Mr. Lund’s Fifth Grade Class
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Can You Solve This? – Wedding Cake
Being sick this week caused me to miss posting the weekly “Can You Solve This” math problem! On Friday, I was absent from school to attend my brother-in-law’s wedding in Madison, Wisconsin. The weather was gorgeous and the wedding was equally as nice! Sitting at the reception, my mind wandered to my missed math post. It was about this time that the bride and groom went for the traditional cake photos and I had my idea for this week’s math problem!
Brian and Caryn’s wedding cake has three square tiers; each is 8cm high. The tiers have lengths of 60cm, 48cm, and 36cm. What is the surface area to be covered by frosting? There is no frosting between layers. To receive credit this week, you must post your answer (How you solved it & why you did those steps) by the start of school on Thursday. – Mr. Lund
4 Responses to “Can You Solve This? – Wedding Cake”
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November 11th, 2009 at 8:12 pm
Well first I know that the height is 8cm high. Then for the smallest one the edges are 36cm for each side of a square. So then I did 36 times 4 and got 144. Then for the medium tire. I know that the height is 8cm high and the sides of the 4 sides are 48cm. So I did 48 times 4 and got 192cm. Then I know that for the biggest tire. The height is 8cm high and I know that each side of the square is 60cm. So I did 60 times 4. Then I got 240cm. Finally,I did 192cm plus 240cm plus 144cm and got 576cm of icing as my final answer.
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November 11th, 2009 at 11:59 pm
First, since I knew the height was 8cm and the sides on the top tier was 36cm. So since the bottom layer was not going to be covered I did 36 times 36 which equals 1,296 because the top layer is a square too. Then I did (36*8)*4 which equals 1,152 because 8 is the height and to find out the area of one side you have to multiply 36 by 8. But I multiplied 4 onto it because there are 4 sides on the whole tier,not including the bottom. Then I went to the second tier. This time the sides were 48cm, but the height was still the same, 8cm. So I did 48 times 48 minus 36 times 36 which is 1,008cm squared. I did that to find out the area between the second tier square and the first tier square. Then I multiplied 48 times 8 times 4 which equals 1,536 to find out the area of the remaining sides of the second tier. I followed the same procedure for the third tier. The numbers that I used for the third one were: (60*60)- (48*48)= 1,296, (60*8)*4=1,920.
Now I added up all the totals for the three tiers.
1,296+1,152+1,008+1,536+1,296+1,920=8,208cm squared is the total area to be frosted.
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November 12th, 2009 at 5:50 pm
Well Mr. Lund this math problem is very hard. I might get it wrong but this was the first thing I thought would be right. I did 8 times 60 then got 48. Next I did 8 times 48 and got 384. I added 48 plus 384 and got 432. Now I did 8 times 36 and got 288. I added 432 plus 288 and got 720. My answer is 720cm. Hope I get this right. Can’t wait till the next math challenge.
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November 12th, 2009 at 7:10 pm
The biggest square has a total area of 120 cm, the middle has 96 cm and the top has an area of 72 cm.
Bottom: 60×8=480 divided by 4=120cm
Middle: 48×8=384 divided by 4=96cm
Top: 36×8=288 divided by 4= 72cm
All together:72+96+120=288cm sq
I did this because multiplying will give me the area, but then I will have to divide by 4 to get the actual shapes frosting cover. The total frosting cover is 288 cm sq.
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