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

9 күн бұрын

Excellent! Thanks for sharing ❤️

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

8 күн бұрын

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

@davidseed2939

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

9 күн бұрын

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Very beautiful video nice information thanks for sharing❤

@PreMath

8 күн бұрын

So nice of you 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

9 күн бұрын

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Very good aproach!!

@PreMath

8 күн бұрын

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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

9 күн бұрын

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Thank you!

@PreMath

8 күн бұрын

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

@PreMath

8 күн бұрын

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

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

@PreMath

9 күн бұрын

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👍👍👍

@PreMath

9 күн бұрын

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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#

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

8 күн бұрын

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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

9 күн бұрын

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

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!

S=39,2 square units

@PreMath

9 күн бұрын

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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

8 күн бұрын

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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

9 күн бұрын

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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

8 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

8 күн бұрын

Well done! Thanks for sharing ❤️

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

9 күн бұрын

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

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

@PreMath

9 күн бұрын

Супер! Спасибо

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

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

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

(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

8 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

8 күн бұрын

Excellent! Thanks for sharing ❤️

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

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

9 күн бұрын

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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

8 күн бұрын

Thanks for the feedback ❤️

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

@PreMath

9 күн бұрын

👍😀 Excellent! Thanks for the feedback ❤️

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

8 күн бұрын

Excellent! Thanks for sharing ❤️

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

(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

8 күн бұрын

Thanks for the feedback ❤️

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

8 күн бұрын

Excellent! Thanks for sharing ❤️

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

@PreMath

9 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

9 күн бұрын

Amazing!👍 Thanks for sharing ❤️

196/5

@PreMath

8 күн бұрын

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

@PreMath

8 күн бұрын

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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

9 күн бұрын

Excellent! Thanks for the feedback ❤️

245

I'll do it in CAD, much easier

@PreMath

8 күн бұрын

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Самый простой способ.

Какой же нудный этот индус!