Can you find area of the Yellow shaded Square? | (Triangle) |

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Learn how to find the area of the Yellow shaded Square inscribed in the right Triangle. Important Geometry and Algebra skills are also explained: similar Triangles; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Комментарии : 99

@himo3485 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@erdemakca433 2 месяца назад

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

@unknownidentity2846 3 месяца назад

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 3 месяца назад

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

@RAG981 3 месяца назад

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

@PreMath 3 месяца назад

Thanks for the feedback ❤️

@santiagoarosam430 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@ОльгаСоломашенко-ь6ы 3 месяца назад

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

@PreMath 3 месяца назад

Супер! Спасибо

@quigonkenny 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@eduardozappi6397 3 месяца назад

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 3 месяца назад

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

@davidseed2939 2 месяца назад

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

@is7728 Месяц назад

Note that ∆ADE and ∆CDF are similar triangles, the the side length ratio of 1 : 2. Let DE = DF = x, so that AE = 1/2 (DF) = 1/2 x. x^2 + (1/2) x^2 = 7^2⠀(Pyth. thm.) x^2 = 39.2 And that's the area of the yellow part!

@hongningsuen1348 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@AmirgabYT2185 3 месяца назад

S=39,2 square units

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@devondevon4366 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@LuisdeBritoCamacho 3 месяца назад

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 3 месяца назад

Amazing!👍 Thanks for sharing ❤️

@PrithwirajSen-nj6qq 3 месяца назад

🔺 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 3 месяца назад

Excellent! Thanks for sharing ❤️

@juergenilse3259 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@Alishbavlogs-bm4ip 3 месяца назад

Very beautiful video nice information thanks for sharing❤

@PreMath 3 месяца назад

So nice of you Thanks for the feedback ❤️

@MrPaulc222 3 месяца назад

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 3 месяца назад

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

@marcgriselhubert3915 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@gelbkehlchen Месяц назад

a = side of the square. Similarity of the triangle EDA and FCD: a/7 = FC/14 ⟹ a/7 = √(14²-a²)/14 |*7*14 ⟹ 14a = 7*√(14²-a²) |()² ⟹ 196a² = 49*(196-a²) ⟹ 196a² = 9604-49a² |+49a² ⟹ 245a² = 9604 |/245 ⟹ a² = 9604/245 = 39,2 = area of the yellow square

@stevetitcombe939 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@MrPaulc222 3 месяца назад

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 3 месяца назад

Well done! Thanks for sharing ❤️

@GhdVhs 3 месяца назад

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

@phungpham1725 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@kranik1 2 месяца назад

Какой же нудный этот индус!

@cyruschang1904 3 месяца назад

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

@sergeyvinns931 3 месяца назад

(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 3 месяца назад

Excellent! Thanks for sharing ❤️

@georgebliss964 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@МаргаритаМ-в5й 2 месяца назад

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

@iveswidmer1027 2 месяца назад

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

@brettgbarnes 3 месяца назад

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 3 месяца назад

Excellent! Thanks for sharing ❤️

@ermajisetiawan4019 2 месяца назад

Simpel ...👍👍

@yalchingedikgedik8007 3 месяца назад

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

@PreMath 3 месяца назад

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

@HamedAhmadi-tj6vf 2 месяца назад

Solving 21/(a+sqrt(7^2 - a^2))=14/a for a a = 39.2E^2 I used geometry

@AndreasPfizenmaier-y7w 3 месяца назад

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

@wackojacko3962 3 месяца назад

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

@PreMath 3 месяца назад

👍😀 Excellent! Thanks for the feedback ❤️

@tlmotorscbb 2 месяца назад

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

@preetjitsingh328 3 месяца назад

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#

@JSSTyger 3 месяца назад

196/5

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@rabotaakk-nw9nm 2 месяца назад

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

@jamestalbott4499 3 месяца назад

Thank you!

@PreMath 3 месяца назад

You are very welcome! Thanks for the feedback ❤️

@Tom-zu2yc 3 месяца назад

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!

@marcelowanderleycorreia8876 3 месяца назад

Very good aproach!!

@PreMath 3 месяца назад

Glad to hear that! Thanks for the feedback ❤️

@tijanimaths6006 3 месяца назад

👍👍👍

@PreMath 3 месяца назад

Excellent! Thanks for the feedback ❤️

@giuseppemalaguti435 3 месяца назад

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

@PreMath 3 месяца назад

Excellent! Thanks for sharing ❤️

@himadrikhanra7463 3 месяца назад

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 3 месяца назад

Thanks for the feedback ❤️