Can you find area of the Green shaded Triangle? | (Quadrilateral) |

[@jamestalbott4499]

Thank you! Thank you for breaking the problem down, solving by using a ratio.

[@PreMath]

Happy to help!😀 You are very welcome! Thanks for the feedback ❤️

[@AndreyDanilkin]

S(ACB)=S(ACF)+S(AFB)=11s+9s; S(ACF)=S(CEF)+S(EFA)=3t+7t=11s -> t=11s/10; 20s=3t+25 -> s=250/167; S(CEF)=3t=3*(11/10) * 250/167 = 3*275/167 = 825/167

[@PreMath]

Excellent! Thanks for sharing ❤️

[@himo3485]

(3x+7x)(11x+9x) - 3x*11x = 200x² - 33x² = 167x² = 25 x² = 25/167 area of the Green triangle : 33x² = 825/167(cm²)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@連-x5h]

Here is another solution that does not require trigonometric functions. This use the "area ratio is equal to the base ratio". You can make an auxiliary line from E to B. Then Set triangle EAB to a. a + (3a/7 * 9/20) = 25 a = 3500/167 triangle CEF =3a/7 * 11/20 = 825/167

[@maroonshaded]

That's similar to my solution, I also used ratios, but with a little different steps.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sorourhashemi3249]

Thanks so challenging

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@santiagoarosam430]

Si AEF=a---> ECF=3a/7 . Si BFA=b---> FCA=11b/9 ---> a+b=25---> b=25-a ---> a+(3a/7)=11b/9=11(25-a)/9 ---> a=1925/167 ---> Área verde EFC =(3/7)(1925/167) =825/167 cm². Gracias y saludos

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Ibrahimfamilyvlog2097l]

Thanks so much sir

[@PreMath]

You are very welcome! Thanks ❤️

[@sorourhashemi3249]

Area of ∆CEF=16.5x^2 and area of ∆ABC= 100x^2,so 16.5 x^2+25=100x^2, ===>x=0.547 and area of green is 3(0.547)×11(0.547)/2= 4.94

[@PreMath]

That's a great approach, I like how you broke it down step by step! Thanks for sharing ❤️

[@MrPaulc222]

I'm not totally certain of my method here, but will give it a go: 10x * 20x = 25 + G (for Green area). 3x * 11x = G 200x^2 - 33x^2 = 167x^2 167x^2 = 25 G = 25*(33/167), so about 5. Comes out at 4.94 cm^2 I suspect the angle at C isn't 90, but it may not matter as it's about proportions. Well, it looks about right, visually, but I'll look at the video now. Wow! I can't believe I got that right, and I did it more simply, too. Wonders never cease.

[@PreMath]

Excellent! Keep rocking👍 Thanks for sharing ❤️

[@PrithwirajSen-nj6qq]

If ABF =9x, AFC=11x CEF=11x*3/10=33x/10 AEF =11x *7/10=77x/10 Then 9x +77x/10=25 >167x =250 >x =250/167 Area of 🔺 CEF =33x/10 =33*250/10*167= 33*25/167=825/167 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johnbrennan3372]

Can also be done by joining A to F and working out the ratios of two pairs of triangles having the same heights . Let area of triangle AFB= t, so area of triangleAFE = 25-t and let area of green triangle =g etc.

[@PreMath]

Thanks for the feedback ❤️

[@marcgriselhubert3915]

Area of green triangle CEF = (area of big triangle ABC).((3.x)/(10.x)).((11.x)/(20.x)) = (33/200).(area of big triangle ABC) So, by difference, area of AEFB = (167/200).(area of big triangle ABC) = 25. That gives: area of big triangle ABC = 25.(200/167), and finally: area of green triangle CEF = 25.(200/167).(33/200) = 25.(33/167) = 825/167. (Very easy.)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

16.5÷ 3.34=4.94

[@PreMath]

Excellent! Thanks for sharing ❤️

[@imetroangola17]

*Outra forma de ver a questão:* [ABC] - [CFE] = [AEFB] sen θ × (200x² - 33x²)/2 = 25 167(sen θ × x² )/2 = 25 *(sen θ × x² )/2* = 25/167 [CFE] = 33(sen θ × x² )/2 [CFE] = 33 × 25/167 *[CFE] = 825/167 cm²*

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ Label G and A as the areas of the green and ABC triangles respectively. We have: G = 33/2 . sq x . sin. (C) A = 100 . sqx. sin (C) --> G/A=33/200 --> G/33=A/200= (A-G)/(200-33) =25/167 -> G = 33. 25/167= 4.94 sq cm😅😅😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@misterenter-iz7rz]

Not difficult but need plenty of computation. 😢

[@PreMath]

Sure! Thanks for the feedback ❤️

[@TimothyCizadlo]

It seems to me that once you define the area of ABC as (1/2)*(AC)*(BC)*(sinθ) you end up with 25 = 100x²(sinθ) or 1/4 = x²(sinθ). You can then substitute this x²(sinθ) value into the law of sines for EFC which is 1/2*(3x)*(11x)*(sinθ), which simplifies to (33/2)*x²(sinθ) or 33/8, which is a different answer than presented. I think that the quadrilateral AEFB does not need to be found, and causes some extra errors. Would someone please confirm? Edit: I see where my error was. I took the area of the large triangle to be 25 not the area of the quadrilateral to be 25.

[@PreMath]

Thanks for the feedback ❤️

[@cabasantbab]

16.5 sqx

[@wackojacko3962]

Too really sharpen your math skills, ignore the caution that the presented diagram is not drawn to scale. Fire away and solve. Then circle back and do it the right way. Too learn maths you gotta do maths! 😊

[@PreMath]

That's a great point! 😀 Thanks for the feedback ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The areas of the triangles ABC and CEF can be calculated as follows: A(CEF) = (1/2)*CE*CF*sin(∠ECF) = (1/2)*(3x)*(11x)*sin(∠ECF) = (33x²/2)*sin(∠ECF) A(ABC) = (1/2)*AC*BC*sin(∠ACB) = (1/2)*AC*BC*sin(∠ECF) = (1/2)*(AE+CE)*(BF+CF)*sin(∠ECF) = (1/2)*(7x+3x)*(9x+11x)*sin(∠ECF) = (1/2)*(10x)*(20x)*sin(∠ECF) = (200x²/2)*sin(∠ECF) Now we are able to obtain the area of the green triangle CEF: A(ABFE) = A(ABC) − A(CEF) 25cm² = (200x²/2)*sin(∠ECF) − (33x²/2)*sin(∠ECF) 25cm² = (167x²/2)*sin(∠ECF) ⇒ A(CEF) = (33x²/2)*sin(∠ECF) = (33/167)*(167x²/2)*sin(∠ECF) = (33/167)*(25cm²) = (825/167)cm² ≈ 4.940cm² Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️