Can you calculate area of the Yellow shaded region? | (Triangles) |

[@jamestalbott4499]

Thank you. Another good example of finding a special triangle and using it to solve the problem.

[@PreMath]

I'm glad you found it helpful! 👍 Thanks for the feedback ❤️

[@marioalb9726]

1) Yellow shaded area, is equal to area of triangle CDF 2) Angle ECB = 90° - Angle ADB , Then F is midmoint of hypotenuse EC 3) Minor leg of right triangle 3-4-5 is 3 cm A = ½b.h = ½*4*(½3) A = 3 cm² ( Solved √ )

[@jaimeyomayuza6140]

Ecuacion de EC y=4x/3 + 4 Ecuacion de BD y=-1.3333x Resolviendo x= -1.5 y= 2 Área triángulo DEF =. 3 Cordial saludo desde Bogotá D.C. Colombia

[@PreMath]

Excellent! Thanks for sharing ❤️ Love and prayers from the USA! 😀

[@ChuzzleFriends]

By the Pythagorean Triples (3, 4, 5), DE = 3. Find the area of △CDE. A = (bh)/2 = (3 * 4)/2 = 12/2 = 6 Because △ABD ≅ △CDE by SSS, The area of △ABD must be 6 as well. Let α & β be the measures of complementary angles. Let m∠A = α. Then m∠C = α by CPCTC. Therefore, m∠ADB = m∠CED = β by similar logic. And since ∠ADC is a right angle, m∠BDC = α. So, △CFD & △DFE are isosceles triangles & CF = DF = EF. Thus, DF is a median of △CDE that divides the area in half. Therefore, the area of △DFE is 3. Quadrilateral ABFE Area = △ABD Area - △DFE Area = 6 - 3 = 3 So, the area of the yellow shaded region is 3 square units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

The overlapping area is an isosceles triangle whose base is 3. The isosceles triangle can be divided by its height into two 3-4-5 triangles, so its height = 2 The area of the isosceles triangle = 3(2)/2 = 3 Yellow area = (3)(4)/2 - 3 = 3

[@quigonkenny]

As the hypotenuse of each triangle is 6 and the given leg length of each triangle is 4, then each triangle is a 3-4-5 Pythagorean triple right triangle and the remaining leg length is 3. Thus each triangle on its own has an area of Aᴛ = bh/2 = 3(4)/2 = 6. The yellow area Aʏ is the difference in area between Aᴛ and the overlap of the two triangles, Aᴏ. That overlap is represented by triangle ∆DFE. As the two triangles ∆DBA and ∆EDC are given as congruent, then angles ∠CED and ∠EDB are congruent. This means that ∆DFE is an isosceles triangle and DF = FE. Drop a perpendicular from F to T on DE. As ∆DFE is an isosceles triangle, then FT is a perpendicular bisector and divides ∆DFE into two congruent right triangles ∆FTD and ∆ETF. Therefore ET = TD. As FT and DC are parallel, then ∠TFE and ∠DCE are corresponding angles and thus congruent. As ∠ETF = ∠EDC = 90°, then congruent triangles ∆ETF and ∆FTD are similar to congruent triangles ∆EDC and ∆DBA. As ET = TD = ED/2, then ∆ETF and ∆FTD are each 1/2² = 1/4 the area of ∆EDC or ∆DBA. Thus triangle ∆DFE is 2/4 = 1/2 the area of ∆EDC or ∆DBA. Triangle ∆DFE: Aᴏ = Aᴛ/2 = 6/2 = 3 Quadrilateral ABFE: Aʏ = Aᴛ - Aᴏ = 6 - 3 = 3 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@michaelkouzmin281]

I dropped the height [FG] of the triangle EFD from F down to [ED]. Triangles EFG and ECD are similar => FG/CD = EG/ED => FG = CD*EG/ED = 4*1.5/3 =2 Area of the triangle EFD = ED*FG/2= 3*2/2 =3; Ayellow = A(ABD) - A(EFD) =6-3=3 sq units.

[@nandisaand5287]

Too bad I can't take the easy road like Prof did. I calculated it using analytic geometry: Assign D (0,0) Line EC: Y=4/3X+4 Line BD: Y=-4/3X Coordinate F: 4/3X+4=-4/3X 8/3X=-4 X=-3/2 Y=-4/3(-3/3)=2 Area ABFE=🔺️ABD-🔺️EFD =½•3•4-½•3•2 =6-3 =3 Put a box around it. How exciting.

[@PreMath]

No worries... Thanks for the feedback ❤️

[@misterenter-iz7rz]

Complicated figures to make one confused, but clarify your mind to reach a very simple and nice solution. This 1/2 3 4-1/2 3 4/2=6-3=3.😍😍😍😍😍😍😍

[@RK-tf8pq]

The length of a perpendicular from point F to CD is half of ED, since F is mid point of the line CE. Thus this perpendicular length is 1.5. So the area of the triangle CDE is 4x1.5/2 = 3, which is equal to area of the yellow region.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Son dos triángulos congruentes solapados, entonces las áreas no solapadas de ambos triángulos deben ser iguales. P es la proyección ortogonal de F sobre DC---> BD=DE=3---> FD=FE---> FP=DE/2 =3/2---> CDF=4(3/2)/2 =3 =Área sombreada amarilla ABFE. Gracias y saludos

[@PreMath]

Excellent! Thanks for sharing ❤️

[@alexundre8745]

Bom dia Mestre

[@PreMath]

Hello dear😀 Thanks ❤️

[@giuseppemalaguti435]

4*3/2-2*3/2=6-3=3

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sorourhashemi3249]

Thanks. Challenging ❤

[@PreMath]

Excellent! Keep practicing! 💪 Thanks for the feedback ❤️

[@ekoi1995]

My crazy solution using Analytic Geometry and Integrals from Calculus: We can construct 3 lines: y = (4/3)x + 4 y = (3/4)x + (15/4) y = -(4/3)x Finding Intersections: Intersection of y = (3/4)x + (15/4) and y = -(4/3)x: (-1.8, 2.4) Intersection of y = (4/3)x + 4 and y = -(4/3)x: (-1.5, 2) Finding x-Intercepts: x-intercept of y = (3/4)x + (15/4): (-5, 0) x-intercept of y = (4/3)x + 4: (-3, 0) Forming the Definite Integrals: To calculate the area under the curves: For y = (3/4)x + (15/4) and y = -(4/3)x: integral(from -5 to -1.8)((3/4)x + (15/4))dx + integral(from -1.8 to -1.5)(-(4/3)x)dx = 4.5 For y = (4/3)x + 4: integral(from -3 to -1.5)((4/3)x + 4)dx = 1.5 Calculating the Yellow Area: The area of the yellow region is: 4.5 - 1.5 = 3 square units

[@nenetstree914]

3

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

A to b is 3.904344 nat 4

[@wackojacko3962]

The occult! ...@ 6:13 , another example of "What Pythagoras Wrought" #333(message of an angel) or a (Crank Mathematician)! 🙂

[@PreMath]

😀 Thanks for the feedback ❤️

[@pralhadraochavan5179]

Good night sir

[@PreMath]

Thanks dear❤️ Take care

[@LuisdeBritoCamacho]

The answer is : The Yellow Shaded Area is equal to Half of the Triangle [ABD] Area = 3 Square Units. 01) In a Cartesian Plane the Coordinates of Point F are : F = (7/2 ; 2) 02) So, Triangle [DEF] Area is equal to (3 * 2) / 2 = 6 / 2 = 3 sq un 03) Traingle [ABD] Area = (4 * 3) / 2= 12 / 2 = 6 sq un 04) YSD = 6 - 3 = 3 sq un. That's All!!

[@PreMath]

Excellent work! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... First of all we calculate the missing side lenghts of the congruent triangles ABD and CDE. Since these triangles are right triangles, we can apply the Pythagorean theorem: BD² = DE² = CE² − CD² = 5² − 4² = 25 − 16 = 9 ⇒ BD = DE = 3 Let's add point G on AD such that ABG and BDG are right triangles. Now we calculate the height BG of the triangle ABD according to the base AD: A(ABD) = (1/2)*AB*h(AB) = (1/2)*AD*h(AD) (1/2)*AB*BD = (1/2)*AD*BG ⇒ BG = AB*BD/AD = 4*3/5 = 12/5 By applying the Pythagorean theorem to the right triangle ABG we obtain: AG² = AB² − BG² = 4² − (12/5)² = 16 − 144/25 = 400/25 − 144/25 = 256/25 ⇒ AG = √(256/25) = 16/5 Now let's assume that A is the center of the coordinate system and that AD is located on the x-axis. Then we obtain the following coordinates: xA = 0 yA = 0 xB = AG = 16/5 yB = BG = 12/5 xC = 5 yC = 4 xD = 5 yD = 0 xE = AE = AD − DE = 5 − 3 = 2 yE = 0 The lines BD and EC are represented by the following functions: BD: y = (yB − yD)*(x − xD)/(xB − xD) + yD = (12/5 − 0)*(x − 5)/(16/5 − 5) + 0 = (12/5)*(x − 5)/(16/5 − 25/5) = (12/5)*(x − 5)/(−9/5) = (−4)*(x − 5)/3 EC: y = (yC − yE)*(x − xE)/(xC − xE) + yE = (4 − 0)*(x − 2)/(5 − 2) + 0 = 4*(x − 2)/3 These two lines intersect at point F: (−4)*(xF − 5)/3 = 4*(xF − 2)/3 −(xF − 5) = xF − 2 5 − xF = xF − 2 7 = 2*xF ⇒ xF = 7/2 ⇒ yF = (−4)*(xF − 5)/3 = (−4)*(7/2 − 5)/3 = (−4)*(7/2 − 10/2)/3 = (−4)*(−3/2)/3 = 2 (⇒ yF = 4*(xF − 2)/3 = 4*(7/2 − 2)/3 = 4*(7/2 − 4/2)/3 = 4*(3/2)/3 = 2 ✓) Since yF is equal to the height of the triangle DEF according to the base DE, we are able to calculate the area of the yellow region: A(yellow) = A(ABD) − A(DEF) = (1/2)*AB*BD − (1/2)*DE*yF = (1/2)*4*3 − (1/2)*3*2 = 6 − 3 = 3 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergioaiex3966]

Solution: In ∆ CDE: DE = 3 (Pythagorean Triples) ∆ ABD is congruent to ∆ CDE ∆ ABD A = α B = 90° D = β ∆ CDE C = α E = β D = 90° In vertex D, α + β = 90°, therefore the ∆ DEF and CDF are an isosceles triangles In ∆ DEF, EF = FD = 5/2 Let's suppose a point P as midpoint to DE, forming the triangle PDF, such that PD = 3/2 PF = 4/2 = 2 DF = 5/2 Area ∆ DEF = ½ DE . PF Area ∆ DEF = ½ 3 . 2 Area ∆ DEF = 3 Yellow Shaded Area (YSA) = ∆ ABD Area - ∆ DEF Area YSA = ½ 3 . 4 - 3 YSA = 3 Square Units ✅

[@marcgriselhubert3915]

We use an orthonormal center D and first axis (DA). We have two 3-4-5 triangles: EDC and DBA, so D(0; 0) E(3; 0) A(5; 0) C(0; 4) VectorCE (3; -4), Equation of (CE): (x -3).(-4) - (y).(3) = 0 or 4.x + 3.y -12 = 0 t = angleADB, in triangle DBA we have tan(t) = AB/BD = 4/3. Equation of (DB): y = tan(t).x or y = (4/3).x Intersection of (CE) and (DB): 4.x + 3.y -12 = 0 and y = (4/3).x, so 8.x = 12 and x = 3/2 and y = (4/3).(3/2) = 2, so F(3/2; 2) The triangle DFE has DE = 3 for basis and the ordinate of F = 2 for height, its area is then (1/2).(3).(2) = 3 The area of triangle ABD is (1/2).AB.BD = (1/2).(4).(3) = 6, finally by difference the yellow area is 6 - 3 = 3.

[@PreMath]

Excellent! Thanks for sharing ❤️