Can you find area of the Green shaded Quadrilateral? | (Square) | #math #maths | #geometry | #viral

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Can you find area of the Green shaded Quadrilateral? | (Square) | #math #maths | #geometry | #viral

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[BBMathTutorials]

Great video sir🎉Thank you so much🙋🏻‍♂️

[stanbest]

A nice problem. You need to see that the triangles are similar. ,

[himo]

AB=BC=CD=DA=3a
(2a)²+(3a)²=(2√13)² 13a²=52 a=2 3a=3*2=6
AF=a*2/3=2a/3

Greeen shaded area = ABCD - (AEF + DCE) = 6*6 - (2*4/3*1/2 + 4*6*1/2) = 36 - 4/3 - 12 = 24 - 4/3 = 72/3 - 4/3 = 68/3

[Mediterranean]

a=2
Side of square is 6
AFE is similar to EDC
Scale factor is 1/3
Area of EDC = 1/2*4*6=12
Area of AFE = 12*(1/3)^2=4/3
So the green area is = 6^2-12-4/3

[shalinisuryavanshi]

Hey pre math can you make a video about the cpncept similarity and congruency

[cyruschang]

(2a)^2 + (3a)^2 = (2✓13)^2
13a^2 = 4(13)
a = 2
The two white triangles are similar.
Area = square area - combined area of the two triangles = (3a)^2 - (2a)(3a)/2 - a(2a/3)/2 = (9 - 3 - 1/3)a^2 = (6 - 1/3)(4) = 24 - 4/3 = 23 - 1/3 = 22 + 2/3

[quigonkenny]

As ABCD is a square, all side lengths are equal and all internal angles are 90°. Therefore, DC = CB = BA = AD = a+2a = 3a.

Let ∠DCE = α and ∠CED = β, where α and β are complementary angles that sum to 90°. As ∠FEC = 90°, then ∠AEF = 180°-90°-β = 90°-β = α. As ∠FAE = 90°, then ∠EFA = β. Therefore, by AAA, ∆EDC and ∆FAE are similar triangles.

Triangle ∆EDC:
ED² + DC² = CE²
(2a)² + (3a)² = (2√13)²
4a² + 9a² = 52
13a² = 52
a² = 52/13 = 4
a = √4 = 2

Triangle ∆FAE:
FA/AE = ED/DC
FA/a = 2a/3a = 2/3
FA = 2a/3 = 2(2)/3 = 4/3

The area of green quadrilateral BFEC is equal to the area of square ABCD minus the areas of the two triangles ∆FAE and ∆EDC.

Green quadrilateral BFEC:
Aɢ = (3a)² - a(2a/3)/2 - 3a(2a)/2
Aɢ = 9a² - a²/3 - 3a² = (27-1-9)a²/3
Aɢ = 17a²/3 = 17(2)²/3 = 68/3 = 22.666... sq units

[Waldlaeufer]

(2a)² + (3a)² = (2√13)²
4a² + 9a² = 4 * 13
13a² = 4 * 13
a² = 4
a = 2

3a : 2a = a : x
3a * x = 2a²
x = 2a² / 3a = 2 * 4 / 6 = 4/3

A(green) = (3a)² - 1/2 (2a * 3a + 4/3 a)
A(green) = 36 - 1/2 (6 * 4 + 4/3 * 2)
A(green) = 36 - 1/2 (24 + 8/3)
A(green) = 36 - (12 + 4/3)
A(green) = 36 - (36/3 + 4/3)
A(green) = 36 - 40/3 = 108/3 - 40/3 = 68/3 ≈ 22.67 square units

[marcgriselhubert]

*In EDC: EC^2 = ED^2 + DC^2, so 52 = 4.a^2 + 9.a^2 =13.a^2, so a^2 = 4 and a = 2.
*Area of the big square: (3.a).(3.a) = 9.a^2 = 9.4 = 36
*Area of triangle EDC: (1/2).(2.a).(3.a) = 3.a^2 = 12
*TrianglesaEDC and AFE are similar (same angles), and AE/DC = 1/3 . The side lengths of AFE are 1/3 of the side lengths of EDC, so its area is 1/9 of the area of EDC, so it is 12/9 = 4/3.
*Finallt the green area is 36 = 36 - 12 - 4/3 = 68/3.

[MrPaulc]

Observing the lower triangle, perhaps a can be calculated by (2a)^2 + (3a)^2 = (2*sqrt(13)^2
4a^2 + 9a^2 = 52
13a^2 = 52
a^2 = 4
a=2
Therefore, the square is 6*6 = 36 un^2
AFE and EDC are similar
Calculate AF: a/(AF) = 6/4
2/(AF) = 6/4
The usual is to cross multiply, but a quick inspection reveals that AF is one third of 4
This gives the relevant sides of the small triangle as 2 and 4/3, and the larger one as 6 and 4
The triangles areas are 4/3 and 12 un^2
Subtract those from 36 for 22 and 2/3.
A decimal approximation is 22.67 un^2

Yes, we went the same route, except for the very end part.
I managed to do much of this one in my head and only wrote some parts to verify it.

[santiagoarosam]

(2a)²+(3a)²=(2√13)²→ a²=4→ a=2→ Razón de semejanza entre triángulos blancos, s=a/3a=1/3→ Razón entre áreas =s²=1/9→ Área verde =36-12-(12/9)=68/3 ud².
Gracias y saludos.

[erdemakca]

İt is easy. This shape is square. So DC= 3a. Then EC²= ED²+DC². 52=9a²+4a². 52=13a². So a²=4. Then a=2. Then we have to find AF. SO We have to set resemblance with ADC triangle with AEF triangle. After then we can find what shaded area is.

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

01) (2a)^2 + (3a)^2 = (2*sqrt(13))^2 ; 4a^2 + 9a^2 = 4 * 13 ; 13a^2 = 52 ; a^2 = 52/13 ; a^2 = 4 ; (Negative Solution : a = - 2) or a = 2
02) Square [ABCD] Side = 3 * (2) = 6 lin un
03) Square [ABCD] Area (SA) = 6 * 6 ; SA = 36 sq un
04) AE = 2 lin un
05) DE = 4 lin un
06) Angle (FEC) = 90º
07) Triangle [CDE] Area = (3a * 2a) / 2 = (6 * 4) / 2 = 24 / 2 = 12 sq un
08) Triangles [AEF] and Triangle [CDE] are Similar.
09) AF / a = 2a / 3a ; AF / 2 = 4 / 6 ; AF / 2 = 2 / 3 ; AF = 4/3 lin un ; AF ~ lin un
10) Triangle [AEF] Area = (2 * 4/3) / 2 = 4/3 sq un
11) Green Shaded Area (GSA) = 36 - (12 + 4/3) ; GSA = 36 - 40/3 ; GSA = (108 - 40) / 3 ; GSA = 68 / 3 sq un ; GSA ~ sq un

Therefore,


OUR BEST ANSWER is :

Green Shaded Area is equal to 68/3 Square Units or approx. equal to 22, 67 Square Units.


P.S. - Best Regards from The Islamic Institute of Mathematical Sciences.

[unknownidentity]

Let's find the area:
.
..
...
....


The side length of the square is obviously s=AE+DE=a+2a=3a. By applying the Pythagorean theorem to the right triangle CDE we obtain:

CD² + DE² = CE²
(3a)² + (2a)² = (2√13)²
9a² + 4a² = 2²*13
13a² = 2²*13
a² = 2²
⇒ a = 2

Now let's have a look at the interior angles of the right triangles CDE and AEF:

CDE: ∠CDE = 90° ∠DCE = α ∠CED = β
AEF: ∠EAF = 90° ∠AEF = α ∠AFE = β

Therefore these two triangles are similar and we can conclude:

AF/AE = DE/CD
AF/a = (2a)/(3a)
⇒ AF = 2a/3

Now we are able to calculate the area of the green quadrilateral:

A(BCEF) = A(ABCD) − A(CDE) − A(AEF) = s² − (1/2)*CD*DE − (1/2)*AE*AF
A(BCEF) = (3a)² − (1/2)*(3a)*(2a) − (1/2)*a*(2a/3) = (9 − 3 − 1/3)a² = (18/3 − 1/3)a² = 17a²/3 = 17*2²/3 = 68/3

Best regards from Germany

[wackojacko]

I do this developing fantasy baseball diamonds for video games for my non Euclidean universes. Mo fun livin hyperbolicly. 🙂

[sergioaiex]

Solution:

a/EF = 3a/2√13
EF . 3a = a . 2√13
EF = a . 2√13/3a
*EF = 2√13/3*

3a/2a = a/AF
3a . AF = 2a . a
AF = 2a . a/3a
AF = 2a/3

A1 = ½ (a. 2a/3)
A1 = a²/3

A2 = ½ 2a . 3a
A2 = 3a²

ASquare = 3a . 3a
ASquare = 9a²

Green Shaded Area (GSA) = ASquare - (A1 + A2)
GSA = 9a² - (a²/3 + 3a²)
GSA = 9a² - (a²/3 + 3a²)
GSA = 9a² - 10a²/3
GSA = 17a²/3 ... ¹

(2a)² + (3a)² = (2√13)²
4a² + 9a² = 52
13a² = 52
a² = 52/13
a² = 4
a = 2

GSA = 17 (2)²/3

GSA = 68/3 Square Units

ou

GSA ~= 22, 67 Square Units

[Birol]

My way of solution is ▶
consider the ΔEDC:
∠EDC= 90°
∠DCE= α
∠CED= β
⇒
α+β = 90°

consider the ΔFAE:
∠CED= β ⬆
∠FEC= 90°
⇒
∠AEF= α
∠FAE= 90°
∠EFA= β
⇒
ΔEDC ~ ΔFAE
DC/AE= CE/EF= ED/FA
DC= 3a
ED= 2a
AE= a
⇒
3a/a= 2a/x
x= 2a/3

if we apply the Pythagorean theorem for the ΔEDC
(2a)²+(3a)²= (2√13)²
4a²+9a²= 4*13
13a²= 52
a²= 4
a= 2

Agreen= (3a)²-A(ΔEDC) - A(ΔFAE)
Agreen= 9a² - 6a²/2 - a*(2a/3)*(1/2)
Agreen= 9a²-3a²- 2a²/6
Agreen= 6a² - a²/3
Agreen= 6*2²- 2²/3
Agreen= 24- 4/3
Agreen= 68/3
Agreen ≈ 22, 7 square units