Can you calculate area of the Blue Parallelogram? | (Rectangle) | #math #maths | #geometry

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Can you calculate area of the Blue Parallelogram? | (Rectangle) | #math #maths | #geometry

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[jonchester]

I always start my day by solving your problem of the day. It's one of the brightest spots of my day. Keep up the great work.

[marioalb]

Labelling: b = DB; h = CG
sinα = 8/8√5= 1/√5 --> cosα= 2/√5
h = 8 / cosα = 4√5 cm
A = ½b.h = ½ 8√5. 4√5
A = 80 cm² ( Solved √ )

[alexundre]

Bom dia Mestre
Vou pausar p tentar resolver
Grato pela aula

[santiagoarosam]

EFB=EAD=GHD=GCB.
AB=√[(8√5)²-8²]=16 =EA+EB---> EB=16-EA =16-EF
En el triángulo EFB: EF²=(16-EF)²-8²---> EF=6---> ED=10.
Área GBED=16*8-6*8 =80 u².
Gracias y saludos

[alanthayer]

Nice work so THANKU for daily exercises sir !

[murdock]

Nice, many thanks, Sir! BG = k; HB = DF = CD = AB = 16; BC = BF = HD = AD = 8 → AC = BD = 8√5
ABD = DBH = δ; GBC = θ = 3φ - 2δ; sin⁡(δ) = √5/5 → cos⁡(δ) = 2√5/5 →
sin⁡(2δ) = 2sin⁡(δ)cos⁡(δ) = 4/5 = cos⁡(θ) = 8/k → k = 10 → 8k = 80 = area BGDE

[phungpham]

1/ DF= 16
Label the angle BDF= alpha. We have tan(alpha)=1/2
2/ The 4 white small triangles are congruent —-> DG=GB=BE=ED —> the quadrilateral DGBE is a diamond.
—> DB is the bisector of angle GDE—> angle BEF=angle GDE= 2 (alpha)
—-> tan (BEF) = 1/(1-1/4)=4/3
—> the triangle BEF is a 3-4-5 triples
—-> BE= 10
Area of the blue diamond = 80 sq units 😅😅😅 0:04

[prossvay]

Blue shaded area=(16)(8)-2(1/2)(6)(8)=80 square units.❤❤❤

[zawatsky]

Основание прямоугольника за а, тогда Поворот отсекает подобные треугольники с теми же пропорциями - 1 к 2, значит это будут стороны 8 и 4. На основание параллелограмма остаётся 10, значит площадь 8*10=80.

[cynthiastandley]

How do you know the rectangles are identical? (P.S., my last math class was 50 years ago. 😊

[cyruschang]

Rectangle length = √[(8√5)^2 - 8^2] = 8√[(√5)^2 - 1] = 16 = a + b
8^2 + a^2 = b^2 = (16 - a)^2
32a = 16^2 - 8^2
2a = 16 - 4
a = 6, b = 10
Blue area = 8 x 16 - 8 x 6 = 80

[unknownidentity]

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


Let a and b=8 be the side lengths of the rectangle and let d be the length of its diagonal. Now we apply the Pythagorean theorem to the right triangle ABC:

AB² + BC² = AC²
a² + b² = d² ⇒ a² = d² − b² = (8√5)² − 8² = 8²*5 − 8² = 8²*(5 − 1) = 8²*4 = 8²*2² = 16² ⇒ a = √(16²) = 16

The right triangles ADE and BEF are congruent (∠DAE=∠BFE=90° ∧ ∠AED=∠BEF ∧ AD=BF=b). Therefore we know: AE=EF. Now we apply the Pythagorean theorem to the right triangle ADE:

DE² = AD² + AE²
(DF − EF)² = AD² + AE²
(a − EF)² = b² + EF²
a² − 2*a*EF + EF² = b² + EF²
a² − b² = 2*a*EF
⇒ EF = (a² − b²)/(2*a) = (16² − 8²)/(2*16) = 16²/(2*16) − 8²/(2*16) = 8 − 2 = 6
⇒ DE = DF − EF = a − EF = 16 − 6 = 10

Now we are able to calculate the area of the blue region:

A(BEDG) = DE*h(DE) = DE*BF = 10*8 = 80

Best regards from Germany

[marcgriselhubert]

Something much different (I understand it is too long and too complicated!)
The diagonal of each rectangle is 8.sqrt(5), the length is 8, so the height is sqrt(((8.sqrt(5))^2 - (8^2)) = sqrt(320 - 64 ) = sqrt(256 = 16.
Be t = AngleFDB = tan-1(8/16) = tan-1(1/2). Let's imagine a rectangle DUVF of same dimansions than rectangle HDFB situated just under HDFB, then tne rectangle DABC is the image of DUVF in the rotation center D and angle 2.t.
We use a orthonormal center D and first axis (DF), the analytic expression of this rotation is x' = cos(2.t).x - sin(2.t).y and y' = sin(2.t).x + cos(2.y).y
Let's calculate cos(2.t): cos(2.t) = 2.((cos(t))^2) -1. With (cos(t)^2) = 1/(1 + (tan(t))^2) = 1/(1 + ((1/2)^2)) = 1/(5/4) = 4/5, so cos(2.t) = 2.(4/5) - 1 = 3/5
and with (cos(2.t))^2 + (sin(2.t))^2 = 1 we get sin(2.t) = 4/5 (sin(2.t ) and cos(2.t) are both positive as 0<2.t<90°, as 0<t<45°)
We then have the analytic expression of the rotation: x' = (3/5).x -(4/5).y and y' = (4/5).x + (3/5)y.
Point C is the image of point F(16; 0) in this rotation, so we have C((3/5).16; (4/5).16) or C(48/5; 64/5). VectorDC is colinear to VectorZ(3; 4) and the equation od (DC) is then y = (4/3).x. Its intersection with (HB) whose equation is y = 8, is G(6; 8), so HG = 6 and GB = 16 - 6 = 10.
The dimensions of the blue parallelogram are then 10 and 8 and its area is 80.

[devondevon]

80
8 sqrt 5)^2-64= c^2
c=16 the length of the rectangle
rectangle area = 16 x 8 =128
n^2=256 +NxN -32n + 64
n=10
Hence, the triangle is a 3-4-5 scaled up by 2
Hence, the other sides are 6 and 8

Hence, the area of the triangle = 24
since there are two = 48 ( 2x 24)
shaded = 128 -48 = 80 answer

[wackojacko]

@ 1:29, Paganism at it's finest! @ 2:34, observe the nature of the universe is binary. I rethunk this problem too make it easier on myself so I can go back to bed. I let BF=0 so that the angles of the parallelogram GBFD : Angle D=0° Angle B=0° and Angle G=180° Angle F=180° . Only problem I have now is too reset my alarm clock. 😊

[sergioaiex]

Solution:

Since the two rectangles are identicals, we have:

BF = DH = BC = AD = 8
AC = 8√5

Applying Pythagorean Theorem in ∆ ABC:

AB² + BC² = AC²
AB² + (8)² = (8√5)²
AB² + 64 = 320
AB² = 320 - 64
AB² = 256
AB = 16
CD = 16
DF = 16
BH = 16

∆ ADE is congruent to ∆ BED, therefore in ∆ EFB, we have the following lengths

DF = 16
EF = a
BE = DE = 16 - a

Applying, once again, Pythagorean Theorem in ∆ BEF

a² + (8)² = (16 - a)²
a² + 64 = 256 - 32a + a²
64 = 256 - 32a
32a = 192
a = 6

∆ BEF Area = ½ b h
∆ BEF Area = ½ 6 × 8
∆ BEF Area = 24

∆ DGH Area = 24 (because ∆ DGH is congruent to ∆ BEF)

Rectangle BFDH Area = length × width

Rectangle BFDH Area = DF × BF = 16 × 8

Rectangle BFDH Area = 128

Blue Shaded Area (BSA) = Rectangle BFDH Area - ∆ BEF Area - ∆ DGH Area

BSA = 128 - 24 - 24

BSA = 80 Square Units ✅

[LuisdeBritoCamacho]

MY BEST ANSWER IS :

Blue Shaded Area is equal to 80 Square Units!!


01) (8sqrt(5))^2 - 8^2 = 320 - 64 = 256 ; sqrt(256) = 16 lin un

02) So, we have : Rectangle Diagonal = (8sqrt(5)) lin un ; Rectangle Width = 8 lin un ; Rectangle Length = 16 lin un

03) The given Rectangle is made of two Squares with Side = 8 lin un

04) As the Sides of the Blue Paralelogram are equal they can be defined in two different ways :

05) DF = DE + EF ; 16 = (16 - X) + X ; NOTE : EF = X lin un

06) DE = (16 - X)

06) EB = sqrt(X^2 + 8^2)

07) DE = EB

08) (16 - X) = sqrt(X^2 + 64)

09) (16 - X)^2 = X^2 + 64 ; 256 - 32X + X^2 = X^2 + 64 ; 256 - 32X = 64 ; 32X = 256 - 64 ; 32X = 192 ; X = 192 / 32 ; X = 6 lin un

10) EF = 6 lin un

11) DE = 10 lin un

12) Blue Paralellogram Area BPA) = (8 * 10) = 80 sq un or BPA = [(8 * 16) - (2 * 24)] ; BPA = (128 - 48) ; BPA = 80 sq un


Therefore,

As told before, the Blue Shaded Area, beyond any reasonable doubt, is equal to 80 Square Units.