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!
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.
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
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.
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
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
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!
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#
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😊
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.
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
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.
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
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
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
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!
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.
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.
7 : 14 = 1 : 2 ED=DF=x FC=2x
x²+(2x)²=14² 5x²=196
Yellow Area = x*x = x² = 196/5 = 39.2
Excellent!
Thanks for sharing ❤️
I solve at the same way. I found the same.
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
Excellent! You are the best!
Glad to hear that!
You are very welcome!
Thanks for sharing ❤️
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
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!
Very beautiful video nice information thanks for sharing❤
So nice of you
Thanks for the feedback ❤️
Very good aproach!!
Glad to hear that!
Thanks for the feedback ❤️
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.
Excellent!
Thanks for sharing ❤️
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
Excellent!
Thanks for sharing ❤️
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
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.
Excellent!
Thanks for sharing ❤️
👍👍👍
Excellent!
Thanks for the feedback ❤️
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
Excellent!
Thanks for sharing ❤️
(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.
Excellent!
Thanks for sharing ❤️
Thank you!
You are very welcome!
Thanks for the feedback ❤️
Thanks Sir
Thanks PreMath
Very nice and useful
We are learning more about Math.
Good luck with glades
❤❤❤❤
So nice of you, dear
You are very welcome!
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
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
Excellent!
Thanks for sharing ❤️
Simpel ...👍👍
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!
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#
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😊
Excellent!
Thanks for sharing ❤️
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.
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
Три подобных треугольника. Немного по другому решала. Но тоже через подобие.
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
3:33-6:33 ΔAED ~ ΔDFC (AA) =>
=> ED/AD=FC/DC => FC=a•14/7=2a
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.
👍😀
You are very welcome!
Thanks for the feedback ❤️
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
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
Well done!
Thanks for sharing ❤️
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
Excellent! You are the best!👍
Thanks for sharing ❤️
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
S=39,2 square units
Excellent!
Thanks for sharing ❤️
Intercept theorem says FC/14=a/7, so FC= 2a, then finish as you did.
Thanks for the feedback ❤️
@ 6:59 , I absolutely love filling in the blanks of the Pagan Formula a² + b² = c². Life is good. 🙂
👍😀
Excellent!
Thanks for the feedback ❤️
arccos(l/7)=arcsin(l/14)...√(1-l^2/49)=l/14...l^2=196/5
Excellent!
Thanks for sharing ❤️
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!
Thanks for the feedback ❤️
(L' Aire ) /Le petit carré = 30,8 .le petit triangle = 11,76. Le grand triangle = 47,05 M² . Sur la base 3 , 4, 5 .
Thanks for the feedback ❤️
🔺 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
Excellent!
Thanks for sharing ❤️
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
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.
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
Solving 21/(a+sqrt(7^2 - a^2))=14/a for a
a = 39.2E^2
I used geometry
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.
Amazing!👍
Thanks for sharing ❤️
Triangle AED is similar to DCF, so all their sides are proportional. DC is twice AD so FC is 2a.
196/5
Excellent!
Thanks for sharing ❤️
14²/5=39.2
49/1,25 (1,25=1^2+0,5^2)
Thanks for sharing ❤️
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.
Excellent!
Thanks for the feedback ❤️
I think ED is NOT DF.
245
36 dim²
I'll do it in CAD, much easier
Thanks for the feedback ❤️
196 ?
AD=DF, so 7*7=49, chearS
Самый простой способ.
Какой же нудный этот индус!
여기저기 정치판검사들이 판치는 군요