Math mini-contest problem for Day 20 on D.Buzz for February 2021 π
Math problem for Day 20 π
Turner placed a square canvas roof 16 square meters big above level ground as shade. However, one of the roof's corners was placed 1 meter higher than the others. If the sun is shining directly above, what area of the ground is shaded by the roof?
0
0
0.000
My guess is that there is no solution. There is not enough information to solve the question in my opinion. Without the dimensions of the roof, I feel that I can't find out the angle and the correct proportions of of the shaded area versus the roof.
Though it might just because I spent 10 minutes and don't feel like working anymore.
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The roof has the dimensions 16 square meters, and is a square. Therefore, it has a side measure of 4 meters. π€ The roof was placed just above ground, so it has a height of 0 meters. π€
You might want to look at the solution (after I post it which is soon). π
Answer 15.48331 sq. m.
Steps: It's a kite shape (adj. edges equal len). Solve for short edges using pythag:
4^2=X^2+1 => X=sqrt(15)
. And hypot of4x4 = sqrt(32)
Then solve forarea of iso tri with base sqrt(32), sides sqrt(15)
. Thenadd half of 4*4 sq. m.
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edit: lol, reading this again I think my assumption was wrong: it is bent, creating a projection of two triangles, the one across the three poles on the same level (8m2) plus the one with one pole 1m up as isosceles triangle with b=sqrt(32) amd a=sqrt(15), making a total area of 15.48 m2
I think there is some information missing. Assuming the 4x4m roof is not bent in z direction, there are multiple ways the other corners could be:two corners on levela
, two corners on levela+1m
one corner on levela
, one corner on levela+1m
, the other two corners on levelsa + x
anda + 1m - x
(with0<x<1m
).For the first case, the area would be4m * sqrt(15)m = 15.49 m2
.sqrt(15)
comes via pythagoras:1^2 + x^2 = 4^2
<=>x = sqrt(15)
.The second case would create a non-square shade projection. I can't get my mind around this atm...Congrats, you were upvoted from this account because you were in Top 25 engagers yesterday on STEMGeeks .
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I made the same assumptions and looks like we solved in the same way: a kite shape is basically two isosceles triangles in area. Ahhh Mathland where material doesn't stretch, sag, crease or flap in the breeze.
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ah, indeed!
yep, there might be multiple ways to get there, but the result is unambiguous :D
Answer for Day 20 Math Problem
Solution
The original problem did not directly specify how far above the ground the square canvas roof is. Since the sun is overcast and very far away, moving the roof up several meters will not have a noticeable difference on the shadow formed (by the way, clothes on a clothesline should be placed higher to "dry faster because they are nearer the sun"). π Therefore, we can assume that the 3 corners of the roof are attached to the ground (so the heights of the 3 corners are 0 meters).
Other information that are not explicitly shown in the given but are logically true for the Math problem:
The tarpaulin is not elastic. ("mathland" according to @eturnerx-dbuzz)However, the problem is that it is not specified whether the tarpaulin is can be bent or folded. Therefore, just any person capable of solving Math challenge problems can make his or her own scenario, then formulate an answer based on his or her preferred details.
We should use variables to refer to the points on the square.
Our goal here is to be able to represent a kite representing the shade's produced shadow. The surface area of a kite is equal to the product of its diagonals divided by 2. The corners of the kite (arranged counterclockwise) should be ABCE, where AC and BE are the diagonals.
From the given data, we can now derive the following.
We now have a right triangle BEF where F is the right angle. We know that line segment BE is 4β2 meters and EF is 1 meter, so the missing length is line segment BF. We should use the Pythagorean Theorem to get BF, which is BF = β((BE)Β² - (EF)Β²) = β((4β2)Β² - 1Β²) = β31 meters. If the canvas cannot be folded, corners A and C will be pulled to D. It is up to the solver to continue solving from this.
Winner: @jfang003 π
1 HIVE has been sent to @jfang003's Hive account. π°
Special awardee: @minus-pi ποΈ
For a fast and accurate reply for my solution, @minus-pi has also been awarded 1 HIVE. However, according to the rules, @minus-pi should have been active on D.Buzz for at least 7 days within the last 14 days to qualify as winner (next time), unless a user active on D.Buzz vouches for them.
Mentions: @ahmadmanga (@ahmadmangazap) and @holovision π€
Special mentions: @chrisrice and @jancharlest π€―
I think I disagree :) Let me know if I miss something...
I think the direct link from B to E is not
4*sqrt(2)
, because the triangle ABC is completely on the ground. So BF is not sqrt(31), but 2*sqrt(2) + the red part in my image.Your understanding of the original problem is that the roof can bend like the one you showed on your picture. My solution didn't allow bending. π€
I see that the problem really lacks enough details for an unambiguous answer. π Thanks for sharing! π
hmm, I see what you mean. But if you don't allow the roof to be bent, then the angle between AB and BC is not 90 degree anymore, so AC cannot be 4*sqrt2?
I now understand what you mean there. I'll make a less ambiguous Math problem next time! π
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Don't worry, it was really fun :D
I have updated my solution and the official answer for the problem! π
https://peakd.com/@savvyplayer/re-savvyplayer-qovwpq
hehe, I still believe your approach can be calculated as well, but that would be quite a bit more difficult then...
Thanks a lot for the 1 HIVE, highly appreciated! :D
I'm sorry, I didn't notice the dbuzz rule - can you point me to the rules of your contests? I found you via stemgeeks
The rules for my Math mini-contest are at https://peakd.com/@savvyplayer/official-rules-for-my-math. ππ
Thanks! OK, then I'm not eligible for at least a couple of days. My acc was just created around a day ago and this is my first buzz - another frontend to learn :D
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