Can you find area of the Yellow shaded Trapezoid? | (Two Methods) |

[@jarikosonen4079]

Very well proven...

[@PreMath]

Glad you think so! Thanks for the feedback ❤️

[@Mastero-dj4uw]

Thanks for the sharing this problem ! I would like to share a shortcut that starts from here : 1:50 From point H drop a perpendicular line segment on AB at point R We get the right triangle HRF and since the side length of the square is a+5 , we get the following lengths of triangle HRF HR = a + 5 RF = a - 5 HF = 50 Using Pythagorean theorem we get a = 35 , means that the side length of the square is 40 square area = 1600 Half of it is 800 which is the answer

[@gaylespencer6188]

Extended FP to DC. New line is 50. Then drew identical parallel line to the right +5 on line AB. Used PT, 50^2=(a-10)^2+a^2. a=+/- 40. Reject -40, then solve.

[@alexniklas8777]

Let's denote the side of the square as 'x'. Area CDEG= x^2/2. According to the Pythagorean theorem: (a-10)^2+a^2=(25+25)^2; a^2-10a-1200=0; a= 40. S=40^2/2= 800 Thanks sir😊

[@calvinmasters6159]

FGP is a 45deg triangle, therefore sides are 1 - 1 - sqrt2. hypoteneuse is 25 * sqrt2. Thin wedge triangle side a^2 = (25 * sqrt2)^2 - 5^2 a = 35 Side of major triangle = 35 + 5 Area of major triangle = 40 * 40 = 1600 Half of that = 800

[@calvinmasters6159]

Correction : major square

[@ChuzzleFriends]

Extend segment FP to a point on side CD. Label the point H. Connect vertices E, F, & G together to form a triangle △EFG. Sides EF & FG are hypotenuses of right triangles △EAF & △FBG (Both are right triangles by definition of squares). Label the side length of the blue square as s. So, AE = BF = s - 5. So, △EAF ≅ △FBG by SAS. By CPCTC, EF = FG. Now look at △EPF & △GPF. They are also right triangles and they share leg FP. So, △EPF ≅ △GPF by HL. EP = GP by CPCTC. We will now focus on quadrilaterals BFPG & DHPE. We know the following info about both: • ∠B & ∠D are right angles. • BG = DE = 5. • ∠BGE ≅ ∠DEG by the Alternate Interior Angles Theorem (because ABCD is a square, so opposite sides are parallel) • EP = GP. • ∠EPH & ∠FPG are right angles. So, BFPG ≅ DHPE by ASASA. Therefore, by definition of corresponding parts in congruent figures, FP = HP = 25 & BF = DH = s - 5. Finally, compare trapezoids ADHF and DCGE (the yellow trapezoid). We know THIS info about both: • AF = DE = 5. • ∠A, ∠D, & ∠C are right angles. • AD = CD = s. • CG = DH = s - 5. So, ADHF ≅ DCGE by SASAS. So, EG = FH = 50. But EP = GP, so the segments are 25 units long each. By definition of isosceles right triangles, EF = 25√2. Use the Pythagorean Theorem on △EAF. a² + b² = c² (s - 5)² + 5² = (25√2)² (s - 5)² + 25 = 1250 (s - 5)² = 1225 s - 5 = 35 s = 40 We must substitute the missing side lengths for trapezoid DCGE before doing the final step. So, CD = 40 & CG = 35. Finally, find the area of the trapezoid. A = [(a + b)/2]h = 1/2 * (5 + 35) * 40 = 20 * 40 = 800 So, the area of the yellow trapezoid is 800 square units.

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@cleiberrocha1449]

If we draw the diagonal BD of the square, we can observe that we will have the triangle above adding a value to the yellow area. However, below we will have the DEP triangle decreasing exactly the same value as the area. Therefore, it will remain at the same value. So, we just need to calculate half the area of ​​square ABCD A = 1&__(a + 5) (a + 5) A = (35 + 5) (35 +5) A = 40 x 40 = 1/2 (1600) A = 800

[@blobfish1112]

a^2 + 5^2 = (25sqrt(2))^2 --> a=35. A= (a+5)^2/2=800.

[@user-xz7sv2dn9g]

So i do See it🎉

[@harikatragadda]

Rotating EG by 90° anti-clockwise makes it coincide with FH because of Symmetry, hence EG = FH = 2*25

[@santiagoarosam430]

El cuadrilátero de la figura es un cuadrado y está dividido en cuatro trapezoides iguales---> el área amarilla es la mitad del cuadrado ---> (25√2)²-5²=b²---> b=35---> Área amarilla =(35+5)²/2=800 ud². Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@naveenkumarjha5428]

At first there should be condition that we are working in a square field. In yellow part there is no issue but in white part there is an issue. AB may or may not be parallel and equal to DC. Only by observation it is looking that all corners are at right angles. But there is no symbol given. Any way let us start considering the whole shape is square. Let us assume blue line is x. Side of the square will be x+5. Now area of the yello trapezoid will be 1/2 ( sum of parallel sides)×perpendicular distance between parallel sides = 1/2 (x+5)(x+5) And length of the slant side of the yellow trapezoid is ((x+5)^2+(x-5)^2)^0.5 = (x^2+10x+25+x^2-10x+25)^0.5= (2x^2+50)^0.5 Connect F to E and G. It will form a triangle Area of this triangle EFG will be 1/2 (EG×PF)= 1/2 × ((2x^2+50)^0.5)×25 Area of triangle AEF= area of triangle FBG= 1/2 *5x Now area of the white part will be same as yellow part. Thus area of triangle EGG in this way will be 1/2 (x+5)^2 - 5x =(x^2+25)/2 This area should be equal to area calculated earlier I.e. 1/2 × ((2x^2+50)^0.5)×25 =(x^2+25)/2 Or, ((2x^2+50)^0.5)×25 =(x^2+25) Squaring on both sides we get 2(x^2+25)*625 =(x^2+25)^2 Or, x^2+25=1250 Or, x^2=1250-25=1225 Or, x=35 Thus area of trapezoid is 1/2 (x+5)^2 =1/2 *(35+5)^2=1/2*40*40=800

[@michaelkouzmin281]

One more method: 1) Let a = DC = BC; 2) Ayellow = (a-5+5)*a/2 = a^2/2; 3) A(EABG) = A(EDCG); 4) A(EAF) = A(FBG) = 5*(a-5)/2; 5) A(EFG) = EG*25/2; EG^2=a^2+(a-5-5)^2 => EG = sqrt(a^2+(a-10)^2); 6) 2*5*(a-5)/2 + 25* sqrt(a^2+(a-10)^2)/2 = a^2/2; 2*5*(a-5) + 25* sqrt(a^2+(a-10)^2) = a^2; 25* sqrt(a^2+(a-10)^2) = a^2 - 2*5*(a-5); (25* sqrt(a^2+(a-10)^2))^2 = (a^2 - 2*5*(a-5))^2; 625*(a^2+a^2-20*a+100) = (a^2-10*a+50)^2; 1250*(a^2-10*a+50) = (a^2-10*a+50)^2; 7) x = a^2-10*a+50; 1250*x = x^2; x1=0; x2=1250; 8) a^2-10*a+50 = 0; No real roots; a^2-10*a+50 = 1250; a1 = -30 rejected; a2 = 40 9) Ayellow = a^2/2= 40^2/2 = 800 sq units.

[@giuseppemalaguti435]

√(l^2+(l-10)^2)=√((l-5)^2+5^2-625)+√((l-5)^2+5^2-625)...√(2l^2-20l+100)=2√(l^2-10l-575).. 2l^2-20l+100=4(l^2-10l-575)...2l^2-20l-2400=0...l^2-10l-1200=0...l=5+√1225=40... Ayellow=(5+35)40/2=40*40/2=800

[@marcgriselhubert3915]

We use an orthonormal center D and first axis (DC). E(2; 5) G(c; c - 5) F(5; c), with c the side length of the square. VectorEG(c; c -10) Equation of (EG): (x).(c - 10) - (y - 5).(c) = 0 or (c - 10).x - c.y + 5.c = 0 Distance from F to (EG) : abs((c -10).(5) - c^2 +5.c)/ sqrt((c -10)^2 + c^2) = abs(-c^2 +10.c -50)/sqrt(2.c^2 -20.c +100) This distance is 25. We square and obtain: c^4 +100.c^2 +2500 -20.c^3 +10.c^2 -1000.c = 1250.c^2 - 12500.c + 625.. or c^4 -20.c^3 -1050.c^2 + 11500.c -60000 = 0 or (c -40).(c^3 + 20.c^2 -250.c +1500) = 0 c = 40 is the unique positive solution, as if we note f(c) = c^3 + 20.c^2 -250.c +1500 then f has a minimun strictly positive for x = (-20 + 5.sqrt(46))/3 Finally c = 40 Then the yellow area is 40.((5 +35)/2) = 800.

[@PreMath]

Thanks for sharing ❤️

[@Waldlaeufer70]

We can construct a black cross in the middle with arms 25. Now let's build a right triangle PF(middle of side a of square): (a/2)² + (a/2 - 5)² = 25² a²/4 + a²/4 - 5a + 25 = 625 2a²/4 - 5a + 25 = 625 a²/2 - 5a + 25 = 625 a² - 10a + 50 = 1250 a² - 10a + (50 - 25) = 1225 (a - 5)² = 1225 a - 5 = +- 35 (negative solution rejected!) a = 35 + 5 = 40 A(yellow) = a²/2 = 40²/2 = 800 square units (since the yellow area is half of the square due to symmetry)

[@toninhorosa4849]

I solved a little different: I have a Square ABCD AB = BC = DC = AD = x AF = BG = ED = 5 From point F we draw a line parallel to line AD until we find point Q on line DC. We extend line FP until we find point H on line DC. We have now: AF= DQ= HC = 5 FP= PH = 25 FH = FP+PH = 25+25 = 50 QH = x - DQ - HC QH = x - 5 - 5 = (x - 10) ∆FQH applying Pythagoras: FH^2 = FQ^2 + QH^2 50^2 = x^2 + (x - 10)^2 2500= x^2 + x^2 - 20x + 100 2x^2 - 20x - 2400 = 0 (÷2) x^2 - 10x - 1200 = 0 x = (10+-(√4900))/2 X1 = (10+70)/2 = 40 accepted X2 = (10-70)/2 = - 30 rejected Then: x = 40 Area Trapezoid= ((B+b)* h)/2 = ((35+5)*40)/2 Area Trapezoid EDCG = 800 units Square.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johnbrennan3372]

Draw triangle EFG. Area of triangle EFG is (1/2)25 |eg|. Let side of square= x. Draw eh parallel to dc .In the triangle EGH , x^2 +(x-10)^2=|eg|^2. Therefore |eg|= sqroot of 2(x ^2)-20x+100. Area ofABGE = 1/2(5+x-5)x =1/2 (x^2).Area of ABGE is also equal to 25/2 by sqroot of 2(x^2)-20x+100 +2(1/2)5(x-5). Putting that equal to 1/2(x)^2 we get 25 by sqroot 2(x^2)-20x+100=(x)^2-10x +50. Divide both sides by sqroot of x^2-10x+50. From that we get 25 by sqroot 2= sqroot x^2-10x+50. (We also get sqroot of x^2-10x+50=0 but that doesn’t have real roots).Squaring both sides of 25sqroot2=sqroot (x^2-10x+50 )we get x^2 -10x -1200=0 which gives x=40.So shaded area =800.

[@ΙΩΑΝΝΗΣΣΤΑΥΡΟΥΛΑΚΗΣ-ο5π]

I have already mentioned that this is called trapezium (τραπεζιον). Trapezoid is something else. Please correct it.

[@davidstecchi9501]

Once you solved for a, there was no need to calculate the area of the square ABCD as you had all the information needed to use the formula for the area of a trapezoid

[@PreMath]

Thanks for the feedback ❤️

[@rabotaakk-nw9nm]

=> ΔEAF=ΔFBG(SAS) => EF=FG;

[@sergioaiex3966]

Solution: Yellow Area (YA) = ½ h (a + b) YA = ½ (a + 5) (a + 5) YA = (a + 5)²/2 A1 = ½ a . 5 A1 = 5a/2 A2 = The same area A1 A2 = 5a/2 A3 = ½ b . h A3 = ½ 50 . 25 A3 = 625 White Area (WA) is equal to Yellow Area, then: WA = A1 + A2 + A3 (a + 5)²/2 = 5a/2 + 5a/2 + 625 (×2) to remove the fration a² + 10a + 25 = 5a + 5a + 1250 a² + 10a + 25 = 10a + 1250 a² = 1225 a = √1225 a = 35 Thus the side of the square is 35 + 5 = 40 Then, YA = ½ Square YA = ½ (40)² YA = 1600/2 YA = 800 Square Units ==================

[@PreMath]

Excellent! Thanks for sharing ❤️

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ AB= DC which is a square ! AE= x also FB= GC = x DC= x+5 ⇒ Area of the square, A(ABCD) A(ABCD)= (x+5)² A(EDCG)= (x+5)*(x+5)/2 A(EDCG)= (x+5)²/2 ⇒ A(EGBA)= (x+5)²/2 ⇒ A(EDCG)= A(EGBA) both trapezoids have the same area ! Let's build the triangle ΔEGF, which is equal to the area of: A(EGBA) - A(ΔEFA) - A(ΔGBF) ⇒ A(ΔEGF)= A(EGBA) - A(ΔEFA) - A(ΔGBF) A(ΔEFA) = 5*x/2 A(ΔGBF)= 5*x/2 both triangles have the same area. A(ΔEGF)= (x+5)²/2 - 2*(5x/2) A(ΔEGF)= (x²+10x+25)/2 - 5x A(ΔEGF)= x²/2 +5x + 25/2 -5x A(ΔEGF)= x²/2 + 25/2 A(ΔEGF)= EG*FP/2 x²/2 + 25/2 = EG*25/2 EG= 1+ x²/25 which is the base of this triangle ! EK // DC= 5+x KG= x-5 According to the Pythagorean theorem: (1+x²/25)²= (x-5)²+ (5+x)² 1+ 2x²/25 + x⁴/625 = x²-10x+25 + 25+10x+x² 1+ 2x²/25 + x⁴/625 = 2x²+50 x²/25 = u x²= 25u ⇒ 1+ 2u + u²= 50u+ 50 u²-48u-49=0 Δ= 48²-4*1*(-49) Δ= 2500 √Δ= 50 u₁= (48+50)/2 u₁= 49 x²/25 = 49 x²= 25*49 x₁= √25*49 x₁= 5*7 x₁= 35 length units u₂= (48-50)/2 u₂= 1 x²/25 = 1 x²= 25 x₂= √25 x₂= 5 ❌ x₂ > 5 ⇒ x= 35 length units A(EDCG)= (x+5)*(x+5)/2 A(EDCG)= (35+5)*(35+5)/2 A(EDCG)= 40²/2 A(EDCG)= 800 square units !

[@PreMath]

Thanks for sharing ❤️

[@prossvay8744]

Connect P to H (H on CD that CH=5) and FP=PH=25 and connect F to I (l on CD) Let side of the square is x So FI=AD=x In∆ FIH FI^2+HI^2=FH^2 x^2+(x-10)^2=50^2 So x=40 AE=BF=CG=40-5=35 So yellow shaded area=40^2-1/2(5+35)(40)=800 suuare units.❤❤❤

[@PreMath]

Thanks for sharing ❤️

[@nenetstree914]

800

[@PreMath]

Thanks for sharing ❤️

[@wackojacko3962]

Alternative method is just more work. I don't work, I just get paid. 🙂

[@LuisdeBritoCamacho]

😂😂😂😂😂😂😂😂😂😂

[@PreMath]

Fair enough! Thanks for the feedback ❤️

[@LuisdeBritoCamacho]

The fastest way to solve this. Let Square Side = 2X 25^2 = X^2 + (X - 5)^2 625 = X^2 + X^2 - 10X + 25 625 = 2X^2 - 10X + 25 2X^2 - 10X - 600 = 0 X^2 - 5X - 300 = 0 (-5X = -20X + 15X and -20 * 15 = -300) X = -15 or X = 20 2X = 40 40 * 40 = 1600 Square Units Trapezoid [CDEG] = 1600 / 2 = 800 sq un Yellow Shaded Area equal 800 Square Units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

@@PreMath , All I post here is for free! I don't get paid for nothing!! It's "Pro Bono"!!

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let's assume that D is the center of the coordinate system and that DC is located on the x-axis. With s being the side length of the square we obtain the following coordinates: A: ( 0 ; s ) B: ( s ; s ) C: ( s ; 0 ) D: ( 0 ; 0 ) E: ( 0 ; 5 ) F: ( 5 ; s ) G: ( s ; s−5 ) P: ( xP ; yP ) Now we have a closer look at the triangles AEF and BFG: AEF: ∠EAF = 90° AE = s−5 AF = 5 BFG: ∠FBG = 90° BF = s−5 BG = 5 Therefore these two triangles are congruent and we can conclude: EP = GP ⇒ xP = (xE + xG)/2 = (0 + s)/2 = s/2 ∧ yP = (yE + yG)/2 = [5 + (s − 5)]/2 = s/2 From the known length of the line FP we obtain: FP² = (xP − xF)² + (yP − yF)² 25² = (s/2 − 5)² + (s/2 − s)² 25² = (s/2 − 5)² + (−s/2)² 625 = s²/4 − 5*s + 25 + s²/4 0 = s²/2 − 5*s − 600 0 = s² − 10*s − 1200 ⇒ s = 5 ± √(5² + 1200) = 5 ± √(25 + 1200) = 5 ± √(1225) = 5 ± 35 Since s>0, the only useful solution is s=5+35=40. Now we are able to calculate the area of the yellow trapezoid: A(CDEG) = (1/2)*(DE + CG)*CD = (1/2)*[5 + (s − 5)]*s = s²/2 = 40²/2 = 1600/2 = 800 Best regards from Germany

[@devondevon4366]

800