I used a combination of geometry and trigonometry. Step 1 is to calculate the smallest angle in the right triangle at the bottom of the array of squares. It's just the arctan of 1/3. Step 2 is to subtract this angle from 45 degrees to get the smallest angle in the red triangle (it's about 26.565 degrees). Step 3 is to calculate the base of the red triangle using the Pythagorean Theorem: √40. Step 4 is to calculate the height of the red triangle (the opposite side) using the tangent function, since we know the angle and the adjacent side (the base). Step 5 is just to compute the area of the red triangle since we now have its base and height.
Beautiful method. At 4:17, a perpendicular line could have been drawn from CD to vertex A, forming two triangles, ACP and ADP, in which ACP is similar to BCF. If CP= n, then DP =4-n, but since ACP is similar to BCF, and the relationship between the base and height is one is three times the others, the 4-n = 3n Hence, n =1 Hence, 4-n =3 (height Hence area = 3 (height) * 4 (base) * 1/2 = 6 6 + 4 = 10 answer
Another way to solve this is to set up cartesian space & solve for coordinates of point A. Assign point B as (0,0). Equation for AB is Y = -X. Slope of CB is -1/3, so slope of CA is 3. Equation for CA is Y = 20 + 3X. Solving intersection we get Y = 5 & X = -5. From this we get length CA = √10 & CB = √40. Area ABC = (1/2)(√10)(√40) = 10.
Second solution was very nice, thought about it for a little while but couldn't do it without Pythagoras. I used a coordinate system (origin at point F) to find the length of AC as the height of the triangle (with BC as base): Gradient of AB=gradient of BD=-1. AB through point B(6,0) gives AB:y=-x+6 BC has gradient -1/3 --> gradient of AC is 3. AC through point C(0,2) gives AC:y=3x+2 AB=AC --> -x+6=3x+2 --> x=1, y=-1+6=5 --> point A(1,5) AC=√((1-0)²+(5-2)²)=√10 BC=√(6²+2²)=2√10 Area triangle=1/2*BC*AC=1/2*2√10*√10=10 Happy 2025
The lower right corner = a The right-side corner of the red triangle = 45° - a Sin(a) = 1/√10 Sin(45° - a) = Sin(45°) x Cos(a) - Cos(45°) x Sin(a) = (1/√2)(3/√10) - (1/√2)(1/√10) = 1/√5 Cos(45° - a) = 2/√5 Red area = (2√10 x √10)/2 = 10
I solved it using sines and cosines. First, by Pithagoras BC = 2√10. Let the angle CBF be theta. So sin theta = 1/√10 and cos theta = 3/√10. The angle DBF is 45º, therefore the angle DBC is 45º - theta. Then I calculated cos (45 - theta) = cos 45. cos theta + sin 45.sin theta = 2/√5, and AB = BC / (2/√5) = 5√2. Then AC² + (2√10)² = (5√2)², therefore AC = √10. Finally, the area is AC.BC/2 = √10. 2√10 / 2 = 10.
Треугольник вправо от С прикладываем к квадрату с верхним левым углом В, но тут же вспоминаем, что столько же придётся отнять внизу (малый влево от В). В среднем квадрате остаётся закрашенной только половина. Поскольку мы уже маленький посчитали, смело прибавляем ещё половину квадрата с диагональю DB - получаем целый квадрат 4. Ещё равная площадь 4, подсчитанная сверху, всего 8. Кстати, такой же результат можно получить движением т. В влево, на стык оснований первого и второго квадратов, по теореме о сохранении площади при движении вершины параллельно основанию, получить диагонально построенный квадрат двойной площади.
Don't do that. Drop a perp from A onto CD. Since CA has gradient 3, it goes 1 to the right as it goes 3 up. Since CE has gradient -1, it goes 5 to the left as it goes 5 up. I.e. these two lines intersect at A. ABE has height 3 and base 4 gives area 6, and BEC has base 4 and height 2, gives area 4. Thus answer is 10.
Happy New Year! I really love the problems you present. I did it a little different... I came up with formulas for the lines that go through each side of the triangle, assuming your point F is at (0,0): y = 3 x + 2 y = -1/3 x + 2 y = -x + 6 From this, I was able to get the corners as (0,2), (6,0), and then compute the third corner as (1,5). That makes it easy to find the lengths of the two sides adjacent to the right triangle, sqrt(10) and sqrt(40), making the area = 10.
Si dibujamos el triángulo ABC en una matriz de 3*4 celdas cuadradas de lado 2, vemos que AC es la hipotenusa de un triángulo rectángulo de catetos de longitudes 1 y 3→ AC=√10 y CB=2√10→ Área ABC =√10*2√10/2 =10 u². Gracias y un saludo cordial.
Thanks for sharing!❤ Could you help me with something unrelated: I have a okx wallet with some money, and I have the seed phrase: iron observe slam major mad decorate feed photo awesome vast kitchen faint. What's the best way to send them to Binance?🤔
10 The area of the rectangle =12 ( 2 *6) The area of the triangle with bases 2 and 6 = 6. Let's call it triangle ABC The area of the triangle inside the square = 2 The area of the red shaded INSIDE the triangle = 6- 2 =4 ********* The area of the remaining red shaded above the RECTANGLE : Notice that one of its angles = 45 degrees due to the 45-degree triangle inside the square Draw a perpendicular line from the base of the red shaded above the RECTANGLE to its vertex, forming a 45- 45-90 degree triangle, and another triangle, STU Let the side to the left of the height =n, Hence, the other side =4-n, Hence, the height also = 4-n and hence the base triangle STU is n, and its height is 4-n Notice that the triangle with bases 2 and 6 is similar to the triangle STU Since the height of triangle ABC is three times its base. Hence 4-n = 3n Hence 4= 4n Hence, n=1 Hence, the height of the shaded triangle ABOVE the rectangle = 3 (4-n or 4-1) And the base = 4 Hence, the area = 3 * 4 * 1/2 = 6****** Hence the total area of the red shaded = 6 + 4 = 10 (something square)
It's interesting that the length x in this geometric construction is always 1/2 the length of the small square. How about a geometric proof of this proposition. No algebra, no numbers, nothing but geometric proof using proven theorems.
Why would you make far too complex a solution to a problem that has a much simpler solution. Find the length CB of hypotenuse of triangle CBF using Pythagorean theorem, You have the base of 6 and the height of 2, [root 40]. Also find angle CBF. You know angle DBF is 45 degrees. Subtract angle CBF from angle DBF and this gives you angle ABC [26.56]. With base CB and angle ABC you can find length AC. Now having height AC and base CB, you have all you need to calculate for area of triangle ACB. The units are not mentioned. Inches or centimetres? But the answer is 10 units sq.
CB squared is equal to the sum of CF squared and FB squared. This is very much Pythagoras theorem. Your statement that we would avoid Pythagoras theorem is not true.
Hi! First of all, thank you for your comment and for sharing your thoughts! I want you to know that I truly respect Pythagoras and his monumental contributions to mathematics. The crossed image in the thumbnail was never intended to be disrespectful-it was simply a metaphorical way to spark curiosity and highlight how we approach or challenge classical ideas in new and creative ways. Trust me, no "harm" was intended to the great Pythagoras! I deeply value your feedback and will keep this perspective in mind for future thumbnails.
@@ThePhantomoftheMath For my part, I positively evaluate your personality, as well as your knowledge of mathematics. I am subscribed to your channel, and I follow you. I appreciate your kind response (despite my own aggression). I am Greek, and I assume you understand my special appreciation, and sympathy for Pythagoras. These for now... I wish you a Happy New Year...
@@AthSamaras Thank you so much for your kind words and for being a subscriber to my channel! I truly appreciate your support and understanding. It means a lot to me, especially knowing how deeply Pythagoras and his legacy resonate with you as a Greek. Your feedback helps me grow, and I’m grateful for the opportunity to connect with passionate viewers like you. Wishing you a very Happy New Year as well! May it bring you joy, success, and plenty of mathematical inspiration. 😊
I used a combination of geometry and trigonometry. Step 1 is to calculate the smallest angle in the right triangle at the bottom of the array of squares. It's just the arctan of 1/3. Step 2 is to subtract this angle from 45 degrees to get the smallest angle in the red triangle (it's about 26.565 degrees). Step 3 is to calculate the base of the red triangle using the Pythagorean Theorem: √40. Step 4 is to calculate the height of the red triangle (the opposite side) using the tangent function, since we know the angle and the adjacent side (the base). Step 5 is just to compute the area of the red triangle since we now have its base and height.
@@j.r.1210 Really nice! Great job! 👍
Meglio : arctan1/3=15 45-15=30 i cateti sono : 2V10 e V10 A=10 ! (triang. rettang. con ang 30° cateti 1:2 )
Same here.
Beautiful method.
At 4:17, a perpendicular line could have been drawn from CD to vertex A, forming
two triangles, ACP and ADP, in which ACP is similar to BCF.
If CP= n, then DP =4-n, but since ACP is similar to BCF, and the relationship between the base
and height is one is three times the others, the 4-n = 3n
Hence, n =1
Hence, 4-n =3 (height
Hence area = 3 (height) * 4 (base) * 1/2 = 6
6 + 4 = 10 answer
@@devondevon4366 That work like a charm! Nicely done! 👏🏼
Another way to solve this is to set up cartesian space & solve for coordinates of point A. Assign point B as (0,0). Equation for AB is Y = -X. Slope of CB is -1/3, so slope of CA is 3. Equation for CA is Y = 20 + 3X. Solving intersection we get Y = 5 & X = -5. From this we get length CA = √10 & CB = √40. Area ABC = (1/2)(√10)(√40) = 10.
@@bpark10001 Really creative! I love it! 👏🏼
Δ ABC ~ Δ CBE (all angles equal)
(in the notations V stands for square root)
=> AC/CE = CB/EB
=> AC = CE*CB/EB = 2V2*2V10/4V2 = V10
A = CB*AC/2 = 2V10*V10/2 = 10
I used coordinate geometry and fact that perpendicular lines have slopes that multiply to -1.
Me too, like JR13751
5:00 ΔAEC ~ ΔCEB (AA) =>
AE/CE=CE/BE => AE=vʼ2; AB=5vʼ2
[ABC]=½AB•CE=½(5vʼ2)2vʼ2=10 sq.un.
Second solution was very nice, thought about it for a little while but couldn't do it without Pythagoras.
I used a coordinate system (origin at point F) to find the length of AC as the height of the triangle (with BC as base):
Gradient of AB=gradient of BD=-1. AB through point B(6,0) gives AB:y=-x+6
BC has gradient -1/3 --> gradient of AC is 3. AC through point C(0,2) gives AC:y=3x+2
AB=AC --> -x+6=3x+2 --> x=1, y=-1+6=5 --> point A(1,5)
AC=√((1-0)²+(5-2)²)=√10
BC=√(6²+2²)=2√10
Area triangle=1/2*BC*AC=1/2*2√10*√10=10
Happy 2025
That works! Nice! Happy 2025 to you as well!
The lower right corner = a
The right-side corner of the red triangle = 45° - a
Sin(a) = 1/√10
Sin(45° - a) = Sin(45°) x Cos(a) - Cos(45°) x Sin(a) = (1/√2)(3/√10) - (1/√2)(1/√10) = 1/√5
Cos(45° - a) = 2/√5
Red area = (2√10 x √10)/2 = 10
I solved it using sines and cosines. First, by Pithagoras BC = 2√10. Let the angle CBF be theta. So sin theta = 1/√10 and cos theta = 3/√10. The angle DBF is 45º, therefore the angle DBC is 45º - theta. Then I calculated cos (45 - theta) = cos 45. cos theta + sin 45.sin theta = 2/√5, and AB = BC / (2/√5) = 5√2. Then AC² + (2√10)² = (5√2)², therefore AC = √10. Finally, the area is AC.BC/2 = √10. 2√10 / 2 = 10.
That work perfectly!
Треугольник вправо от С прикладываем к квадрату с верхним левым углом В, но тут же вспоминаем, что столько же придётся отнять внизу (малый влево от В). В среднем квадрате остаётся закрашенной только половина. Поскольку мы уже маленький посчитали, смело прибавляем ещё половину квадрата с диагональю DB - получаем целый квадрат 4. Ещё равная площадь 4, подсчитанная сверху, всего 8. Кстати, такой же результат можно получить движением т. В влево, на стык оснований первого и второго квадратов, по теореме о сохранении площади при движении вершины параллельно основанию, получить диагонально построенный квадрат двойной площади.
Отлично сделано!
10
Don't do that. Drop a perp from A onto CD. Since CA has gradient 3, it goes 1 to the right as it goes 3 up. Since CE has gradient -1, it goes 5 to the left as it goes 5 up. I.e. these two lines intersect at A. ABE has height 3 and base 4 gives area 6, and BEC has base 4 and height 2, gives area 4. Thus answer is 10.
Happy New Year! I really love the problems you present. I did it a little different... I came up with formulas for the lines that go through each side of the triangle, assuming your point F is at (0,0):
y = 3 x + 2
y = -1/3 x + 2
y = -x + 6
From this, I was able to get the corners as (0,2), (6,0), and then compute the third corner as (1,5). That makes it easy to find the lengths of the two sides adjacent to the right triangle, sqrt(10) and sqrt(40), making the area = 10.
That's a really nice use of analytical geometry! Awesome! Happy New 2025!
Si dibujamos el triángulo ABC en una matriz de 3*4 celdas cuadradas de lado 2, vemos que AC es la hipotenusa de un triángulo rectángulo de catetos de longitudes 1 y 3→ AC=√10 y CB=2√10→ Área ABC =√10*2√10/2 =10 u².
Gracias y un saludo cordial.
d=s√2 это частный случай Теоремы Пифагора - не считается.
Thanks for sharing!❤ Could you help me with something unrelated: I have a okx wallet with some money, and I have the seed phrase: iron observe slam major mad decorate feed photo awesome vast kitchen faint. What's the best way to send them to Binance?🤔
Who WAS (!) Pythagoras? Now he is dead, but his philosophy is still "alive" in our minds (at least in most of ours)!
He will always be "alive and present"...we can't do much without him...
φ = 30°; ∎ABCD → AD = BC = 2 = AB/3 → AB = 6 = CD = CE + DE = 2 + 4;
∆ ABD → sin(DAB) = 1; BD = 2√10; ∆ BFD → sin(FDB) = 1; ABD = δ; DBF = θ →
δ + θ = 3φ/2 → tan(δ) = 1/3 → tan(θ) = 1/2 → area ∆ BFD = 10
Very nice!
10
The area of the rectangle =12 ( 2 *6)
The area of the triangle with bases 2 and 6 = 6. Let's call it triangle ABC
The area of the triangle inside the square = 2
The area of the red shaded INSIDE the triangle = 6- 2 =4 *********
The area of the remaining red shaded above the RECTANGLE :
Notice that one of its angles = 45 degrees due to the 45-degree triangle inside the square
Draw a perpendicular line from the base of the red shaded above the RECTANGLE
to its vertex, forming a 45- 45-90 degree triangle, and another triangle, STU
Let the side to the left of the height =n,
Hence, the other side =4-n, Hence, the height also = 4-n
and hence the base triangle STU is n, and its height is 4-n
Notice that the triangle with bases 2 and 6 is similar to the triangle STU
Since the height of triangle ABC is three times its base.
Hence 4-n = 3n
Hence 4= 4n
Hence, n=1
Hence, the height of the shaded triangle ABOVE the rectangle = 3 (4-n or 4-1)
And the base = 4
Hence, the area = 3 * 4 * 1/2 = 6******
Hence the total area of the red shaded = 6 + 4 = 10 (something square)
@@devondevon4366 Excelent job! 🤟
@@ThePhantomoftheMath Thank You
CD = x + 3x = 4x
CD = 2 + 2 = 4
4x = 4
x = 1
AC² = x² + (3x)²
AC² = 1² + 3²
AC² = 1 + 9
AC² = 10
AC = √10
BC² = CF² + BF²
BC² = 2² + 6²
BC² = 4 + 36
BC² = 40
BC = 2√10
[ABC] = (1/2)·BC·AC
[ABC] = (1/2)·(2√10)·(√10)
[ABC] = 10
@@brettgbarnes Excellent! 👍
It's interesting that the length x in this geometric construction is always 1/2 the length of the small square. How about a geometric proof of this proposition. No algebra, no numbers, nothing but geometric proof using proven theorems.
Why would you make far too complex a solution to a problem that has a much simpler solution. Find the length CB of hypotenuse of triangle CBF using Pythagorean theorem, You have the base of 6 and the height of 2, [root 40]. Also find angle CBF. You know angle DBF is 45 degrees. Subtract angle CBF from angle DBF and this gives you angle ABC [26.56]. With base CB and angle ABC you can find length AC. Now having height AC and base CB, you have all you need to calculate for area of triangle ACB. The units are not mentioned. Inches or centimetres? But the answer is 10 units sq.
CB squared is equal to the sum of CF squared and FB squared. This is very much Pythagoras theorem. Your statement that we would avoid Pythagoras theorem is not true.
Субтитры мешают!
You may be a genius in mathematics, but you have no right to put (X) in the face of Pythagoras. It is disrespectful..!!
Hi! First of all, thank you for your comment and for sharing your thoughts! I want you to know that I truly respect Pythagoras and his monumental contributions to mathematics. The crossed image in the thumbnail was never intended to be disrespectful-it was simply a metaphorical way to spark curiosity and highlight how we approach or challenge classical ideas in new and creative ways. Trust me, no "harm" was intended to the great Pythagoras! I deeply value your feedback and will keep this perspective in mind for future thumbnails.
@@ThePhantomoftheMath For my part, I positively evaluate your personality, as well as your knowledge of mathematics. I am subscribed to your channel, and I follow you. I appreciate your kind response (despite my own aggression). I am Greek, and I assume you understand my special appreciation, and sympathy for Pythagoras.
These for now... I wish you a Happy New Year...
@@AthSamaras Thank you so much for your kind words and for being a subscriber to my channel! I truly appreciate your support and understanding. It means a lot to me, especially knowing how deeply Pythagoras and his legacy resonate with you as a Greek. Your feedback helps me grow, and I’m grateful for the opportunity to connect with passionate viewers like you.
Wishing you a very Happy New Year as well! May it bring you joy, success, and plenty of mathematical inspiration. 😊