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
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.
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
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.)
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.
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😅😅😅
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.
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
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! 😊
Thank you! Thank you for breaking the problem down, solving by using a ratio.
Happy to help!😀
You are very welcome!
Thanks for the feedback ❤️
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
*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*
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
(3x+7x)(11x+9x) - 3x*11x = 200x² - 33x² = 167x² = 25 x² = 25/167
area of the Green triangle : 33x² = 825/167(cm²)
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.
Thanks so challenging
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
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
Thanks so much sir
You are very welcome!
Thanks ❤️
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
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.)
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.
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😅😅😅
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.
16.5÷ 3.34=4.94
Not difficult but need plenty of computation. 😢
16.5 sqx
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
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! 😊