Can you find area of the Pink shaded Quadrilateral? | (Right Triangles) | #math #maths | #geometry

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Can you find area of the Pink shaded Quadrilateral? | (Right Triangles) | #math #maths | #geometry

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[yalchingedikgedik]

That’s very difficult ..
But the explanation is good .
thanks PreMath
Thanks prof.
With my respects
❤❤❤

[yxkrurq]

Using A as the origin (0, 0), Let's drop a perpendicular FG to the X-axis and a perpendicular FH to the y-axis. Line AD is y= (1/2)x and Line EC is y = (9/8)x - 45 and they intersect at F(72, 36). The pink area is made up of a 48 by 36 rectangle(FGBH), a 32 by 36 right triangle (EFG) and a 48 by 24 right triangle (DFH) so the area is 48*36 + (1/2)*32*36 + (1/2)*24*48 = 2880 square units.

[santiagoarosam]

Areas de triángulos: → Ambos triángulos tienen igual área→ Restándoles el solapo quedan dos triángulos sueltos de igual superficie→ AEF=CDF=a→ DBF=(60/30)a=2a=(80/40)a=EBF→ ABD=3600=5a→ a=720→ EBDF=4a=4*720=2880 ud².
Bonito rompecabezas. Gracias y un saludo cordial.

[marcgriselhubert]

Let's use an orthonormal center E and first axis (EA)
We have B(0; 0) E(80; 0) A(120; 0) D(0; 60) C(0; 90)
We obtain easily the equation of (CE): 9.x +8.y -720 = 0
and of (AD): x +2.y -120 = 0,
and their intersection F(48; 36)
Then, area of triangle DBF = (1/2).DB.(abscissa of F)
= (1/2).60.48 = 1440
And area of triangle BFE: (1/2).BE.(ordinate of F)
= (1/2). 80. 36 = 1440
Finally, pink area = 1440 + 1440 = 2880.

[jaimeyomayuza]

B=(0, 0)
1) Hallar las coordenadas de F
2) Trazar el segmento FB, que divide al cuadrilátero EFDB en 2 triangulos T1= EFDB. T2= FBD
3) Hallar las AREAS de los triangulos
4) área cuadrilátero =
área T1+área T2 =2*1440 MI RESPUESTA
DESDE Bogotá D.C. COLOMBIA

[hongningsuen]

Method using similar triangles, area ratios of 4 triangles in trapezoid divided by diagonals:
1. Join DE. Triangles BED and BAC are similar (ratio of 2 sides, inclusive angle)
(equal side ratios of 2 pairs of corresponding sides are 80:(80+40) and 60:(60+30) = 2:3)
Hence ED//AC and ED:AC = 2:3
2. Quadrilateral ACDE is trapezoid as ED//AC with top side ED:bottom side AC = 2:3
Area ratios of the 4 triangles in the trapezoid DEF:AFE:CDF:ACF = 4:6:6:9
Actual area of trapezoid = area of triangle BAC - area of triangle BED = 5400 - 2400 = 3000
Hence actual area of triangle DEF = [4/(9+6+6+4)](3000) = 480
3. Area of pink region = area of triangle BED + area of triangle DEF = 2400 + 480 = 2880

[himo]

ABD=120*60/2=3600 EBC=90*80/4=3600
40 : 80 = 1: 2 30 : 60 = 1: 2
AEF=x FEB=2x CDF=y BFD=2y
x+2x+2y=3600 y+2y+2x=3600 3x+2y=2x+3y x=y
5x=3600 x=720 y=720

Pink Area = 2x+2y = 2(x+y) = 2*1440 = 2880

[allanflippin]

I found a way which is simpler, but it uses analytic geometry. Here are the steps:
1) Set point A as the origin (x=0, y=0)
2) Make equations for lines AD and EC
3) Solve the two equations to find x and y.
4) The y value is the same as your "a". Your "b" value is 120-x
5) Add up the areas of the EFB and BFD triangles
For line AD, y = x/2. For line EC, y = 9x/8 - 45 (using point/slope method). Setting the two equations equal, x = 72. Since y = x/2, y = 36. Your "a" value is 36 and "b" is 120-72 = 48. The two triangle areas added are 36*80/2 + 48*60/2 = 2880.

[Waldlaeufer]

line 1: y = 60/120 x + 60 = 1/2 x + 60
line 2: y = 90/80 x + 90 = 9/8 x + 90

y = y
1/2 x + 60 = 9/8 x + 90
(9/8 - 1/2) x = 60 - 90
(9/8 - 4/8) x = - 30
5/8 x = - 30
x = -30 * 8/5 = - 6 * 8 = - 48

y = 1/2 * - 48 + 60 = - 24 + 60 = 36

A(red) = 1/2 (80 * y + 60 * |x| = 1/2 (80 * 36 + 60 * 48) = 40 * 36 + 30 * 48 = 2880 square units

[unknownidentity]

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


Let's assume that B is the center of the coordinate system and that BA and BC are located on the x-axis and y-axis, respectively. Then we obtain the following coordinates from the sketch:

A: ( −120 ; 0 )
B: ( 0 ; 0 )
C: ( 0 ; 90 )
D: ( 0 ; 60 )
E: ( −80 ; 0 )
F: ( xF ; yF )

The lines AD and EC are represented by the follwing functions:

AD: y = (yD − yA)*(x − xA)/(xD − xA) = (60 − 0)*(x + 120)/(0 + 120) = (x + 120)/2
EC: y = (yC − yE)*(x − xE)/(xC − xE) = (90 − 0)*(x + 80)/(0 + 80) = 9*(x + 80)/8

These two lines intercept at F. Therefore we obtain:

(xF + 120)/2 = 9*(xF + 80)/8
4*(xF + 120) = 9*(xF + 80)
4*xF + 480 = 9*xF + 720
−5*xF = 240
⇒ xF = −48

yF = (xF + 120)/2 = (−48 + 120)/2 = 72/2 = 36
yF = 9*(xF + 80)/8 = 9*(−48 + 80)/8 = 9*32/8 = 9*4 = 36 ✓

A(pink)
= A(ABD) − A(AEF)
= (1/2)*AB*h(AB) − (1/2)*AE*h(AE)
= (1/2)*AB*BD − (1/2)*AE*yF
= (1/2)*120*60 − (1/2)*40*36
= 3600 − 720
= 2880

A(pink)
= A(BCE) − A(CDF)
= (1/2)*BC*h(BC) − (1/2)*CD*h(CD)
= (1/2)*BC*BE − (1/2)*CD*(−xF)
= (1/2)*90*80 − (1/2)*30*48
= 3600 − 720
= 2880 ✓

Best regards from Germany

[georgebliss]

Dropping perpendicular from point F onto EB to point G.
Let length FG = h.
Let length EG = x.
Triangles AFG & ADB are similar.
So h / (40 + x) = 60 /120.
h = (40 + x) / 2.
Triangles EFG & ECB are similar.
So x / h = 80 / 90.
h = 9x / 8.
Equating 2 values for h.
(40 + x) / 2 = 9x / 8.
18x = 320 + 8x.
10x =320
x = 32.
h= 9x / 8 (as previously calculated) = 9 * 32 / 8 = 36.
Area triangle AEF = 1/2 * 40 * 36 = 720.
Area triangle ABD = 1/2 * 120 * 60 = 3600.
Pink area = 3600 - 720 = 2880.

[LuisdeBritoCamacho]

Let the Coordinates of Point B = (0 ; 0)

Let the Coordinates of Point F = (x ; y)

Equation of Straight Line AD : x / 120 + y / 60 = 1
Equation of Straight Line EC : x / 80 + y / 90 = 1

Solving the following System of 2 Linear Equations with 2 Unkowns :

x / 120 + y / 60 = 1
x / 80 + y / 90 = 1

Solutions : x = 48 and y = 36

2 * Area = (60 * x) + (80 * y) ; 2 * Area = (60 * 48) + (80 * 36) ; 2 * A = 2.880 + 2.880 ; A = 2.880

Pink Area = 2.880 Square Units

[adept]

AE : EB = CD : DB = 1 : 2, ⇛, AC//DE, ACDE - тrapezoid.
Additional construction: FH ⟂ AC, FK ⟂ DE. FK : FH = DE : AC = 100 : 150 = 2 : 3.
S(ABC) = 5400, S(BDE) = 2400, S(ACDE) = 3000. HK = 3000/[0, 5(AC + DE) = 3000/[0, 5(150 + 100) = 24.
FK = (2/5) × HK = (2/5) × 24 = 9, 6. S(EFD) = 1/2(DE × FK) = 1/2(100 × 9, 6) = 480.
S(pink) = S(BDE) + S(EFD) = 2400 + 480 = 2880.

[phungpham]

1/ Label FT=a and ET=x
Area = Area [EBD]+Area[EFD]= Area[EBD]+Area[AED]-Area[AEF]
=2400+1200-Area [AEF]
=3600-Area[AEF]
2/The 2 triangles AFT and ABD are similar so:
FT/ET=DB/AB=60/120=1/2
—-> a/(40+x) =1/2—> x=2a-40 (1)
2/ The 2 triangles EFT snd ECD are similar so
x/a= EB/BC=80/90= 8/9 (2)
From (1) and (2) we have:
(2a-40)/a = 8/9—> a=36
—-> Area= 3600 - 1/2 . 36. 40 = 3600- 720= 2880 sq units

[nandisaand]

Assign point B (0, 0)
Eqn for Line AD: Y=(90/80)X+90 or
Y=9/8X+90
Eqn for Line EC: Y=(60/120)X+60 or
Y=1/2X+60
Find intersection (pt F):
1/2X+60=9/8X+90
X=(-48); Y=36
Area=
Rectangle (48x36)=1728
Triangle 1/2(80-48)(36)=576
Triangle 1/2(48)(60-36)=576
Total Area: 2880

[MarieAnne.]

Here's how I solved:
Join B to F to form 4 triangles.
△AEF and △BEF have same height (altitude from F to AB)
△AEF has base = 40 and △BEF has base = 80, so it has twice area of △AEF
Area(△AEF) = x, Area(△BEF) = 2x
△BDF and △CDF have same height (altitude from F to BC)
△CDF has base = 30 and △BDF has base = 60, so it has twice area of △CDF
Area(△CDF) = y, Area(△BDF) = 2y
Area(△ABD) = Area(△AEF) + Area(△BEF) + Area(△BDF)
1/2(120)(60) = x + 2x + 2y
*3x + 2y = 3600 (1)*
Area(△BCE) = Area(△BEF) + Area(△BDF) + Area(△CDF)
1/2 (80)(90) = 2x + 2y + y
*2x + 3y = 3600 (2)*
Adding (1) and (2) we get
5x + 5y = 7200
*x + y = 1440*
Area of pink quadrilateral
= Area(△BEF) + Area(△BDF)
= 2x + 2y = 2(x+y) = 2(1440)
*= 2880*

[solimana-soli]

once you know that Area ABD = Area EBC = 3600 you can easily show that area AEF = area FDC, then
1/2*40*a = 1/2*30*b
b = 4/3 a
triangles AFT and ABD are similar then
a : 60 = (120 - 4/3a) : 120
120a = 7200 - 240/3a
a = 36
Pink area = 3600 - 1/2*40*36 = 3600 - 720 = 2880

[valeriykotlov]

It’s so simple. Areas of AEF and BEF 1 : 2, also CDF:BDF 1 : 2. Area of ABD = area of BCE = 3600. So ABD consists of x + 2x + 2x, x = 3600/5 = 720, pink = 4 * 720 = 2880.

[albertpaquette]

There’s a much easier solution. Join pb as shown, but notice that triangles AFE and EFB have the same height but their bases are in a ratio of 2 to 1. Therefore their areas are also in the same ratio. Likewise for the other pair of triangles at the top right of the diagram. Using these facts you can quickly determine the pink area.

[quigonkenny]

Draw BF. By observation, it is apparent that the pink area is equal to the sum of the areas of the triangles ∆BFE and ∆DFB. Drop perpendiculars from F to M on AB and N on BC. Let FM = y and FN = x.

Triangle ∆ABD:
A₁ = ∆ABF + ∆DFB
120(60)/2 = 120y/2 + 60x/2
3600 = 60y + 30x ---- [1]

Triangle ∆EBC:
A₂ = ∆BFE + ∆FBC
90(80)/2 = 80y/2 + 90x/2
3600 = 40y + 45x ---- [2]

Multiply [2] by 3/2 and subtract from [1].

3600 - 3600(3/2) = 60y + 30x - (40y+45x)(3/2)
3600 - 5400 = 60y + 30x - 60y - 135x/2
-1800 = -75x/2
75x = 3600
x = 3600/75 = 48

3600 = 60y + 30(48) <--- [1]
60y = 3600 - 1440 = 2160
y = 2160/60 = 36

Pink quadrilateral BDFE;
Aₚ = ∆BFE + ∆DFB
Aₚ = 40y + 30x = 40(36) + 30(48)
Aₚ = 1440 + 1440 = 2880 sq units