7 : 14 = 1 : 2 ED=DF=x FC=2x x²+(2x)²=14² 5x²=196 Yellow Area = x*x = x² = 196/5 = 39.2

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

@erdemakca433

10 күн бұрын

I solve at the same way. I found the same.

sin²β + cos²β = 1 ------ sin(β) = a/14 ----- cos(β) = a/7 ---- (a²/196) + (a²/49) = 1 ---- a² = 39.2 ----- yellow area = 39.2 square units I love your channel

@PreMath

23 күн бұрын

Excellent! You are the best! Glad to hear that! You are very welcome! Thanks for sharing ❤️

@davidseed2939

16 күн бұрын

θ=smaller angle s=sinθ, c=cosθ consider sides of the square 14s=7c 2s=c 4ss=cc=1-ss ss=1/5 Area =(14s)²= 196/5=39.2

Los triángulos AED y DFC son semejantes→ Razón de semejanza s=7/14=1/2→ Si ED=b→AE=b/2→ b²+(b/2)²=7²→ b²=4*49/5=196/5=39,20 ud². Gracias y un saludo cordial.

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

Very beautiful video nice information thanks for sharing❤

@PreMath

23 күн бұрын

So nice of you Thanks for the feedback ❤️

Very good aproach!!

@PreMath

23 күн бұрын

Glad to hear that! Thanks for the feedback ❤️

Method using similar triangles and Pythagoras theorem: 1. Let side of yellow square be 2a. 2. Triangles ADE and DCF are similar, by corresponding sides proportionality equations, AE = a, CF = 4a 3. Hence AB = 3a and BC = 6a 4. In triangle ABC, by Pythagoras theorem, (7+14)^2 = (3a)^2 + (6a)^2 Hence a^2 = 49/5 5. Area of yellow square = (2a)^2 = 4a^2 = 196/5

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

Thanks Sir Thanks PreMath Very nice and useful We are learning more about Math. Good luck with glades ❤❤❤❤

@PreMath

23 күн бұрын

So nice of you, dear You are very welcome! Thanks for the feedback ❤️

Thank you!

@PreMath

23 күн бұрын

You are very welcome! Thanks for the feedback ❤️

39.2 The triangles are similar Let the side of the square = n Let the base of the the triangle on the right = p, then n/7 = p/14 14n= 7p 2n = p Therefore, the longest base of each triangle is TWICE the shortest base. Therefore, the length of the base of the big triangle = 3n (2n + n) Hence, the shortest base for the triangle on top is 0.5n. Hence, the length of the base of the big triangle = 1.5n (n + 0.5n) Hence, the sides of the big triangle are 1.5n , 3n and 21 (14+ 7) Let's employed Pythagorean Theorem (1.5n)^2 + (3n)^2 = 21^2 2.25n ^2 + 9n^2 = 441 11.25n^2 = 441 n^2= 39.2

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

As the big and the small triangle are similar and 7 is the half of 14, AE is half DF. So (1/2a)^2+a^2=7^2 => 1.25.a^2=49 =>a^2=39.2

The video ephasizes how many different paths you can dive into looking for your solution. Always something to learn from. However, reading the comments from so many viewers it is hard not to get the impression that the video is missing the obvoius ratio 7:14 staring at you even before you start the video. And that ratio makes the problem so easy, that most viewers find the solution in their heads. Maybe next time it would make sense to change the angles a little, so finding the ratio actually requires a pen and paper for most. Anyway, great work!

👍👍👍

@PreMath

24 күн бұрын

Excellent! Thanks for the feedback ❤️

Because the triangles CDF and DEA are similar with a length scaling of 2 then we can see that the smaller right angle triangle DEA comprises a hypotenues of length 7 and base side and height side lengths of lengths "a" and "1/2a" respectively. Using pythag we see that 7^2=a^2 + (1/2a)^2. Expanding out we see that 49 = a^2 + 1/4 a^2 = 5/4 a^2 Rearranging we see that a^2 (which also happens to be the area of yellow square = (4 . 49)/5 =39.2 units^2 Simple

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

Triangles AED and DFC are similar, FC/ED = 14/4 = 2, so FC = 2.c with c the side length of the square. Then in triangle DFC DC^2 = DF^2 + FC^2, or 14^2 = 4.c^2 + c^2. So c^2 = 14^2/5 The area of the square is c^2 = 14^2/5 = 196/5.

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

@ 6:59 , I absolutely love filling in the blanks of the Pagan Formula a² + b² = c². Life is good. 🙂

@PreMath

24 күн бұрын

👍😀 Excellent! Thanks for the feedback ❤️

Taking the secant of the shared triange of the smallest triangle and the biggest one as the same. Let length of 🟨 = a 7/ ( 49-a^2)^.5 = 21/ (a+(49-a^2)^.5 Divide by 7 and cross miltiply A + (49- a^2)^.5 = 3(49-a^2)^0.5 Remove the extra (49-a^2)^0.5 A = 2 (49- a^2) ^ 0.5 Square both sides A^2 = 4 ( 49- a^2) A^2 on one side 5a^2= 196 A^2= 196/5#

Три подобных треугольника. Немного по другому решала. Но тоже через подобие.

@PreMath

24 күн бұрын

Супер! Спасибо

If the square's sides are x, then FC = 2x due to the 7:14 ratio. By the same principle, AB is one and a half x so (3/2)x, making AE ((1/2)x Although the triangles are similar, it looks like I need an additional parameter from somewhere. The base is twice the height. tan(-1)(1/2) is 26.57deg so want ED/7 = cos(26.57) 7*cos(26.57) = 6.26... Square it for 39.19 un^2 (rounded) I have now looked. Your way was cleaner, not least because it gave an exact answer rather than relying on the close approximations of trigonometry. Thank you.

@PreMath

24 күн бұрын

👍😀 You are very welcome! Thanks for the feedback ❤️

Triangle AED is similar to DCF, so all their sides are proportional. DC is twice AD so FC is 2a.

Intercept theorem says FC/14=a/7, so FC= 2a, then finish as you did.

@PreMath

24 күн бұрын

Thanks for the feedback ❤️

No .1 similarity 2. Summation of area of triangles and square by assuming sides x,y and a little bit manipulation of sides length.. 3. Formula: ab / a + b = x. Delta ( abc) = x^2!

@PreMath

23 күн бұрын

Thanks for the feedback ❤️

The 📐 above the ⬛ and the 📐 to the right of the ⬛ are similar If each side of the ⬛ is x The 3 sides of the small 📐 are x/2, x, 7 The 3 sides of the large 📐 are x, 2x, 14 x^2 + 4x^2 = 14^2 area of the square = x^2 = (14^2)/5 = 196/5 = 39 + 1/5

Angle ADE = angle DCF. Cos DCF (ADE) = a / 7. Sin DCF = a /14. Tan = Sin / Cos. Tan DCF = (a / 14) / (a / 7) Tan DCF = (a /14) x (7 / a) Tan DCF = 1/2 = 0.5. Tan -1, DCF = 26.565 degrees. Sin 26.565 = a / 14. a = 14 sin 26.565 = 6.261. Area= 6.261^2 = 39.2.

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

I missed a trick here. With x as the square's side length, I could have gone for (3x)^2 + ((3/2)x)^2 = 21^2 9x^2 + (9/4)x^2 = 441 (45/4)x^2 = 441, ---> 45x^2 = 1764 ---> x^2 = 1764/45 = 39.2

@PreMath

23 күн бұрын

Well done! Thanks for sharing ❤️

(b-x)/7=x/14, b=3x/2, (a-x)/14=x/7, a=3x, a^2+b^2=21^2, (3x)^2+(3x/2)^2=441, 45x^2/4=441, x^2=441*4/45, x^2=39,2. Area of the shaded Square = 39,2.

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

3:33-6:33 ΔAED ~ ΔDFC (AA) => => ED/AD=FC/DC => FC=a•14/7=2a

S=39,2 square units

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

Let a be the side of the square. The two triangles AED and ABC are similar--> ED/BC=AD/AC=7/21=1/3-a/BC=1/3--> BC=3a -->FC=2a Consider the triangle DFC Sqa+Sq (2a)=sq14 Sqa=sq14/5 Area of the yellow square=196/5=39.2 sq units😊

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

CF/ED = 14/7 CF= 2ED DF = ED CF² + DF² = 14² (2ED)² + ED² = 14² (2a)² + a² = 14² 4a² + a² = 196 5a² = 196 a² = 196/5

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

Fairly simple. Answer I came up with in my head: 196/5 sq units Now let's see if I'm right: Let s be the side length of square BEDF, so BE = ED = DF = FB = s. Let ∠BAC = α and ∠ACB = β, where α and β are complementary angles that sum to 90°. As ∠DEA = 90°, ∠ADE = 90°- α = β, and as ∠EDF = 90°, ∠FDC = 180°-90°- β = α, so ∆DEA and ∆CFD are similar to ∆ABC and to each other. BA/FD = AC/DC BA/s = 21/14 = 3/2 BA = 3s/2 CB/DE = AC/AD CB/s = 21/7 = 3 CB = 3s BA² + CB² = AC² (3s/2)² + (3s)² = 21² 9s²/4 + 9s² = 441 45s²/4 = 441 s² = 441(4/45) = 49(4/5) = 196/5 = 39.2 sq units

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

Let x the side length of the square. The triangle right of the square and the triangle topof the square are similar..The hypothenuse of the triangle top of the squareis hallf the lengthof zje hpothhenuse of the square right of the square. So the length of the legs of the triangle right of the square are x and 2x. Accordng to pythagoras, we have the equation x^2+(2x)^2=14^2 x^2+4x^2=196 5x^2=196 x^2=39,2 That is also the area of the square. It is unnecessary to calculate the length of BC,because we can get the length of FC directly from the similarity of the 2 triangles rigthof the square and top of the square.

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

Let's find the area: . .. ... .... ..... The right triangles ADE and CDF are obviously similar. So with s being the side length of the square we can conclude: AE/DF = DE/CF = AD/CD AE/s = s/CF = 7/14 = 1/2 AE/s = 1/2 ⇒ AE = s/2 ⇒ AB = AE + BE = s/2 + s = 3*s/2 s/CF = 1/2 ⇒ CF = 2*s ⇒ BC = BF + CF = s + 2*s = 3*s The triangle ABC is also a right triangle. Therefore we can apply the Pythagorean theorem in order to obtain the area of the yellow square: AB² + BC² = AC² AB² + BC² = (AD + CD)² (3*s/2)² + (3*s)² = (7 + 14)² 9*s²/4 + 9*s² = 21² = 3²*7² s²/4 + s² = 7² (5/4)*s² = 49 ⇒ A(BEDF) = s² = 4*49/5 = 196/5 = 39.2 Best regards from Germany

@PreMath

24 күн бұрын

Excellent! You are the best!👍 Thanks for sharing ❤️

arccos(l/7)=arcsin(l/14)...√(1-l^2/49)=l/14...l^2=196/5

@PreMath

24 күн бұрын

Excellent! Thanks for sharing ❤️

(L' Aire ) /Le petit carré = 30,8 .le petit triangle = 11,76. Le grand triangle = 47,05 M² . Sur la base 3 , 4, 5 .

@PreMath

23 күн бұрын

Thanks for the feedback ❤️

a/14 = sinα a/7 = cosα tgα = sinα/cosα = (a/14)/(a/7) = 1/2 b = AE = a*tgα = a/2 a² + b² = 7² a² + a²/4 = 49 5a²/4 = 49 a² = 4*49/5 = 39.2 Keep It Simple

Let AE be x. (a+x):a=21:14=3:2. Hence 1/2a^2+a^2=49. Finished! a^2=39,2

Let's make it quicker Sin(Thida) = X/14 = sqrt(49-X^2)/7 7X = 14 sqrt(49-X^2) X = 2.sqrt(49-X^2) X^2 = 4(49-X^2) X^2 =196-4X^2 5X^2 =196 X^2 = 39.2

🔺 ABC BC Ii ED Hence AE/EB=7/14=1/2 AE/ED=1/2 (as EB =ED) ED=2 AE 🔺 AED AE^2 +ED^2=49 AE^2+(2AE^2)=49 AE=7/√5 2AE=14/√5 Area =(14/√5)^2=196/5 sq units Comment please

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

49/1,25 (1,25=1^2+0,5^2)

@PreMath

23 күн бұрын

Thanks for sharing ❤️

196/5

@PreMath

23 күн бұрын

Excellent! Thanks for sharing ❤️

STEP-BY-STEP RESOLUTION PROPOSAL : 01) BE = BF = FD = ED = X 02) FC = Y 03) 7 / X = 14 / Y 04) As : DC = 14 and AD = 7, 14 = (2 * 7); one can easily see that FC = 2X, and AE = X / 2 05) X^2 + (2X)^2 = 196 ; 5X^2 = 196 ; X^2 = 196 / 5 ; X^2 = 39,2 06) (X/2)^2 + X^2 = 49 ; X^2 / 4 + X^2 = 49 ; X^2 + ^2 = (49 * 4) ; 5X^2 = 196 ; X^2 = 196 / 5 ; X^2 = 39,2 07) It seems to me that the Yellow Area is Equal to 39,2 Square Units. Best Regards from the Department of Ancient (Indo-Arabic and Persian) Mathematical Thinking, Knowledge, and Wisdom. AL ANDALUS DISTRICT.

@PreMath

24 күн бұрын

Amazing!👍 Thanks for sharing ❤️

196 ?

245

I'll do it in CAD, much easier

@PreMath

23 күн бұрын

Thanks for the feedback ❤️

Самый простой способ.

Interesting but easy puzzle, (3s)^2+(3/2 s)^2=45/4 s^2=21^2, s^2=4×21^2/45=4×49/5=4×49×5/25, s=14/5 sqrt(5), bit the answer is simply 39.2.

@PreMath

24 күн бұрын

Excellent! Thanks for the feedback ❤️

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