Can YOU Find the Red Triangle’s Area? | Geometry Puzzle

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  • Опубликовано: 3 фев 2025

Комментарии • 67

  • @chrishelbling3879
    @chrishelbling3879 Месяц назад +2

    Solution #2 = beautiful.

  • @bpark10001
    @bpark10001 Месяц назад +7

    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.

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад +1

      @@bpark10001 Really creative! I love it! 👏🏼

    • @jassit
      @jassit Месяц назад +1

      Cool method.
      Also once you solve for A(-5,5) and C is (-6,2) and B(0,0) => Area by Determinant method is 1/2 Det |(0,0,1),(-5,5,1),(-6,2,1)| = 1/2 *20 = 10

  • @hcgreier6037
    @hcgreier6037 2 дня назад

    Nice problem! What about simple reasoning?
    Stack 3 of the same "4"-squares upon the first given lower left square, getting an "L" shape of 6 squares. The lower leg of the triangle is √(6² + 2²) = √40 = 2√10. And so is the diagonal of the 3 "stacked" squares, and the upper vertex of the triangle lies in the middle of that diagonal (= the midpoint of the second "stacked" square), so the left leg is 2√10/2 = √10. Area of the triangle is therefore A = (1/2)·2√10·√10 = 10 sq. units.

  • @j.r.1210
    @j.r.1210 Месяц назад +5

    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.

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад +1

      @@j.r.1210 Really nice! Great job! 👍

    • @sertorio1040
      @sertorio1040 Месяц назад +1

      Meglio : arctan1/3=15 45-15=30 i cateti sono : 2V10 e V10 A=10 ! (triang. rettang. con ang 30° cateti 1:2 )

    • @nonec5246
      @nonec5246 Месяц назад

      Same here.

    • @angeluomo
      @angeluomo 16 дней назад

      Also my method. Exact same approach and logic.

  • @vojtodrozdik4755
    @vojtodrozdik4755 21 день назад +1

    If you map the red triangle to grid of 9 squares (3 X 3) than it is evident, it lies on diagonal of this grid. The diagonal of s small square with area 4 is root(8). Let use “d”. It is evident from this grid that this red square has its hight “d” and divides this red square into triangles T1 and T2. Area(T1) is 2”d” * “d” * ½. Area(T2) is ”d” * ½ ”d” * ½. (Similar triangles T1 and T2, 2 to 1, and 1 to * ½.) The area of red triangle is therefore: 2,5 “d” * “d” * ½. (2,5 * root (8) * root (8) * ½ = 2,5 * 8 * ½ = 10)

  • @davidsousaRJ
    @davidsousaRJ Месяц назад +2

    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.

  • @devondevon4366
    @devondevon4366 Месяц назад +2

    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

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад

      @@devondevon4366 That work like a charm! Nicely done! 👏🏼

  • @JR13751
    @JR13751 Месяц назад +3

    I used coordinate geometry and fact that perpendicular lines have slopes that multiply to -1.

  • @gregorymagery8637
    @gregorymagery8637 Месяц назад

    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

  • @slytherinbrian
    @slytherinbrian Месяц назад

    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.

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад

      That's a really nice use of analytical geometry! Awesome! Happy New 2025!

  • @AbeIJnst
    @AbeIJnst Месяц назад

    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

  • @zawatsky
    @zawatsky Месяц назад

    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.

  • @Rajeev_Walia
    @Rajeev_Walia Месяц назад

    We can also solve it using Trigonometry (and Geometry). One of the acute angles (call it c) of the red triangle which is inside the 3rd square on the right is given by c = (45 - a) where a is the acute angle in the right triangle that is half of the rectangle formed by the 3 squares. This right triangle has two of its sides as 2 and 6 and so tan(a) = 2/6 = 1/3. Now, tan(45 - a) can be found by using tan(A-B) formula which comes out to be 1/2. Now, the side of the red triangle that is almost vertical is d tan(c) where d is the diagonal of the rectangle formed by the 3 squares and d = sqrt(40). Finally, the area of the red triangle is 1/2 * d * d tan(c) = 1/2 * d^2 * tan(c) = 1/2 * 40 * 1/2 = 10.

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад +1

      @@Rajeev_Walia Yes! That works perfectly! Great job! 👏🏼

  • @jassit
    @jassit Месяц назад

    Let angle CBF = x => angle ABC = 45-x
    tan x = 1/3 and BC = 2.sqrt(10)
    tan (45-x) = (1-tan x)/(1+tan x) = 1/2
    Area ABC = 1/2*BC* AC = 1/2 * BC*BC *tan(45-x)
    = 1/2 * 40 * 1/2 = 10

  • @rabotaakk-nw9nm
    @rabotaakk-nw9nm Месяц назад +1

    5:00 ΔAEC ~ ΔCEB (AA) =>
    AE/CE=CE/BE => AE=vʼ2; AB=5vʼ2
    [ABC]=½AB•CE=½(5vʼ2)2vʼ2=10 sq.un.

  • @santiagoarosam430
    @santiagoarosam430 Месяц назад

    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.

  • @GiovanniRusso-d6g
    @GiovanniRusso-d6g 2 дня назад

    very fast with a Cartesian reference

  • @gelbkehlchen
    @gelbkehlchen Месяц назад

    Solution:
    s = side of a square = √4 = 2,
    A = top left corner of the left square,
    B = bottom right corner of the right square,
    C = top corner of the red triangle,
    D = top left corner of the right square,
    E = bottom left corner of the left square.
    AB = √(6²+2²) = √40,
    Angle ABE = arctan(2/6) = arctan(1/3),
    Angle CBA = 45°-arctan(1/3),
    AC = AB*tan(angleCBA) = AB*tan[45°-arctan(1/3)]
    Area of ​​the red triangle = AB*AC/2 = AB*AB*tan[45°-arctan(1/3)]/2
    = AB²*tan[45°-arctan(1/3)]/2 = 40*tan[45°-arctan(1/3)]/2
    = 20*tan[45°-arctan(1/3)] = 10

  • @RAG981
    @RAG981 Месяц назад +1

    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.

  • @cyruschang1904
    @cyruschang1904 Месяц назад

    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

  • @zawatsky
    @zawatsky Месяц назад

    d=s√2 это частный случай Теоремы Пифагора - не считается.

  • @nenetstree914
    @nenetstree914 Месяц назад

    10

  • @84com83
    @84com83 Месяц назад

    Who WAS (!) Pythagoras? Now he is dead, but his philosophy is still "alive" in our minds (at least in most of ours)!

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад

      He will always be "alive and present"...we can't do much without him...

  • @murdock5537
    @murdock5537 Месяц назад

    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

  • @devondevon4366
    @devondevon4366 Месяц назад

    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)

  • @brettgbarnes
    @brettgbarnes Месяц назад

    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

  • @neillawrence4198
    @neillawrence4198 Месяц назад

    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.

  • @Teapot69
    @Teapot69 14 дней назад

    The 4 lies in the white portion, if for the whole square it should be in both parts of that square.

  • @nonec5246
    @nonec5246 Месяц назад

    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.

  • @shankarlalsaraswat6283
    @shankarlalsaraswat6283 Месяц назад

    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.

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад

      @@shankarlalsaraswat6283 Method 2 is without Pythagoras, not Method 1. 😉

  • @wasimahmad-t6c
    @wasimahmad-t6c Месяц назад

    6.666666

  • @Andrej-v9i
    @Andrej-v9i Месяц назад

    Субтитры мешают!

  • @משהנאון-י6מ
    @משהנאון-י6מ 13 дней назад

    לא נכון

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  12 дней назад

      תודה על ההערה! אשמח אם תוכל להסביר למה אתה מתכוון כדי שאוכל לבדוק את זה. 😊

    • @משהנאון-י6מ
      @משהנאון-י6מ 12 дней назад

      קודם כל היכן כתוב זו משולש ישר זוית 90 מעלות לפי זה קודם כל הדרך שלך נהרס לחלוטין

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  12 дней назад

      תודה על ההערה! אני ממליץ לבדוק שוב את הסרטון בסביבות 0:30, שם אני מציין במפורש שהקו המדובר ניצב לאלכסון שנבנה. בנוסף, זה מוצג גם בתמונה המקדימה (thumbnail). כך שזה כן תנאי שניתן במסגרת השאלה. תודה שצפית וששיתפת את מחשבותיך!

  • @AthSamaras
    @AthSamaras Месяц назад

    You may be a genius in mathematics, but you have no right to put (X) in the face of Pythagoras. It is disrespectful..!!

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад +4

      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.

    • @AthSamaras
      @AthSamaras Месяц назад

      @@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...

    • @ThePhantomoftheMath
      @ThePhantomoftheMath  Месяц назад +2

      @@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. 😊

  • @משהנאון-י6מ
    @משהנאון-י6מ 12 дней назад

    @משהנאון-י6מ
    לפני 0 שניות
    קודם כל היכן כתוב זו משולש ישר זוית 90 מעלות לפי זה קודם כל הדרך שלך נהרס לחלוטין