It's the long awaited sequel to my first Geometry Problem style video! Hope you all enjoy and let me know if you find any other methods to solving the problem 🧠
@@AngusCalvert The (12 + y)² is a binomial formula and gets solved like this: (a + b)² = a² + 2ab + b² a in this case is 12 and b is y. The full way to solve it is: (a + b)² = (a + b) * (a + b) Here you solve it via the 'FOIL' method, aka 'First, Outer, Inner, Last - you multiply first the two first variables of the respective brackets, then the two outer variables, then the two inner variables, and then the last two variables of each respective bracket: (a + b) * (a + b) = a² + a*b + a*b + b² You can sum up the two [a*b] parts and get a² + 2ab + b² Now in our specific case, it'd go as follows: x² = (12 + y)² = 12² + 2*12*y + y² = 144 + 24y + y²
The (12 + y)² is a binomial formula and gets solved like this: (a + b)² = a² + 2ab + b² a in this case is 12 and b is y. The full way to solve it is: (a + b)² = (a + b) * (a + b) Here you solve it via the 'FOIL' method, aka 'First, Outer, Inner, Last - you multiply first the two first variables of the respective brackets, then the two outer variables, then the two inner variables, and then the last two variables of each respective bracket: (a + b) * (a + b) = a² + a*b + a*b + b² You can sum up the two [a*b] parts and get a² + 2ab + b² Now in our specific case, it'd go as follows: x² = (12 + y)² = 12² + 2*12*y + y² = 144 + 24y + y²
208 + 64 + 16 = 288 (total area of square without the green one) Now we try squaring whole numbers 17²= 289 (illogical to be our full square) 18²=324 (is probably the area of our full square) 19²=361 (might be) So we take our added up area and subtract it from our potential full square 324 - 288 = 36 (the area of our green square) Now we know the blue is 16 in area And Red is 64 Obviously the green is somewhere in between and 36 is definitely between them so it's 36
there's always a possibility that the green square is not a square that has sides with integer values (for example 5.5, 5.9, etc). proving it with a solution will always be the best :))
![[ASMR] A Fun Geometry Problem](http://i.ytimg.com/vi/Ff48ReLtoC0/mqdefault.jpg)
![[ASMR] A Fun Geometry Problem](/img/tr.png)
![[ASMR] The Monty Hall Problem](http://i.ytimg.com/vi/Am8uuH7KV6E/mqdefault.jpg)
![[ASMR] Can I Solve this Math Puzzle Before You Fall Asleep?](http://i.ytimg.com/vi/FsRQlQFOw-w/mqdefault.jpg)
![[ASMR] Can YOU Solve This Maths Problem?](/img/1.gif)
It's the long awaited sequel to my first Geometry Problem style video! Hope you all enjoy and let me know if you find any other methods to solving the problem 🧠
I didn't understand why expanding the brackets in the second section resulted in 24y. other than that great video and well explained.
You need more likes im subscribing!
@@AngusCalvert The (12 + y)² is a binomial formula and gets solved like this: (a + b)² = a² + 2ab + b²
a in this case is 12 and b is y.
The full way to solve it is:
(a + b)² = (a + b) * (a + b)
Here you solve it via the 'FOIL' method, aka 'First, Outer, Inner, Last - you multiply first the two first variables of the respective brackets, then the two outer variables, then the two inner variables, and then the last two variables of each respective bracket:
(a + b) * (a + b) = a² + a*b + a*b + b²
You can sum up the two [a*b] parts and get a² + 2ab + b²
Now in our specific case, it'd go as follows:
x² = (12 + y)² = 12² + 2*12*y + y² = 144 + 24y + y²
@@Aghul thank you very much. I now understand double brackets and how to expand them
10:53 why +24y?
The (12 + y)² is a binomial formula and gets solved like this: (a + b)² = a² + 2ab + b²
a in this case is 12 and b is y.
The full way to solve it is:
(a + b)² = (a + b) * (a + b)
Here you solve it via the 'FOIL' method, aka 'First, Outer, Inner, Last - you multiply first the two first variables of the respective brackets, then the two outer variables, then the two inner variables, and then the last two variables of each respective bracket:
(a + b) * (a + b) = a² + a*b + a*b + b²
You can sum up the two [a*b] parts and get a² + 2ab + b²
Now in our specific case, it'd go as follows:
x² = (12 + y)² = 12² + 2*12*y + y² = 144 + 24y + y²
208 + 64 + 16 = 288 (total area of square without the green one)
Now we try squaring whole numbers
17²= 289 (illogical to be our full square)
18²=324 (is probably the area of our full square)
19²=361 (might be)
So we take our added up area and subtract it from our potential full square
324 - 288 = 36 (the area of our green square)
Now we know the blue is 16 in area
And Red is 64
Obviously the green is somewhere in between and 36 is definitely between them so it's 36
I solved the problem since the start of the video and stay to enjoy some good asmr
Awesome!
Enjoyed working on this prob, it was kinda easy but really enjoyed it!
I just saw that the green box was in between them and red is 8x8 and blue is 4x4 so green is 6x6 and I added them all together
not how it really works. could’ve ended up with the wrong answer just by guessing it lol
there's always a possibility that the green square is not a square that has sides with integer values (for example 5.5, 5.9, etc). proving it with a solution will always be the best :))
You whisper is super relaxing and your channel super original! I also do ASMR videos (but not about mathematics ahah)
Dido you should do something regarding physics
Can’t wait for the stream
Me too! Hope to see you there
Can you please do an ASMR preparing you for geometry regions?
Can you make a video about the cos. and sin. rule?
Sure thing! I'll look into a trig video soon
actualy, I've calculated the resoult, before starting the vid, but sill whatched it
Damm i guessed the area of the green square correctly. I feel very smart now
Can you please tutor me in Algebra 2
😴