white piece = 15 x 8 + 16 x 4 : 2 = 152 green = 256 - 152 = 104 @@PreMath
@Alishbavlogs-bm4ip
9 ай бұрын
Mashallah great sharing❤❤❤
@PreMath
9 ай бұрын
Thank you sister❤️
@hcgreier6037
9 ай бұрын
Easy peasy 😂 No use for trig functions here... Complete the outside to a full rectangle, the bottom right point is G. Area is 18·16 = 288. Draw a vertical line through F and intersect it with CD => point H. Now one can see △AGB = △FHE with the area (16·4)/2 = 32. The trapezoid FBCH has the area 8·(16+14)/2 = 120 The green area is then ☐AGCD - 2·△AGB - Trapezoid FBCH = 288 - 2·32 - 120 = 104 square units.
@montynorth3009
9 ай бұрын
Draw a horizontal line from point F to meet line AD at G. Then GD = 16, & GA = 2, because point G is midway between 18 & 14 by symmetry. Drop a perpendicular from point E on to line GF to meet at H. We now have 2 triangles and a rectangle making up the green area. Bottom triangle AGF is similar to right triangle EHF. F is the horizontally mid point of the 16 dimension. So 8 / 2 = 16 / HF. HF = 4. The green area = area of 2 triangles + 1 rectangle. Bottom triangle = 1/2 x 8 x 2 = 8. Triangle on right = 1/2 x 16 x 4 = 32. Rectangle = 16 x 4 = 64. (Width GH = 8 - 4) Total = 104.
@howardaltman7212
9 ай бұрын
I see that we had the same idea.😄
@PreMath
9 ай бұрын
Thanks❤️
@hermannschachner977
9 ай бұрын
white piece = 15 x 8 + 16 x 4 : 2 = 152 green = 256 - 152 = 104
@PreMath
9 ай бұрын
Thanks❤️
@phungpham1725
9 ай бұрын
1/ From F draw a perpendicular line to AD and BC intersecting AD and BC at A' and B' respectively. We have the two right triangles AA'F and BB'F are congruent and AA'=BB' =2 and A"F=B'F= 8 and DA' =16. 2/ From E drop the height EH to A'F. Because AE is also perpendicular to EF so the angles A'FA and HEF are congruent so the two triangles A'FA and HEF are similar----> AA'/AF=HF/HE=1/4----> HF = 16/4 = 4 ----> A'H=4 and ED= 4 3/The area of the blue region = Area of the A'DEF + Area of triangle A'FA = (4+8)/2 x 16 + 1/2 x 2x8 = 96+8= 104 sq units
@Copernicusfreud
9 ай бұрын
That is how I solved the problem.
@PreMath
9 ай бұрын
Thanks❤️
@AmirgabYT2185
3 ай бұрын
S=104 square units
@quigonkenny
5 ай бұрын
Extend CB to G, where G is the point where AG is perpendicular to DA. As DA and BC are both perpendicular to CD, DA and BC are parallel, so AG is perpendicular to GC as well, forming rectangle AGCD. As AGDC is a rectangle, GB = DA-BC = 18-14 = 4, and AG = CD = 16. Triangle ∆AGB: GB² + AG² = BA² 4² + 16² = BA² BA² = 16 + 256 = 272 BA = √272 = 4√17 As AF = FB and AF+FB = AB, AF = FB = (4√17)/2 = 2√17. Let ED = x, so CE will be 16-x. Draw AE and BE. As AF = FB and EF is common, right teiangles ∆AFE and ∆EFB are congruent. Let AE = BE = y. Triangle ∆EDA: ED² + DA² = AE² x² + 18² = y² x² + 324 = y² ---- [1] Triangle ∆BCE: CE² + BC² = BE² (16-x)² + 14² = y² 256 - 32x + x² + 196 = y² x² - 32x + 452 = y² ---- [2] x² - 32x + 452 = (x² + 324)
@marcgriselhubert3915
9 ай бұрын
Let's use an adapted orthonormal. D(0;0) A(18;0) C(0;16) B(14;16) Then VectorAB (-4;16), or Vector U(1;-4) is orthogonal to straight line (EF) Equation of (EF): (x-16).(1) + (-4).(y-8) = 0, or x -4y +16 = 0. Its intersection with (DC), which equation is x =0, is E(0,4) and so we have DE = 4. The area of right triangle ADE is (1/2). AD. DE = (1/2).18.4 = 36. We now have VectorAF(-2;8) and so AF = sqrt (4+64) = sqrt(68) = 2.sqrt(17) and also VectorEF(16;-4) and EF = sqrt(256+16) = sqrt(272) = 4.sqrt(17). The area of right triangle AFE is (1/2).AF.EF = (1/2). (2.sqrt(17).(4.sqrt(17)) = 4.17 = 68. Finally the unknown green area is the sum of the areas of the triangles ADE and AEF, so it is 36 + 68 = 104.
@howardaltman7212
9 ай бұрын
As an alternative, locate K on CE so that segment FK//BC and locate J on AD so that segment BJ//DC. Hence FK is the midsegment of trapezoid ABCD with FK=(14+18)/2=16. ΔAJB is a right triangle with AJ=18-14=4 and JB=16. Since ∠A≅∠KEF, ΔAJB≅ΔEKF by AAS. Hence EK=4 by CPCTC. Green area =[ADKF] - [EKF] = 8(18+16)/2 - 1/2(4)16 = 104.
Just going to throw a random number out there and see if it works...Area = 104
@prossvay8744
9 ай бұрын
Area of the green shaded region=1/2(14+18)(16)-1/2(14)(12)-1/2(2√17)(4√17)=256-84-68=104 square units. ❤❤❤ Thanks
@PreMath
9 ай бұрын
Thanks❤️
@giuseppemalaguti435
9 ай бұрын
104
@PreMath
9 ай бұрын
Thanks❤️
@LuisdeBritoCamacho
9 ай бұрын
Geometric Resolution: The Green Area is inside a Rectangle with Area = 16 * 18 = 288 su. Removing the Triangle Area [A B B'] = (4 * 16) / 2 = 32 su we get a Quadrilateral with Area = 288 - 32 = 256 su Area of the Trapezoide [A D E F'] being F' the intersection of EF to a Line passing the point A and perpendicular to AD. B = 8,5 b = 4 h = 18 Trapezoid Area = (8,5 + 4) * 18/2 Trapezoid Area = 12,5 * 9 Trapezoid Area = 112,5 su Removing the Area of the Triangle [A F F'] with Base = Sqrt(68) and Height = Sqrt(4,25). Triangle Area: (Sqrt(68) * Sqrt(4,25))/2 = (sqrt(289))/2 = 17/2 = 8,5 Green Area = 112,5 - 8,5 = 104 Answer: Green Area equal to 104 su.
@PreMath
9 ай бұрын
Thanks❤️
@Lord_Volkner
8 ай бұрын
You sure went about that the hard way.
@wackojacko3962
9 ай бұрын
@ 11:04 ...I can't help myself and need too throw some politics into the discussion here and just say that a lot of inner cities schools have been stupefied and math has become racist and have no idea of how to simply add 36 + 68 ! So sad. @ 😢
@PreMath
9 ай бұрын
Thanks for the feedback❤️
@maxforsberg8852
9 ай бұрын
You can simplify the calculation by looking at angle DAB and realizing that tan of said angle is 4. This angle then reappears if you draw a vertical line from F to bisect line CD to a point G located 8 units from both C and D. Now DAFG form a trapezoid with base 8 and heights 18 and 16. You can calculate the area of this trapezoid as being 136 square units by using the area of trapezoid formula. You now obtain the area of the green region by subtracting the triangle FGE which has a height of 16 units. Because of the 4:1 tangent ratio the base of this triangle becomes 4 and the area becomes 32 square units. Finally area of green region = area of trapezoid DAFG - area of triangle FGE = 136 - 32 = 104 square units.
@PreMath
9 ай бұрын
Thanks❤️
@tombufford136
9 ай бұрын
At a quick glance and some thought, my solution is AF horizontal = AFx = 8 and AF vertical = AFy = 1/2(18-14) = 2.Ratio of AF to AFx = AF/AFx. AF^2 = 8^2 + 2^2. AF/AFx = sqrt(68)/sqrt(64). FE vertical = FEy. ratio FE:FEy = ratio AF: AFx. FEy = 14+ 2 =16. FE = 16 * sqrt(68)/8 . FE = 16.49 . FE horizontal = FEx. FEx^2=16.49^2 - 16^2. FEx = G and AFx = H then the green area is calculated by subtracting the areas of triangles AFH and FEG from area of rectangle AD * 1/2 DC = 18* 8= 144. area AFH = 1/2 * 8 * 2 = 8. area FEG = 1/2 * 16 * 16.49= 132. The Green area = 144-8-132 = 104 Units. Thank you for this puzzle I hope I agree with you !
@robertlynch7520
8 ай бұрын
I don't think anyone approached it quite like this: The line AB has function (𝒚 = ¼𝒙 ⊕ 0), where bottom left is (0, 0). The perpendicular line FE has the function (𝒚 = -4𝒙 + 34). How "34"? because at (𝒙 = 8), this gives ⊕2 rise on the line, which is half way to "4". With that second equation we solve for 18 = -4𝒙 + 34 … 4𝒙 = (34 - 18) = 16 𝒙 = 4 Thus the intercept at the top is (𝒙, 𝒚) = (4, 18). Now there are 3 elements to find the area of. The rectangle starting at (0, 2) to (4, 18) (diagonals) which has an area of Δ𝒙 • Δ𝒚 = 4 × 16 = 64 u²; The next is the △ to the right of the rectangle. It likewise starts at (4, 2), up to (4, 18) and then (8, 2), which being a right △ solves as ½ × 4 × 16 = 32 u²; Last is the bottom little △, which has a 'base' of 2 u, and a width of 8: ½ × 2 × 8 = 8 u²; Add 'em up … 64 + 32 ⊕ 8 = 104 u² And there you are. Another solution. Remembering that the 𝒚 = 𝒎𝒙 + 𝒃 line has a perpendicular of 𝒚' = -1/𝒎 𝒙 + 𝒃' is critical. Didn't need any of the Isoceles △ business either. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
@Copernicusfreud
9 ай бұрын
Yay! I solved the problem.
@PreMath
9 ай бұрын
Great! Thanks❤️
@lukeheatley4148
9 ай бұрын
i went with (16 x 16) - (15 x 8) - (16 x 4)/ 2 = 256 - 120 - 32 = 104
@PreMath
9 ай бұрын
Thanks❤️
@Ibrahimfamilyvlog2097l
9 ай бұрын
Veri nice sharing sir❤❤❤❤
@PreMath
9 ай бұрын
Thanks ❤️
@batavuskoga
9 ай бұрын
I solved it in a completely different way Draw a perpendicular line from line DC to point F. We call this point H. Also draw the line parallel to DC from point A and extend line CB downwards Where these two line come togerther, we call this point G AG²+BG²=AB², AG=CD=16, BG=18-14=4 16²+4²=AB²=272 --> AB=sqrt(272)=4*sqrt(17), AF=BF=AB/2=2*sqrt(17), AF²=68 CH=8, FH=16 (AF=BF), EH=b, DE=8-b AE²=18²+DE²=18+(8-b)²=324+64-16b+b² -->AE²=b²-16b+388 AE²=AF²+EF²=68+EF² EF²=FH²+b² --> EF²=256+b² Combine the three last equations AE²=b²-16b+388, AE²=68+EF², EF²=256+b² b²-16b+388=68+EF² b²-16b+388=68+256+b² --> 16b=388-68-256=64 --> b=4 DE=8-b --> DE=4 EF²=256+b² --> EF²=272 --> EF=sqrt(272)=4*sqrt(17) green area=area triangle ADE+ area triangle AEF area triangle AEF=2*sqrt(17)*4*sqrt(17)/2=68 area triangle ADE=AD*AE/2=18*4/2=36 green area = 36+68=104 First I divided BCEF in two triangles, calculated their areas and subtracted them from the whole area of the trapezium area BEF=2*sqrt(17)*4*sqrt(17)/2=68 area BCE=14*12/2=84 area trapezium=(18+14)*16/2=256 area green area = area trapezium-area BEF-area BCE area green area=256-68-84=104
@PreMath
9 ай бұрын
Thanks❤️
@santiagoarosam430
9 ай бұрын
La horizontal por F corta a AD en G. H es el punto medio de DC》HF=(18+14)/2=16 Los triángulos AGF y EHF son semejantes》GA/GF=EH/HF》2/8=EH/16》EH=4=ED 》 ADEF=GDEF+AGF =[16(8+4)/2]+(2×8/2) =96+8 =104 Gracias y saludos.
@PreMath
9 ай бұрын
Thanks❤️
@robertbourke7935
9 ай бұрын
Got it! Many thanks once again.
@alster724
9 ай бұрын
Yes, I got it! Another great way to start 2024
@PreMath
9 ай бұрын
Thanks❤️
@misterenter-iz7rz
9 ай бұрын
tan x=1/4, sin x=1/sqrt(17), cos x=4/sqrt(17), AF=1/2 4/sin x=2sqrt(17), AG=AFsin x=2, so DG=18-2=16, HF=DG tan x=4, GF=AFcos x=8, DE=GH=8-4=4, therefore the area is 2×8/2+1/2(4+8)×16=8+6×16=8+96=104.😊
@PreMath
9 ай бұрын
Thanks❤️
@misterenter-iz7rz
9 ай бұрын
@@PreMathYou show a clever method that need not trigonometry. 😮
@hermannschachner977
9 ай бұрын
the white piece = 15 x 8 + 16 x 4 : 2 = 152 the green = 256 - 152 = 104
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