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

Are you ready to solve a challenging geometry puzzle? In this video, we dive into an exciting math problem: finding the area of the red triangle. With a twist that might surprise even seasoned problem-solvers, this puzzle will test your understanding of triangles, rectangles, and proportions.

🎥 Chapters:
0:00 Presenting the Puzzle
1:50 Method 1: Using Geometry
7:50 Method 2: NO PYTHAGORAS

🎯 What to expect in this video:
- A detailed breakdown of the solution.
- Insights into key geometry concepts.
- A step-by-step guide to finding the red triangle’s area.

📢 Can YOU solve it before the solution is revealed? Watch now and test your skills! Don’t forget to like, comment, and subscribe for more engaging geometry puzzles and math challenges.

📌 Share your solution or thoughts in the comments below!

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#GeometryPuzzle #RedTriangleArea #MathChallenge #LearnGeometry #ProblemSolving

[bpark]

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.

[j.r.]

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.

[devondevon]

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

[JR]

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

[Rajeev_Walia]

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.

[zawatsky]

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.

[davidsousaRJ]

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.

[rabotaakk-nwnm]

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

[gregorymagery]

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

[AbeIJnst]

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
Happy 2025

[slytherinbrian]

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.

[cyruschang]

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

[gelbkehlchen]

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

[santiagoarosam]

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.

[RAG]

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.

[cRNzvj]

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[zawatsky]

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

[com]

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

[murdock]

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

[devondevon]

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 =
Hence the total area of the red shaded = 6 + 4 = 10 (something square)