Can you find area of the Red and Blue Shaded Trapezoids? | (Trapezium) |

Learn how to find the area of the Red and Blue Shaded Trapezoids. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; similar triangles; triangles; rectangle. Step-by-step tutorial by PreMath.com
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Can you find area of the Red and Blue Shaded Trapezoids? | (Trapezium) | #math #maths | #geometry
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[@stilesstislisky355]

💥 🎉

[@PreMath]

Thanks ❤️

[@Waldlaeufer70]

Another solution: 1) Drop a vertical line through P => You'll get two twin triangles with the same areas (28 cm², 7 cm²) 2) Two congruent triangles form a rectangle => B with 2 * 28 = 56 cm², D with 2 * 7 cm² = 14 cm² 3) You'll get two more rectangles at the other corners. Let's call them A (bottom left) and C (top right) according to the corners. 4) A has half the width, but twice the height, C has half the height, but twice the width. Therefore, A and C must have the same area. 5) We can also state that D : C = A : B 14 : C = A : 56 AC = A² = C² = 14 * 56 = 784 cm² (since A and C have the same area). 6) A = C = √784 cm² = 28 cm² 7) A(blue) = 28 cm² + 28 cm² = 56 cm², C(red) = 7 cm² + 28 cm² = 35 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Waldlaeufer70]

A(red&blue) = 7 + 28 + 28 + 28 = 91 scm A(red) = 7 + 28 = 35 scm A(blue) = 28 + 28 = 56 scm It is rather easy to calculate in your head if you assume that the triangle on the right has a stretch factor of 2 (4 times the area) and use clever measurements to start with: A(top triangle) = 1/2 * 14 * 1 = 7 scm A(bottom triangle) = 1/2 * 28 * 2 = 28 scm Now, it is easy to calculate and add up the two congruent triangles and the rectangles in the red and blue areas.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@Mediterranean81]

the yellow triangles are similar triangles with a scale factor of 2 DEP is similar to ADB by AA theorem we have FB=AE=2x so AD=AE+ED=2x+x=3x so the scale factor is 3 Area of ABD= 7*3^2=63 Blue area = 63-7=56 since ABD and BCD are both halves of the rectangle so their areas are congruent Red Area = 63-28=36

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's face this challenge: . .. ... .... ..... The yellow triangles BFP and DEP are obviously similar. Therefore we can conclude: FP/EP = BF/DE = BP/DP = √[A(BFP)/A(DEP)] = √[(28cm²)/(7cm²)] = √4 = 2 Now we are able to calculate the areas: A(red) = (1/2)*(CD + FP)*DE = (1/2)*(EF + FP)*DE = (1/2)*(EP + FP + FP)*DE = (1/2)*(EP + 2*FP)*DE = (1/2)*(EP + 4*EP)*DE = (1/2)*(5*EP)*DE = 5*(1/2)*EP*DE = 5*A(DEP) = 5*7cm² = 35cm² A(blue) = (1/2)*(AB + EP)*BF = (1/2)*(EF + EP)*BF = (1/2)*(EP + FP + EP)*BF = (1/2)*(2*EP + FP)*BF = (1/2)*(2*EP + 2*EP)*(2*DE) = (1/2)*(4*EP)*(2*DE) = 8*(1/2)*EP*DE = 8*A(DEP) = 8*7cm² = 56cm² Best regards from Germany

[@PreMath]

Super work! Thanks for sharing ❤️🙏

[@hongningsuen1348]

For left upper yellow triangle, let its height be h and width be w. Then hw = 2x7 = 14 From similar yellow triangles area ratio of 1:4 get side ratio is 1:2. Then height of rectangle = 3h and width of rectangle = 3w. Hence area of rectangle = 3h x 3w = 9hw = 9x14 = 126 Area of half of rectangle = 126/2 = 63 Area of blue region = 63 -7 = 56 Area of red region = 63 - 28 = 35.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@kettlebellBob]

You can't have a triangle with side x and area x² The ratio of the areas of similar polygons is square of the ratio of the sides

[@toninhorosa4849]

From point "P" I drew a parallel to the segment DE until reaching the segment DC and marked the point "Q". I formed the triangle DPQ. ∆DEP=∆DQP =~∆PBF Area ∆PBF = 28 cm^2 Area ∆DEP = 07 cm^2 Proportion of Areas∆s = 28/7 = 4 cm^2 Proportion in extension=√4 = 2 If segment EP = b Then segment PF = 2b If segment DE = a Then segment BF = 2a I know (a*b)/2 = 7 Then a*b = 14 Area of Trapezoid: Red Trapezoid = (a*(3b+2b))/2 =5ab/2 = 5*14/2 = 35 cm^2 Blue Trapezoid= (2a (b+3b))/2 Blue Trapezoid= 4*ab = 4*14 Blue Trapezoid= 56 cm^2

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Razón entre áreas de triángulos amarillos =s=28/7=4→ Razón de semejanza =s=√4=2→ DA=h+2h ; DC=b+2b → Área roja =(bh/2)+(2bh) =7+(2*2*7)=7+28=35 ; Área azul =(b2h)+(2b2h/2) =(2*7*2)+28=56. Bonito rompecabezas. Gracias y un saludo cordial.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@quigonkenny]

If ∠EPD = α, and ∠PDE = β, where β and α are complementary angles that sum to 90°, then as ∠FPB is a vertical angle of ∠EPD, ∠FPB = α, and thus ∠PBF = β. Therefore ∆DEP and ∆BFP are similar triangles. As the area of ∆BFP is four times the area of ∆DEP, then BF = 2DE and FP = 2EP. Proof below: BF/FP = k

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergeyvinns931]

DE*EP/2=7, FB*PF=28, PF=2EP, AB=3EP, FB=2DE, AD=3DE, DE*EP=14, 14*9/2 =63, Area Red Nrapezoid=63-28=35. Area Blue Trapezoid=63-7=56.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@RobertHering-tq7bn]

Hello, I did it as following. First two points were added, G and H, connected over P and making GH parallel to AD and BC. With this we get rectangles DEPG with size of 14, DPFC with a size of named B, AHPE with a size of named A and HBFP with a size of 56. With this added we have... Area(RT) = B+7 and Area(BT) = A+28 A:14 = EP*HP : (EP*PG) = HP:PG = PZ*HP : (PZ*PG) = 56:B AB = 14*56 HB:HP = EP:ED HB*ED = EP*HP B = A This is true based on similarty reasons. Now we have... AB = A² = 14*56 = 784 = 7²*4² ==> A = 28 and B = 28 . Finally... Area(RT) = B+7 = 28 + 7 = 35 Area(BT) = A+28 = 28 + 28 = 56

[@marcgriselhubert3915]

Simple when we see taht triangles DEP and PFB are similar, as the area of BFP is four times the area of BFP we conclude that the dimensions of BFP is two times the dimensions of DEP. So we note ED = k and so AE = 2.k, and also EP = m and also PF = 2.m Tha area od triangle DEP is 7, so k.m = 14 The area of the big rectangle ABCD is (3.k).(3.m) =9.(km) = 9.14 = 126. The area of rectangle DEFC is one third of that, so it is 126/3 = 42, and the area of the red trapezoïd is 42 - 7 = 35. The area of rectangle EABF is two thirds of 126, so it is 84, and the the area of the blue trapezoïd is 84 - 28 = 56.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@misterenter-iz7rz]

7:28=1:2^2, therefore the red area is 7+14×2=7×5=35, the blue area is 28+56×(1/2)=56.😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@user-ky7cf6zm3d]

magnifique démonstration Merci

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@batavuskoga]

I solved it completely different Let's say FP=x, BF=y, triangle BFP : area=xy/2 --> xy=56 EP must be x/2, DE=y/2, because area triangle DEP=x/2*y*2*1/2=xy/8=7 AB=CD=3x/2, AD=BC=3y/2 Now let's calculate the blue trapezium area blue=(x/2+3x/2)*y*1/2=2x*y*1/2=xy area blue=56 area red=(3x/2+x)*y/2*1/2=5x/2*y/2*1/2=5xy/8=5*56/8=35 area red=35 I also calculated the area of the rectangle to check my solution, everything was ok.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

1/ The two yellow triangles are similar and the area of the big one is 4 times bigger than the small one, which means that the ratio of the relevant sides is 2 times. So, PF=2PE, and FB=2ED … 2/ From P just draw the PH and PK perpdndicular to AB and CD we czn easily find out that: Blue area= 2((7+7)+28=56 sq units Red area=. (7 + 2x14)= 35 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 1) Draw a Vertical Line passing through Point P. 2) Point between C and D is Point G. 3) Point between A and B is Point H. 4) Now we have 4 Rectangles : 5) Rectangle [DEPG] with Area = 14 sq cm 6) Rectangle [BFPH] with Area = 56 sq cm 7) Rectangle [CFPG] with unknown Area 8) Rectangle [AEPH] with unknown Area 9) 2D Geometrical Ratio (R^2) between the Area of Triangle [BFPH] and the Area of Triangle [DEPG] ; R^2 = 56 / 14 ; R^2 = 4 10) Linear Ratio (R) between Sides of those Triangles ; R = sqrt(4) ; R = 2 11) One must conclude that BH (CG) = 2*AH (DG) and AE (BF) = 2*DE (CF) 12) Now I can divide the Rectangel [ABCD] in 6 different Rectangles, and 12 Different Triangles; and though determine their Areas. 13) Rectangel [CDEF] divided in 3 Rectangles; each one with Area = 14 sq cm 14) Rectangle [ABEF] divided in 3 Rectangles; each one with Area = 28 sq cm 15) AB = 21 cm 16) AD = 6 cm 17) Total Area of [ABCD] = 126 sq cm 18) Rectangle [ABEF] Area = 28 * 3 = 84 sq cm 19) Rectangle [CDEF] Area = 14 * 3 = 42 sq cm 20) Blue Trapezoid Area = (84 - 28) = 56 sq cm 21) Red Trapezoid Area = (42 - 7) = 35 sq cm 22) ANSWER : BTA equal to 56 Square Cm and RTA equal to 35 Square Cm.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@MrPaulc222]

Calling the areas of the red and blue trapezoids r and b respectively, r + 28 = b + 7 due to the diagonal. b - r = 21 Not certain, but I think the large yellow triangle muct be double the size of the small yellow triangle in both directions, so I am tempted to state that the non-hypotenuse side lengths are 2 by 7 and 4 by 14, although it doesn't look that way on the diagram. If that is the case, the rectangle is 21 by 6, making a total area of 126. As the triangles take up 35 of that, the two trapezoids total 91. Back to b - r = 21 and now b + r = 91 Add the two equations for 2b = 112, so b = 56. As b = 56, r = 35 Blue trapezoid = 56 sq un Red trapezoid = 35 sq un I have a feeling that I went wrong because I think it should be more complicated than that. However, either side of the diagonal comes to 63. EDIT: I have returned to this because I wasn't entirely happy that I chose assumed values for the triangles' sides, so I have chosen other lengths to verify my answer. Let the small triangle have side lengths of 12 and 14/12 (or 7/6) because (12*(7/6))/2 = 7. This means that the larger triangle has sides of 24 and 7/3 giving an area of 28. The rectangle would now be 36 by 21/6, giving a rectangle area of 126. Yes, I thought it would be the case that the area is 126 whichever way I slice it, but I needed to verify.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@yalchingedikgedik8007]

Thanks Sir We are learning more and more from like these exercises. ❤❤❤❤❤❤ My respects

[@PreMath]

It's my pleasure, dear🌹 Thanks for the feedback ❤️

[@raya.pawley3563]

Thank you

[@keithwood6459]

Can do it graphically in seconds once similarity is recognized.

[@himo3485]

7 : 28 = 1 : 4 = 1^2 : 2^2 DE=x EP=y BF=2x FP=2y xy/2=7 xy=14 rectangle ABCD = 3x＊3y=9xy=126 Red Trapezoid area = 126/2 - 28 = 35cm^2 Blue Trapezoid area = 126/2 - 7 = 56cm^2

[@PreMath]

Excellent! Thanks for sharing ❤️

[@shreedhanmehta3553]

Noooo... If the areas ratio is 1:4, ie, sides ratio is 9:25... But then the actual sides of the triangle may not necessarily be 1 and 2, then can be 2 and 4 or 10 and 20. So you have to take the sides as x and 2x. ❤❤

[@johnryder1713]

If a shapes length is x and it is a triangle how is x ^2 I thought that would be only true of a square please can you tell me? Or is that only with some triangles?

[@msafasharhan]

Hello,sir what you mean by circle

[@PreMath]

It is just a number with different identity! Thanks for asking❤️

[@professorrogeriocesar]

Show

[@sebasiegrist9341]

By moving the point where triangles touch to the center, I got an area of 35 for a quarter of the rectangle. Anyone else?

[@PreMath]

Thanks for the feedback ❤️