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Can you find area of the Yellow shaded Square? | (Triangle) |
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- Опубликовано: 13 авг 2024
- 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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7 : 14 = 1 : 2 ED=DF=x FC=2x
x²+(2x)²=14² 5x²=196
Yellow Area = x*x = x² = 196/5 = 39.2
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I solve at the same way. I found the same.
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
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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
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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
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
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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
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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.
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Very beautiful video nice information thanks for sharing❤
So nice of you
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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
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Very good aproach!!
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Intercept theorem says FC/14=a/7, so FC= 2a, then finish as you did.
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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.
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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😊
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Три подобных треугольника. Немного по другому решала. Но тоже через подобие.
Супер! Спасибо
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#
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.
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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!
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Thanks Sir
Thanks PreMath
Very nice and useful
We are learning more about Math.
Good luck with glades
❤❤❤❤
So nice of you, dear
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Thank you!
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👍👍👍
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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
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🔺 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
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(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.
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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
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
@ 6:59 , I absolutely love filling in the blanks of the Pagan Formula a² + b² = c². Life is good. 🙂
👍😀
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S=39,2 square units
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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.
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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.
👍😀
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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
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
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Simpel ...👍👍
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!
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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.
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(L' Aire ) /Le petit carré = 30,8 .le petit triangle = 11,76. Le grand triangle = 47,05 M² . Sur la base 3 , 4, 5 .
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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
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!
Triangle AED is similar to DCF, so all their sides are proportional. DC is twice AD so FC is 2a.
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
arccos(l/7)=arcsin(l/14)...√(1-l^2/49)=l/14...l^2=196/5
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196/5
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14²/5=39.2
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.
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36 dim²
49/1,25 (1,25=1^2+0,5^2)
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245
AD=DF, so 7*7=49, chearS
I'll do it in CAD, much easier
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196 ?
Самый простой способ.
Какой же нудный этот индус!
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