This Should Be Easy If You Love Geometry! | Try Solving It

[@bakrantz]

Rotate the small triangle onto the trapezoid. It makes a square of side 5. Thus there would be 5 squares of equal side length, meaning the total square area of 100 is divided by five to make the red area of 20.

[@kimba381]

Good one! Didn.t see that.

[@bakrantz]

The side length of the five equal sized smaller squares does not need to be known (I mistakenly said above that length was 5--thats the angle slicing the smaller squares actually). The only thing that matters is that all five of these squares comprising the larger 10x10 square have equal side length. I said above the length was 5 but really it is 2 root 5, which when squared also equals the answer of 20.

[@niravjhaveri]

Good one

[@babisstafulas7874]

We have 3 triangles: Left big ABC, left up small ADE and left down small BFC. 1) At ABC we have sides 10,5,5*sq(5) (P.T.) 2) ABC similar ADE then AE=2sq(5) and DE=sq(5) (PT) 3) ADE=BFC then FC=DE=sq(5) 4) At the end we have side of tetragon ( square ) EF=AC-AE-FC=5sq(5)-2sq(5)-sq(5)=2*sq(5). Then A=( 2*sq(5))^2==20 s.u. Thank's NGE

[@bpark10001]

There is another simpler way without "out of the box" construction. All triangles are similar to right triangle with legs 5, 10. So all other triangles have legs in 1:2 ratio. If we draw horizontal line through top right corner of square, & vertical line through top left corner of square, a triangle is formed with side of square as hypotenuse & legs in 1:2 ratio. This triangle is congruent to the 4 triangles in the corners so has legs X & 2X, & hypotenuse = 5. X² + (2X)² = 5². 5X² = 25, X² = 5. Square area = (2X)² = 4X² = (4)(5) = 20.

[@ritwikgupta3655]

There is a simpler method. The blue ∆ is similar to the green ∆. So, 10:5=(s+2x)/2x => 2=s/2x +1 Or, s/2x=1, ie 2x=s Now using Pythagoras in green ∆, (2s)^2+s^2=10.10 => 5s^2= 100, ie s^2 = 20 So areas of red square =s^2=20

[@santiagoarosam430]

Si tomamos los cuatro vértices del cuadrado exterior como centros y giramos 270º hacia la derecha el pequeño triángulo rectángulo adyacente, transformamos la figura inicial en una cruz griega compuesta por el cuadrado rojo en el centro y otros cuatro cuadrados blancos en los brazos→ Llamamos "a" al lado del cuadrado rojo→ (a/2)²+a²=5²→ a=2√5→ a²=4*5 =20 u². Interesante puzle. Gracias y un saludo.

[@ThePhantomoftheMath]

¡Excelente trabajo! ¡Muy bien!

[@1ClassicalMusicFan]

Pause at 0:27 and solve this problem quickly by just looking at the graph. (It takes much longer to write my solution in words as follows.) There are 3 different sets of right triangles, with 4 CONGRUENT right triangles in EACH set. Denote “the area of EACH of the right triangles in EACH set” by Al, Am and As, respectively, and denote “the area of the red region by Ar. (l, m, s and r mean large, medium, small and red, respectively.) We can show that Am = 4 · As. (1) The area of the BIG square is 10^2, or 100. (2) Al = (10 · 5)/2 = 25; 4 · Al =100, where 4 · As is counted twice and Ar is excluded. (3) By (2) and (3), 4 · As = Ar: by (1), Am = Ar. Now, looking at the BIG square, we have 100 = (4 · Am) + Ar = 5 Ar, or Ar = 20.

[@ludosmets2018]

The area of ​​the parallelogram between the green lines is equal to the area of ​​the large square (10x10=100) minus the area of ​​the two adjacent right triangles (10x5=50) = 50. The hypotenuse of these triangles is the base of the parallelogram and is equal to sqrt(10^2 + 5^2) = 5sqrt5. The height of the parallelogram is equal to the side of the small square and is 50/5sqrt5 = 10/sqrt5. The area of ​​the small square is (10/sqrt5)^2 = 20.

[@MrPhilippos96]

Thanks for sharing 👍

[@monroeclewis1973]

Hard to describe, but by extending lines outside the large square I created an even larger square rotated to the left, divided into a grid of 9 smaller squares identical to the small red square in area with the small red square centermost in the grid. The new figure looks like a diamond on end. Seeing the red square in the new context, it becomes apparent each square can be divided into 4 identical triangles because of the way diagonal lines cross the sides at the midpoint of the original square and end at the vertexes. When you count those triangles within the original square you will find 20 in all. 4 are in the red square. Therefore, the red square is 1/5 the area of the original large square, or 1/5 of 10 x 10, or 20. If nothing else this explanation proves a picture is worth a thousand words!😊

[@Patrik6920]

indeed true .. a few extra line can turn a difficult problem into an easier one or once u realised x=√5 it fairly easy see that s=2x=2√5 A=(2√5)² OR A=5/9 * ( 3(2√5))² (its just a larger 3 by 3 square rotated)

[@ThePhantomoftheMath]

@@monroeclewis1973 That'a a realy nice way of thinking. I like it. Great job!

[@zawatsky]

Легко и просто: берём одинаковые треугольнички, складываем с одинаковыми же прямоугольными трапециями - получаем крестик из пяти маленьких одинаковых квадратиков, площадь которого равно большому квадрату. Стало быть, верный ответ - пятая часть общей площади, т. е. 10*10/5=10*2=20.

[@ThePhantomoftheMath]

Отличная работа! Здорово!

[@erik5374]

The out of the box method was so obvious, that I knew the answer within 20 seconds.

[@LarryStorm-y5l]

😮The four lines are parallel. The triangles and quadrilaterals can form 4 more squares, identical with the inner square and they cover the original square. The total area is 10x10, so each small square has area 100/5 or 20.

[@goseigentwitch3105]

One of the large triangles (made from one of the lines from a corner of the square drawn to the mid-point of the other side of the square) has an area equal to 5 times that of one of the smaller triangles (with hypotenuse of 10) The red square in the center has an area equal to 4 times that of one of the smaller triangles. Since our square is made up of 4 of these large triangles, their area is 25. 4/5 of 25 is 20.

[@ThePhantomoftheMath]

Great! 🤟

[@Antony_V]

If the lines connect the midpoints of the white square then in all the small and greater right triangles the ratio between the legs is 2:1. Let's call "s" the side of the red square: s^2+(2s)^2=100, so Area=s^2= 20 squ

[@khalidalward3483]

Area = ⅕10² = 20 u²

[@TonyFisher-lo8hh]

I originally solved this when looking for the dissection of a square into 5 equal squares

[@babisstafulas7874]

There is a prove why AC is vertical at DE with the angles of triangles ABC and ADE Thank's again NGE

[@zunkman1]

Love that 2nd method!

[@ThePhantomoftheMath]

Me too!

[@rabotaakk-nw9nm]

All right triangles: (1/2/vʼ5) => A=((2/(2+2+1))(10/2)vʼ5)²=20 sq.un.

[@Gaurav_Yadav_7]

I just used my common sense and forgot maths and just assumed those squares to be congurent by myself and got 5 squares with same sides having area 100 and got answer in just 10 sec😂

[@saucepano6155]

@Gaurav_Yadav_7 its not common sense its called monkey brain

[@lukatolstov5598]

I saw this problem and IK that inner square size is ⅕ of the whole square size, so 10:5=2. 2² Units

[@nenetstree914]

20

[@Aili-kx3kj]

2:04 i dont understand this step of concluding the value of 5

[@shubhamdiwakar8098]

Those inner lines are joined at midpoint of side of square so 10/2 = 5 Side of square= 10 is given in question

[@Aili-kx3kj]

@shubhamdiwakar8098 Then my question is how do we establish that the lines are joined at midpoint

[@shubhamdiwakar8098]

@@Aili-kx3kj it is given in the question

[@shubhamdiwakar8098]

Read question carefully ​@@Aili-kx3kj

[@dlbattle100]

Chatgpt 4o couldn't get this but o1 nailed it!

[@brettgbarnes]

s/5=10/5√5

[@WONGKEANHAOMoe]

2020

[@ConradoPeter-hl5ij]

S=20 cuz A=10² A=4t"+S S=4(25-t") t"=(b"h")/2 t"=[(2b')(2h')/2] t"=[(4b'h')/2] t"=4[(b'h')/2] t"=4t' T=t"+t' T=(4t')+t' T=5t' t'=T/5 t'=[(10×5)/2]/5 t'=25/5 t'=5 t"=4t' t"=4×5 t"=20 S=4(25-t") S=4(25-20) S=4(5) S=20; ( S is the area of red square) =>S=t" t"=z+t'; (when, z is the trapezoid and t' is the smallest triangle) => A=4t"+S => A=4S+S => A=5S => S = A/5

[@ConradoPeter-hl5ij]

That proves if u rotate the smaller triangle onto the trapezoid, u get the same image of red square. And all these images plus the red square have the equal value of the greater square area (A). So, each figure is 1/5 part of A. Then u need to calculate S=A/5. (S is the red square area). S=A/5 S=(10²)/5 S=100/5 S=20

[@ConradoPeter-hl5ij]

Sorry my poor explanation, cuz i'm idiot

[@ThePhantomoftheMath]

You explained great! Nice one!

[@ConradoPeter-hl5ij]

@@ThePhantomoftheMath thank you for being kind to me.

[@frankhooper7871]

Interesting problem? Obvious answer!

[@saucepano6155]

rediculously easy dont be proud of yourself anyone

[@STAIRDROPPER]

🗣️🔥🔥🔥🔥💀

[@saucepano6155]

@ yes