Math POTW #15 - Happy Holidays

Hello all, this will be the final POTW for 2018! We will pick-up with the official POTW #16 when we return from the Winter Break Jan. 7th.

For these weeks, please try all of the questions, including the optional ones. But again, some are optional, so enjoy your Winter Break first and foremost! And...some of the questions this week are great practice for your Spatial Sense Assessment OF Learning this Tuesday.

Grade 7/8 Math POTW Optional Winter Break Question #2 (#1 is down below):

Grade 8 Math POTW #15 Question:

Gr. 7 Math POTW #15 Question (Spatial Sense Practice):

 

Fiona POTW #2 (Question from last week, Week #14) Solution:

This question deals with the Primitive Pythagorean Triples, also known 

as the Pythagorean Triples. All you need to know are the following triples 

and you can solve this problem:

3-4-5

5-12-13

7-24-25

8-15-17

20-21-29

First, we need to look at triangle ABD because we need to know side BD in 

order to find side DC. We could use the pythagorean theorem to solve this,

 but if you notice, both side lengths are divisible by 4. Dividing everything by 4, 

we get:

52/4 = 13

48/4 = 12

This looks familiar because this is ⅔ of one of the triples (5-12-13). All we 

need to do now, is take 5 and multiply it by the common factor, in this case it is 4.

5 x 4 = 20

Now that we know side BD, we can use that to find that triangle BCD is 

also very familiar. It is part of the 20-21-29 triple. Because we already 

have the 20 and the 21, side length CD has to be 29.

Therefore side length CD has to be 29. 

POTW #14 Gr. 7 Solution:

POTW #14 Gr. 8 Solution:

 

 

Math POTW Optional Winter Break Grade 8 Question #1:

Grade 7 Math POTW Optional Winter Break Question #1:

Math POTW #14 - Thanks Fi!

Special thanks to Fiona for sharing a few POTW questions with us. Please find solutions to last week's POTWs and this week's questions below.

Grade 7 POTW #13 Solution:




Grade 8 POTW #13 Solution:



Fiona 7/8 Solution #1:

This question seems more difficult than it is. All we know is that there is an 8 x 8 square 
and the sum of the four corners of 1646. We can use algebra to find the value of the 
top left corner and use that to find the value of the bottom right corner. The top left 
corner has to be a variable, let’s say x. The expression for the top right corner 
would be x + 7 since you need 8 numbers inclusive (including the first and the last). 
The expression for the bottom left would be x + (7 x 24) or x + 168, since from the 
top row to the second row (same column), there is 24 that is added on. Since you 
are moving 7 rows down (for the same reason you added 7 for the top right), you multiply
 the two to get

x + 168. The bottom right uses the expression for the bottom left, and adds an
 extra 7, making the expression x + 175. If I put all of these expressions into an 
equation, we can find the answer.

Let x be the value of the top left corner.

x + (x + 7) + (x + 168) + (x + 175) = 1646

4x + 350 = 1646

4x = 1296

x = 324

Knowing this, we can substitute x with the equation that we put in for the
 bottom right corner.

Bottom right:

x + 175

= (324) + 175

= 499

Therefore, the bottom right corner is 499.

Grade 7 POTW #14 Question:

Grade 8 POTW #14 Question:
 

Grade 7/8 POTW Fiona Question #2:

Question: In the diagram, BD is perpendicular to BC and to AD. If AB = 52, 

and AD = 48, what is the length of DC?

(if the image does not show up on your screen open this 
https://docs.google.com/document/d/1FUNukUoynkVY0RRCGMDfzeaVZmhTf6Tvxwbfujh6qjc/edit )