = 16 464, and three semicircles means 6 462.1 x 3 = 19 386.3. Adding these two products gives us 35 850.3, which is a long way from the textbook's answer of 26 166 inches cubed. Could you please assist Jenny?

Textbook 6B, Practice A, page 39, 4a and b
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Ouch! Sorry about that. In any case, the difference between the volumes of the two shapes we found to be 5192.8  750 = 4442.8, which is still a bit off from the book's answer of 4447.5. Do you think this difference in the answers is significant? 
volume of prisms and cylinders 6B Textbook, Practice A, page 39, #5a
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Thanks for letting me know ... . 
volume of prisms and cylinders 6B Textbook, Practice A, page 39, #5a
Philip Calvert posted a topic in Grade 6  Standards Edition
Hi Jenny, We got the answer to 5b, so our difficulty with 5a surprised us. We calculated the volume of the triangular prism as 10 (base) x 15 (height of triangle) x 20 (height of prism) divided by 2 = 1500 cubic cm. The solid rectangular prism we found as 5 x 10 x 20 = 1000 cubic cm. The circular holes was pi x 4 squared x 10 (height of each hole) = 502.9 x 3 holes = 1508.7 Obviously, we can't subtract the figure we got for the circular holes from the solid rectangular prism. What have we done wrong? 
Textbook 6B, Practice A, page 39, 4a and b
Philip Calvert posted a topic in Grade 6  Standards Edition
Hi Jenny, My son and I had no problem identifying the answer for 4a as "Shape B", but the answer given for 4b indicates that our calculations for shapes A or B or both were incorrect. For shape A, we calculated the volume as 20 (base) x 5 (height of the triangular base) x 15 (height of the prism) and divide by 2 = 225 cubic in. For shape B, we calculated the volume as pi x 10.5 squared x 15 (height of the prism) = 5192.8 cubic in. 4b asks us to find the difference of the volume of these two shapes, which we found as 4967.8; the book's answer is 4447.5 cubic in. What have we done wrong? 
6B Textbook page 27, number 5
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
Indeed, thank you. I see that all one need do to calculate perimeter is to identify the length of curve. 
I'm afraid I missed something in finding the area, so I can't find the perimeter either. I calculated the radius as five squared, multiplied it by pi and divided by 4. I then doubled what I thought was the area, and got 39.2, which isn't even close to the workbook's answer of 57. Could you please help?

Textbook 6A page 149, Ex. 7
Philip Calvert replied to Philip Calvert's topic in Grade 6  Standards Edition
I appreciate the help!