Can you calculate area of the Green shaded region? | (Square) |

[@jamestalbott4499]

Thank you!

[@PreMath]

Thanks for watching! 🙏

[@PrithwirajSen-nj6qq]

In my previous solution I wrote CD = 3sinα And derived an equation 3sinα = 3cos α + 2sinα Now sin^2 α=9cos^2α > 10sin^2α= 9 sin α= 3/√10 3sin α= 9/√10 CD= 9/√10 Area of square = CD^2 = 81/10=8.1 sq units Area of rt 🔺 =3 units Area of green region =8.1-3= 5.1 sq units

[@ИванПоташов-о8ю]

Let DC=3x. CDE and EAF are similar. Then AE=2x, ED=x. ED²+DC²=EC²; (3x)²+x²=3²; 10x²=9. 9x²=8.1 - the square area. 8.1-0.5*3*4=5.1 - green shaded area.

[@user-jm7cx5zc9s]

This is shortcut. Same as mine

[@PreMath]

Thanks for sharing your solution! ❤️

[@ludosmets2018]

Once we know that triangles AEF and EDC are similar we conclude that CD = 3k and AE = 2k. So, ED = 1k. Pyth: (1k)^2 + (3k)^2 = 9. (10k)^2 = 9. k^2 = 9/10. k = sqrt (9/10). Area square = (3k)^2 = 8.1.

[@georgebliss964]

Triangles CDE & EAF are similar. Therefore 3/2 = X/AE. Then AE = 2X/3 So ED = X/3. Triangle CDE. (X/3)^2 + X^2 = 9. (Pythagoras) X^2/9 + X^2 = 9. Multiply both sides by 9. X^2 + 9X^2 = 81. 10X^2 = 81. X^2 = 8.1 = area of square. Area of triangle CEF = 3. So 8.1 - 3 = 5.1.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@imetroangola17]

*Solution:* Note that right triangles AFE and ECD are similar and we easily have: ∠ECD=∠AEF=α. Like this, AE = 2cos α, ED= sin α and DC= 3 cos α. Now, ABCD is a square, therefore: AE + ED = DC 2cos α + sin α = 3 cos α cos α = 3sin α. Using the fundamental theorem of trigonometry, we have: sin² α + cos² α = 1. Give, sin² α = 1/10 → cos² α = 9/10. Consequently, DC² = 9 cos² α, That is, DC² = 81/10 = 8.1, therefore: Green Area = DC² - [FCE] Green Area = 8,1 - 2×3/2 = 8,1 - 3 *Green Area = 5,1 square units*

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sy2008]

Excellent 🎉🎉🎉

[@cyruschang1904]

Upper left and bottom triangles are similar, their linear ratio = 2:3, their area ratio = 4:9 If each side of the square is x, the three sides of the two triangles are upper left: 2x/9 - 2x/3 - 2 bottom: x/3 - x - 3 The three sides of the right side triangle can then be derived as: 7x/9 - x - √13 Area of the green = (4/27 + 1/3 + 7/9)(x^2)/2 = (17/27) x^2 The center white area = (10/27)x^2 = (2)(3)/2 = 3 green area = 3(17/10) = 51/10 = 5.1

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

@ Thank YOU

[@ASHOKKUMAR-fe6jb]

My answer is 6

[@alexundre8745]

Bom dia Mestree Obrigado pela aula

[@PreMath]

Good day😀 You are very welcome! Thanks for the feedback ❤️

[@alexundre8745]

@PreMath Grato

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let's have a look at the interior angles of the triangles AEF and CDE: ∠EAF = 90° ∠AEF = α ∠AFE = 90° − α ∠CDE = 90° ∠DCE = α ∠CED = 90° − α Therefore these two triangles are similar and we can conclude: AF/DE = AE/CD = EF/CE = 2/3 With s being the side length of the square we obtain: AE/CD = 2/3 AE/s = 2/3 ⇒ AE = 2*s/3 ⇒ DE = AD − AE = s − 2*s/3 = s/3 Since CDE is a right triangle, we can apply the Pythagorean theorem: CD² + DE² = CE² s² + (s/3)² = 3² s² + s²/9 = 9 10*s²/9 = 9 ⇒ s² = 81/10 = 8.1 Now we are able to calculate the area of the green region: A(green) = A(ABCD) − A(CEF) = s² − (1/2)*CE*EF = 8.1 − (1/2)*3*2 = 8.1 − 3 = 5.1 This looks very similar to the problem released on November 14th.🙂 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sorourhashemi3249]

Thanks. Easy❤

[@PreMath]

You are very welcome! ❤️ Thanks for the feedback ❤️

[@MrPaulc222]

AFE and EDC are similar, so the side lengths from one to the other are 2/3 or 3/2 depending on perspective. Call the square's sides x. FC = sqrt(13). 3/x = 2/(AE), so (AE) = (2/3)x. As (AE) = (2/3)x, (ED) = (1/3)x. Therefore, (AF) = (2/9)x (due to 2/3 length). (FB) = (7/9)x. x^2 + ((7/9)x)^2 = 13. x^2 + (49/81)x^2 = 13 81x^2 + 49x^2 = 1053 130x^2 = 1053 x^2 = 1053/130 White triangle area = 3, or 390/130 1053 - 390 = 663 Green area = 663/130 un^2 or 5.1 un^2 I've now watched. I could have made it a little easier for myself by reducing 1053/130 to 81/10, but I didn't spot that bit. Thanks again.

[@PreMath]

Excellent! You are very welcome! Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ Let'scheck the right triangle ΔAEF, we have : ∠FAE= 90° ∠AEF= α ∠AEF= β For the right triangle ΔDCE, we have : ∠EDC= 90° ∠DCE= α ∠CED= β ⇒ ΔAEF ~ ΔDCE [EF]/[CE]= [AE]/[DC]= [FA]/[ED] by considering the first two equations, we can write: [EF]/[CE]= [AE]/[DC] [EF]= 2 [CE]= 3 [AE]= x [DC]= a ⇒ 2/3 = x/a x= 2a/3 ⇒ [ED]= a-x [ED]= a - 2a/3 [ED]= a/3 Step-2) Let's consider the righ triangle ΔDCE, by applying the Pythagorean theorem, we can write: [ED]² + [DC]²= [CE]² [ED]= a/3 [DC]= a [CE]= 3 ⇒ (a/3)² + a²= 3² a²/9 + a²= 9 a² + 9a² =81 ⇒ a²= 81/10 A(ΔECF)= 2*3/2 A(ΔECF)= 3 square units Agreen= a² - A(ΔECF) Agreen= 81/10 - 3 Agreen= 51/10 Agreen= 5,1 square units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

.82×1.82÷2=1.1682+2.82×1÷2+2.822÷2=5

[@wackojacko3962]

I guess if I wanna overcome the anxiety of doing maths, I better start studyin! It's weird though cuz the more I know the harder it gets. ...but I don't care, they don't call me Mr "know it all" for nothin. 🙂

[@PreMath]

😀😀 Thanks for sharing ❤️

[@svaypros]

Green shaded area=51/10=5.1 square units

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) AB = AD = BC = CD = X lin un 02) Square [ABCD] Area = X^2 sq un 03) Triangle [CEF] Area = 3 sq un 04) Green Area = (X^2 - 3) sq un 05) AF = Y lin un 06) ED = Z lin un 07) BF = (X - Y) lin un 08) AE = (X - Z) lin un 09) System of Equations : a) X^2 + Z^2 = 9 b) 2X^2 + Y^2 - 2XY = 13 c) (X^2 + Z^2) + Y^2 - 2XZ = 4 ; 9 + Y^2 - 2XZ = 4 ; 5 + Y^2 - 2XZ = 0 10) Solutions : a) X ~ 2,846 b) Y ~ 0,6325 c) Z ~ 0,95 11) GA = (X^2 - 3) sq un ; GA ~ (8,099716 - 3) sq un ; GA ~ 5,099716 sq un Therefore, OUR BEST ANSWER : Green Shade Area approx. equal to 5,099716 Square Units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@nenetstree914]

51/10

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

Fullarea 8-3=5

[@misterenter-iz7rz]

Extremely difficult 😢

[@sergioaiex3966]

Solution: Let's label the segments: AE = a DE = b The ∆ AEF and ∆ CDE are similar, thus we have proportions: 2/a = 3/(a + b) 2 (a + b) = 3a 2a + 2b = 3a 3a - 2a = 2b a = 2b ... ¹ Therefore, the side of square will be 3b Applying Pythagorean Theorem, in ∆ CDE, to calculate "b" (3b)² + (b)² = (3)² 9b² + b² = 9 10b² = 9 b² = 9/10 b = 3/√10 The side of the Square is 3b, therefore we have: 3b = 3 × 3√10 = 9/√10 Side = 9/√10 Green Area = Square Area - ∆ CEF Área Green Area = (9/√10)² - ½ 3 . 2 Green Area = 81/10 - 3 Green Area = 8.1 - 3 Green Area = 5.1 Square Units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Haven't we already seen this problem not so long ago? (Or a very similar one)

[@PreMath]

Thanks for the feedback ❤️

[@unknownidentity2846]

Yes, a very similar problem was presented on this channel on November 14th. Instead of the second side length, the area of the inscribed right triangle was given. Best regards from Germany

[@marcgriselhubert3915]

@@unknownidentity2846 OK, thanks. Greetings from France.

[@misterenter-iz7rz]

Extremely difficult 😢

[@PreMath]

Thanks for the feedback ❤️