Can you find area of the Yellow shaded Square? | (Triangle) | #math #maths | #geometry

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Can you find area of the Yellow shaded Square? | (Triangle) | #math #maths | #geometry

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[is]

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!

[himo]

7 : 14 = 1 : 2 ED=DF=x FC=2x
x²+(2x)²=14² 5x²=196
Yellow Area = x*x = x² = 196/5 = 39.2

[hongningsuen]

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

[Alishbafamilyvlogs-bmip]

Very beautiful video nice information thanks for sharing❤

[santiagoarosam]

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.

[eduardozappi]

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

[gelbkehlchen]

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

[devondevon]

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

[andrepiotrowski]

From behind through the chest into the eye...
CD is double the length of DA, so CF is double the length of ED; thus CF equals 2a.

[georgebliss]

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.

[yalchingedikgedik]

Thanks Sir
Thanks PreMath
Very nice and useful
We are learning more about Math.
Good luck with glades
❤❤❤❤

[Tom-zuyc]

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!

[sergeyvinns]

(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.

[wackojacko]

@ 6:59, I absolutely love filling in the blanks of the Pagan Formula a² + b² = c². Life is good. 🙂

[DarkoMarković-rc]

AD:ED= DC:FC
7:x=14:sqrt (196-x^2)
We get x^2=196/5

[marcgriselhubert]

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.

[stevetitcombe]

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

[iveswidmer]

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

[rabotaakk-nwnm]

3:33-6:33 ΔAED ~ ΔDFC (AA) =>
=> ED/AD=FC/DC => FC=a•14/7=2a

[quigonkenny]

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