POTW #12 Gr. 7 Solution
POTW #12 Gr. 8 Solution:
POTW #13 Grade 7 Question: Spatial Sense!
POTW #13 Grade 8 Question:
POTW #13 Grade 7/8 Question: thanks Fiona (from her CEMC exam, or something similar)
The positive integers from 1 to 576 are written in a 24 x 24 grid so that the
first row contains the numbers 1 to 24, the second row contains the numbers
25 to 48, and so on, as shown. An 8 x 8 square is drawn around 64 of these numbers.
(These 64 numbers consist of 8 numbers in each 8 rows.) The sum of the numbers in
the four corners of the 8 x 8 square is 1646. What is the number in the bottom right
corner of this 8 x 8 square?
first row contains the numbers 1 to 24, the second row contains the numbers
25 to 48, and so on, as shown. An 8 x 8 square is drawn around 64 of these numbers.
(These 64 numbers consist of 8 numbers in each 8 rows.) The sum of the numbers in
the four corners of the 8 x 8 square is 1646. What is the number in the bottom right
corner of this 8 x 8 square?
Grade 8 POTW:
ReplyDelete- A(0,30)
- B(k,50)
- D(40,0)
- Area=1340
Since all of these coordinates are in the first quadrant, k must be positive because of rules and laws and physics (nah, it’s because QI is all positive). Because this is a parallelogram, we know that AB=CD and AB || CD.
To get from Point A to Point B, we need to go up 20 units and go right k units. Repeating this for C and D, we get that Point C is located at (40+k,20).
After drawing a box around the figure and making sure that it is tangent to the parallelogram, we can make some large formula for the area of said box.
(I labelled the box, EFGH with point E at the top left and point H at the bottom left (goes in a clockwise circle).
FG x HG= (AE*EB/2) + (BF*FC/2)+(CG*GD/2)+(DH *HA/2) +1340
50(40+K) = (20*K/2)+(40*30/20)+(20*K/2)+(40X30/2) +1340
2000+50k = 10k+600+10k+600+1340
2000+50k=20k+2540
30=540
k=18
Fantastic, now we can find the coordinates.
B(18,50)
C(58,20)
Oh no the picture doesn't show in the POTW :/
DeleteHey there, sorry I forgot to do last week's POTW. I'll make it up to you with an extra great one today! Today we're going to use a smorgasbord of geometry skills, spatial sense and algebra to determine 2 over-the-top answers to this POTW!
ReplyDeleteMethod 1: Crazy algebra guy
Let's assume the area of this figure is x. Well, we can't get the value of a shape without drawing it, right? Well, kind of. You see, there is a way to tell the number of units from line to line and object to object without drawing it. Let's use the data points (-3,1), (-1,3), and (-3,3) as our points, and we can use the other two points as substitutes. We know that a coordinate is shared between the two former points and the latter, and it goes in a triangle formation. Aha! A triangle! (More specifically, a right triangle because that's where these wack rules apply). Now, we can figure out the area of this triangle by looking at the data points and using the right triangle formula. Now what about the sub points? We use these points to figure out if the triangle goes outside the shape or overlaps with another. (It's a right triangle, so this is pretty simple.) We can do this with quadrilaterals as well. The ideal shapes are linked like this: A quadrilateral [(-3, 1) (-3, 3) (-1, 3) (1, 1)] and a triangle [(-3, 3) (-2, 2) (1, 1)] (this intersects 0, 0.) Now we can add them, right? Not quite. To double check this, we use asymptotes of triangles(?) If you're confused what an asymptote is, it's a line that never gets touched by a function, or in this case, a line that any point on the inside of these shapes never touches. In this case, we'll be using the top of the triangle and the bottom of a quadrilateral since we know they touch vertically via line [(-3, 1) (1, 1)]. As we compare a point getting closer and closer to that asymptote, and another on the other side of it, we can see that the two shapes never intercept. Add up the area of the two (6 units squared plus another 6) and you get yourself a clean, crisp 12 units squared.
Stay tuned for method two!
And the other potw, which I will also solve along with the 7/8 one
Method 2: Triangles, Pentagons, and Tetradecagons
DeleteNow, you may have heard about a shape called a "Pentagon." It's a shape with 5 sides. But have you ever heard of a Tetradecagon? A shape that has 14 sides, and is also rather challenging to calculate an area with. That's why I'll be using it (and some triangles) to tile an irregular pentagon traced out by Vroom. Here we go!
Drawing a tetradecagon in the middle of our shape, we can see that there are approximately 13 triangular spaces needing filling, 11 large, 1 medium and one small to create an approximation of a 2*2 block. Doing this repeatedly and in a smaller degree, (repeat another time to fill in another square, then use two triangles to fill in the rest) gives us an area of 6 units squared for the quadrilateral [(1,1) (-1,3) (-3,3) (-3,1)]. The lower triangle is a bit more tricky, requiring 3 iterations and also 3 area 1/2 right triangles. But then, we run unto a problem. It is impossible to tile a 3*1 triangle with just isosceles right triangles and tetradecagons! I'm sorry, but this mission is...
Not over yet! You see, scaling down the right triangle, we can create an infinite slope of them leading all the way down! Of course, we'll have to repurpose the entire lower half, but so be it!
Finding the space inside is a trickier matter, but we can use integrals to even define seemingly undefinable - and unidentifiable - functions. Let's take our 3*1 triangle and fill it with bars of infinitely small size (0.000...01 is 0) and instead, take the integral of a function t(8-t) and position it in the function as another infinitely small piece. Now find the area of that piece only via subtraction of antiderivatives t=8 and t=(however small you scaled it down to), scale it up and you have yourself a clean, crisp 6 squared units! (I stopped halfway as this method is even more absurd than the last method and also I need to read history before going to bed). Phew! Now, all we need to do is add 6 plus 6 and we have ourselves... 12 square units!
Now just the grade 8 and 7/8 one left...
Thanks for reading!
Error: 3*1 triangle
Deletethis is actually a scalene right triangle, which is 3 units tall and 1 unit wide
Grade 7/8 POTW
DeleteHey everyone! For once, I will use a simple method that involves some algebra so I myself can actually understand the answers I wrote. We will abbreviate the lowest value number of the square as k. We know one of the other corners is k+7, as this is an 8 by 8 square and the two are in a row. We know the opposite corner's value is k+63, as again, and 8 by 8 square contains 64 units (8^2). From that, we know the bottom left corner of the square, which is k+63-7, would sum up to be k+56. Adding these all up, k+k+7+k+56+k+63=1646. Simplifying this equation, we eventually arrive at a nice, crisp equation:
4k + 56 + 7 + 63 = 1646
4k + 126 = 1646
4k = 1646 - 126
4k = 1520
Now, all we have to do is divide 1520 by 4.
1520/4=k
k=380
Bottom right corner = k+63
Bottom right corner = 443
We can double check this by multiplying back by 4, and adding 56, 63, and then 7, and it does equal 1646.
Therefore, the value of the bottom right corner is 443.
yay a non ridiculous solution
To answer this POTW I first created it on the coordinate plane then realizing that the shape is an irregular pentagon, I filled in the missing are to form a rectangle then figured the area of that rectange which is 5 x 4 or 20 units squared then, I subtracted the area of all the triangles I used to fill in the shape and the area was (3 x 1 / 2) + (3 x 3 / 2) + (2 x 2 / 2) = 1.5 + 4.5 + 2 = 8 then, 20 - 8 = 12. Therefore, the area of the irregular pentagon is 12 units squared.
ReplyDeleteSorry kids, the picture for Fiona's Question #13 did not show up for many students, but it does work on my account. Regardless, here is a Google Doc link for the image that is NEEDED to answer the question:
ReplyDeletehttps://docs.google.com/document/d/1FUNukUoynkVY0RRCGMDfzeaVZmhTf6Tvxwbfujh6qjc/edit
Oops this link: https://docs.google.com/document/d/1FUNukUoynkVY0RRCGMDfzeaVZmhTf6Tvxwbfujh6qjc/edit
DeleteThe area of the figure created by Vroom is 12 units squared. Did the work in my math journal but is mainly just put in the coordinates then I split the figure created into two figures. Then, I found the areas of the separate figures and added them together to get the area.
ReplyDeleteGrade 7 POTW
ReplyDeleteFor this POTW, I will draw the figure that Vroom traces on a cartesian plane, and then I will find out the area of the figure. I have drawn my figure on a cartesian plane in my math journal and it looks like a trapezoid on top of a triangle. In order to find the area of the figure, I split it into 3 parts. One is a 2x2 square, anther is a 2x2 triangle and the last is a 4x3 triangle. Now, I will find the area of each of the parts and then add them up to get the total area of the figure.
2x2 square= 4 because 2x2=4
2x2 triangle=2 because 2x2/2=4
4x3 triangle=6 because 4x3=12/2=6
6+4+2=12
The area of the figure that Vroom traced is 12 square units.
I drew it out on grid paper and it gave me a
ReplyDelete2x2 square=4
2x2/2 half a square= 2
and a scalene triangle= 3(height)x4(base)/2=6
4+2+6= 12
in total the area is 12^2 or 12 unit squared.
Grade 7/8 POTW
ReplyDeleteI am just going to go straight into the work. The sum of the four corners is 1646. In other words, a + b + c + d = 1646. To solve this, we need to have only one variable to make it simpler. We will use a. We know the top left corner is A. However, 7 numbers right to A is B or B = A + 7. The bottom left is C which is 7 numbers below A. Each 1 number below is 24 numbers. Therefore, C = A + (24 x 7) = A + 168 D is 7 numbers right of C so it is A + 168 + 7 = A + 175. Now to add.
A + A + 7 + A + 168 + A + 175 = 1646.
4A + 350 = 1646
4A + 350 - 350 = 1646 - 350
4A = 1296
4A/4 = 1296/4
A = 324
B = 324 + 7 = 331
C = 324 + 168 = 492
D = 324 + 175 = 499
To double check, 324 + 331 + 492 + 499 = 1646
Therefore, the bottom right number of the square is 1646.
Yay someone did my POTW :))
DeleteWait, how can the bottom right number of the 8 x 8 square be 1646 if the highest number in the 24 x 24 square is 576?
DeleteWait. Oops, I accidentally wrote 1646 instead of 499. Re saying it properly:
DeleteTherefore, the bottom right number of the square is 499.
Grade 8 POTW
ReplyDeleteI actually found 2 different ways to solve the problem. However, both result in slightly different answers.
1. The first method involves creating a box around the parallelogram. This creates four new right triangles. After this, the values known are used to solve for k which then helps lead to the other coordinates. The top left corner will be labeled E, top right, F, bottom right, G, bottom right, H. The entire square is equal to the areas of the four right triangles + the area of the parallelogram. It would be:
(AE x BE)/2 + (BF x CF)/2 + (CG x DG)/2 + (AH x DH)/2 + 1340 = EF x EH
(20 x k)/2 + (40 x 30)/2 + (20 x k)/2 + (40 x 30)/2 + 1340 = 50 x (40 + k)
20k/2 + 1200/2 + 20k/2 + 1200/2 + 1340 = 2000 + 50k
10k + 600 + 10k + 600 + 1340 = 2000 + 50k
20k + 2540 = 2000 + 50k
20k + 2540 - 20k = 2000 + 50k - 20k
2540 = 2000 + 30k
2540 - 2000 = 2000 + 30k - 2000
540 = 30k
540/30 = 30k/30
18 = k
B(18,50)
40 + 18 = 58
0 + 20 = 20
C(58,20)
Therefore, the B coordinate is (18,50) and the C coordinate is (58,20).
2. The second method I used resulted in a similar answer, but still not the same. It firstly involved the pythagorean theorem (a^2 + b^2 = c^2). The known coordinates are A(0,30), and D(40,0). They both intersect and meet at (0,0) which forms a right triangle. We will call the origin O. AO is equal to 30 units and DO is equal to 40 units. Using the pythagorean theorem, 30^2 + 40^2 = 900 + 1600 = 2500. The square root of 2500 is 50 so AD = BC = 50 units. In order to find the area of a parallelogram, you have to use the formula base x height or b x h. In this case, b is 50 units and the area is 1340 units^2. To solve for the height, we can divide the area by the known side length. 1340/50 = 26.8 units. The height is AB = CD = 26.8 units. To solve for k, we can do the pythagorean theorem in reverse. a^2 + b^2 = c^2.
c^2 - a^2 = b^2
26.8^2 - 20^2 = k^2
718.24 - 400 = k^2
318.24 = k^2
√318.24 = √k^2
17.8392825 = k
B(17.8392825,50)
17.8392825 + 40 = 57.8392825
50 - 30 = 20
20 + 0 = 20
C(57.8392825,20)
Therefore, the B coordinate is (17.8392825,50) and the C coordinate is (57.8392825,20).
17.8392825 ≠ 18
57.8392825 ≠ 58
Both methods resulted in different answers. I will most likely revisit this problem afterwards.
Okay, never mind. The reason why there were different answers was because the side length wasn’t equal to the height which influenced the results. In other words, ignore method 2. Therefore, the coordinate is B(18,50) and C(58,20).
DeleteGrade 7 POTW
ReplyDeleteI will go straight to the math. We can first label each point in order of the lines.
A(0,0)
B(1,1)
C(-1,3)
D(-3,3)
E(-3,1)
F(-2,-2)
(Back to start)
There are many different ways to solve this, but I’ll just use the simple way. First, we can place these points on an actual grid. We can then create the shape which creates an irregular pentagon. Although there are 6 coordinates, A actually lies on the line between B and F. From here, there are many ways in which we can find out how many units^2 the figure is. One way is to divide the pentagon into shapes. There are tonnes of different ways to do this, so I’ll just choose one. We can divide the shape by creating a line segment from point E to point B creating a right trapezoid and a triangle. The area of a trapezoid is ((b1 + b2)/2)h and the area of a triangle is (b x h)/2. The trapezoid’s b1 is 4 units, b2 is 2 units, and h is 2 units. ((4 + 2)/2)2 = (6/2)2 = (3)2 = 6 units^2. The triangle’s b is 4 units, and h is 3 units. (4 x 3)/2 = 12/2 = 6 units^2. 6 + 6 = 12 units^2. Another way to do this is create a shape around the figure and subtract the extra area. By creating a square around the pentagon, we can get 3 right triangles. I will just go straight to the math.
(5 x 4) - ((1 x 3)/2 + (3 x 3)/2 + (2 x 2)/2) =
20 - (3/2 + 9/2 + 4/2 =
20 - (1.5 + 4.5 + 2) =
20 - 8 =
12 units^2
Therefore, the area of the figure Vroom made is 12 units^2.
Hi I forgot about this site until now. (hehehehe)
ReplyDeleteTin's Grade 8 POTW
This might be a bit complicated but... here we go (this might be a bit complicated to understand without a piece of paper)
I'll name the origin (0,0) U
So right off the bat I noticed that the distance from A to D was 50 since UA is 30 and UD is 40, and the Pythagoras theorem works.
Then, it said the parallelogram was 1340 units squared, and since AD (or BC) is 50, I can do 1340/50, which gives me a width of 26.8
Then, I drew a parallel line going through B and C, and labelled where it touched the Y-axis F, and where it touhed the x-axis G (I forgot about E for a second).
Then, I drew a line that went from U going straight and meeting perpendicularly with lines AD and BC (or FG). I labelled where it touched line BC "H" and where it touched line AD "J". I figured out that HJ was 26.8 since it was basically the width of the parallelogram.
Then I used some ratios to figure out that UJ is 24, the math is below;
DU/AD = UJ/UA, since AUD is a right angle, and so is AJU
DU=40, AD = 50, UA=30
40/50=UJ/30
UJ/30=1.25
30/1.25=24
UJ=24
Now I know that UJ is 24, and that JH is 26.8, that means UH is equal to 50.8
Now I use the Pythagorean theorem and some ratios again to measure the big triangle of HUF, and here is my quick maths
UF/UJ=50/40=5/4=1.25
UF=50*(50.8/40)
UF=50*1.27
UF=63.5
Now I drew a horizontal line on (0, 50), and labelled where it met the y-axis "L"
Now all I had to do was find LB, which would be equal to k.
UL=UF-FL
UL=63.5-50
UL=13.5
Now for the last time I use ratio and the Pythagorean theorem to find LB
LB/FL=4/3
LB/13.5=4/3
LB=4*(13.5/3)
LB=4*3.5
LB=18
k=18 B=(18,50)
Now, I find the coordinates of point C
I simply figured that C would be (+18, +20) from D, since the coordinate ratio of D:C = A:B, and B was (18,50), and A was (0,30). Since D was (40,0), that meant C would be at coordinates (58,20)
My answer:
The coordinates of B would be (18,50), and the coordinates of C would be at (58,20).
GOODNIGHT
Hi. I forgot about this website until now. (sad)
ReplyDeleteSo I wrote this previously but then there was some issue with blogger and just as I clicked publish, stuff happened and I was sad. So here it is again (if i fal asleep tmrw in class its bc of this)
HERE WE GO AGAIN (i swear if this gets deleted...)
Tin H (grade 8)
POTW for Week 13.
(you might have trouble reading this without drawing it out on a piece of paper)
First of all, let O be the origin point on the Cartesian plane (0,0)
So first, I noticed that AD was equal to 50, since AO was equal to 30, OD was 40, and that AOD was a 90 degree angle. Then, I drew a line through points B and C, and I marked where it intersected the y-axis "F", and where it intersected the x-axis "G" (i forgot about poor old E). Then, I saw that I could work out the dimensions of the parallelogram, since I already knew the length (AD) and its area (1340). So I simply did 1340/50, which gave me 26.8 as the width of the paralleogram. Then, I drew a line originating from O (0,0), and intersected lines AD and BC (or FG) perpendicularly (90 degrees). So I labelled where the line intersected line AD "J", and where it intersected line BC "H". This meant that HJ was already the height of the parallelogram, which was 26.8, and I needed to find the length of OH, meaning I just needed to find the length of OJ. For this, I used ratios of right triangle AOD to convert into triangle AJO. (angle AJO was also a right angle, so the Pythagorean theorem would work) For this, I simply wrote down;
AO/OJ=AD/DO
30/OJ=50/40
30/OJ=1.25
30/1.25=24
OJ=24.
So there I found that OJ equaled 24, which meant that OH equaled 50.8, now I plotted coordinate (0,50) as "I", since it would be important to mark this down since it is on the same horizontal axis as B while being on the y-axis (0,0). Also, this meant "k" could be calculated as the length of IB (lol international baccalaureate). To do this, I would convert triangle FHO to FIB. For triangle FHO, I again used ratios to find the length of FO (FH is not needed, but FO is since, FO-OI=FI). Below is my quick maths
FO/OH=AD/OD
FO/50.8=50/40
FO=50*(50.8/40)
FO=50*1.27
FO=63.5
So now that I know the value of length FO and OI, I can calculate FI by doing FO-OI=FI, which converts to 63.5=50=13.5 Now I used ratios to find the length of IB as part of triangle FIB using ratios with triangle AOD. Below is more quick maths
IB/FI=OD/AO
IB/13.5=4/3, which converts to
IB=4*(13.5/3)
IB=4*4.5
IB=18, which means
k=18, and the coordinates for B would be (18,50)
now, I find the coordinates of C;
I simply used a strategy where I figure that C is in the same position in correlation with D as B is with A. Thus, since the coordinates of A are (0,30), and coordinates of B are (18,50), I can see that the coordinates of C will be (D+18, D+20), or (58, 20).
Conclusion:
The coordinates of B are (18,50)
The coordinates of C are (58,20)
YAWN
GOODNIGHT
I plotted all the points, and made my image on paper. Then, I broke the image into 3 parts. A 2*2 square, a right triangle with a base of 2 and a height of 2, and an isosceles triangle with a height of 3, and a base of 4. By doing (2*2) + (2*2/2) + (4*3/2) = 4 + 2 + 6 = 12, I know that this figure takes up 12 units squared.
ReplyDeletePOTW:
ReplyDeleteI used grid paper to plot the points given. I divided the image into 3 different parts: A square with dimensions of 2 x 2, a right triangle with b = 2 and h = 2, and another triangle (isosceles) with b = 4, h = 3. I added all the totals up (2 x 2 = 4, 2 x 2 / 2 = 2, and 3 x 4 / 2 = 6). 4 + 6 + 2 = 12. Therefore, the area of the figure than Vroom traced out is 12 units squared.