Can you find area of the Yellow shaded Polygon?

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Can you find area of the Yellow shaded Polygon? | (Math Olympiad) |

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Learn how to find the area of the Yellow shaded Polygon. Important Geometry and Algebra skills are also explained: isosceles Triangles; Interior angles of the polygon formula; triangle area formula. Step-by-step tutorial by PreMath.com
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Пікірлер: 39

@anatoliy3323 Ай бұрын

Nice task about remembering polygons💯👍Thank you so much, Professor. Nice weekend to you:)) 😊

@PreMath Ай бұрын

Excellent! Glad to hear that! Thanks for the feedback ❤️🌹 take care

@skverma7278 Ай бұрын

Draw two parallel lines (parallel to line having length 5 units), from the remaining two corners. (1) Length of the smaller line will be 3√2 units. (2) Length of the trapezium will be 5-3√2 units. Now Total area= (A) Ar.of rectangle 5×2=10 units. (B) Area of top triangle= 0.2×3×3=4.5 units. (C) Area of middle trapezium= 0.5×(5+3√2)× (5-3√2) = 3.5 units. total area= 10+3.5+4.5 units = 18 units

@allanflippin2453 Ай бұрын

As usual, I've come up with a stupid approach, which nevertheless works. The basic steps are: 1) Calculate the distance from point D to line AB 2) Add a horizontal line through D parallel to AB forming a rectangle, with area of 5 * distance of D from AB 3) Calculate area of the two triangles at the upper left and upper right of the rectangle. 4) Subtract from the rectangle area to find area of yellow portion. Clearly, this approach leads to some extra steps, but not as bad as one might expect. Assume we already know that AED and DCB are 135 degrees. So the upper left and upper right corner triangles are both 45 degrees isosceles. Based on this, the distance of point E from AB is 3/sqrt(2) less than the distance from D to AB. Point D to AB is greater than point C to AB by 5 - 3/sqrt(2). The height H can be calculated by adding 2: 7 - 3/sqrt(2). The area of the rectangle is 35 - 15/sqrt(2). The area of the upper left triangle is (3/sqrt(2))^2 / 2 = 9/4. The area of the upper right triangle is (5 - 3/sqrt(2)) ^ 2 / 2. Simplifying, this becomes 59/4 - 15/sqrt(2). Adding in the area of the other triangle, the sum is 17 - 15/sqrt(2). Subtracting this from the area of the rectangle results in 18.

@PreMath Ай бұрын

No worries!😀 Thanks for sharing ❤️

@redfinance3403 Ай бұрын

Draw a perpendicular from D to AB and label the point of intersection as G. Extend the line EA until D and extend the line BC until D. Notice that since the angle on a line is 180 degrees, the interior angle of the right angled triangle is 45 degrees. Therefore, there are two congruent right triangles. Yellow area = Area of two trapeziums Area of trapezium 1 = EA(AG + DG)/2 Area of trapezium 2 = DG(2 + BG)/2 DG = 7 - 3sqrt(2) / 2 EA = h - 3sqrt(2) / 2 AG = 3sqrt(2) / 2 BG = 5 - AG Putting this mess all together, if you substitute it into the area formulas, you get 18 units^2.

@waheisel Ай бұрын

As usual PreMath's solution was much more elegant than mine. I too constructed a bigger right isosceles triangle but made the mistake of DE to AB rather than EA to DC. I got the solution but the algebra was a lot more complex/less elegant. Still another exciting daily math puzzle. Thanks PreMath!

@neofolk3051 23 күн бұрын

I solved it in a complicated way, but it's simple. Thank you.

@ashokkhullar6650 Ай бұрын

I did it in an easier way by drawing a line through E and parallel to AB intersecting BC produced at say P. Required area will be areas of rectangle plus of trapezium and triangle which will be 10, 3.5 and 4.5 respectively.

@MateusMuila Ай бұрын

This method is good but are you sure from E and C can be drawn a paralel line

@quigonkenny Ай бұрын

Extend DE to M and DC to N so that MN is collinear to and parallel with AB. As ∠BAE = ∠EDC = ∠CBA = 90° and ∠AED = 135°: ∠BAE + ∠AED + ∠EDC + ∠DCB + ∠CBA = (n-2)180° 90° + 135° + 90° + ∠DCB + 90° = (5-2)180° = 540° ∠DCB = 540° - 405° = 135° As ∠AED = 135°, ∠MEA = 180°-135° = 45°. As ∠EAM = 90°, ∠AME = 180°-(90°+45°) = 45°. As ∠DCB = 135°, ∠BCN = 180°-135° = 45°. As ∠NBC = 90°, ∠CNB = 180°-(90°+45°) = 45°. As ∠MEA = ∠AME = 45° and ∠BCN = ∠CNB = 45°, ∆EAM and ∆NBC are similar isosceles right triangles, EA = AM = x and BC = NB = 2. Additionally, as ∠AME = ∠CNB = 45°, ∆MDN is also an isosceles right triangle and MD = DN. MD = DN ME + ED = DC + CN √(x²+x²) + 3 = DC + √(2²+2²) √(2x²) + 3 = DC + √8 DC = 3 + √2x - 2√2 = 3 + √2(x-2) Drop a perpendicular from D to P on AB, and drop perpendiculars from E to S on DP and C to T on DP. As ∠DES = 135°-90° = 45° and ∠TCD = 135° -90° = 45, and DP bisects ∠CDE and thus ∠CDP = ∠PDE = 90°/2 = 45°, ∆ESD and ∆DTC are isosceles right triangles, similar to ∆EAM and ∆NBC before. By observation, ES + CT = AB = 5. ES/ED = NB/CN ES/3 = 2/2√2 = 1/√2 ES = 3/√2 CT/DC =NB/CN CT/(3+√2(x-2)) = 3/2√2 = 1/√2 CT = (3+√2(x-2))/√2 = x - 2 + 3/√2 3/√2 + x - 2 + 3/√2 = 5 6/√2 + x = 5 + 2 = 7 x = 7 - 3√2 MD = ME + ED = √2x + 3 MD = √2(7-3√2) + 3 MD = 7√2 - 6 + 3 = 7√2 - 3 The area of the yellow pentagon (Aᴘ) equals the area of ∆MDN (Aᴛ) minus the areas of the constructed triangles ∆EAM and ∆NBC (A₁ and A₂). Aᴛ = bh/2 = (7√2-3)(7√2-3)/2 Aᴛ = (98-42√2+9)/2 = 107/2 - 21√2 A₁ = bh/2 = (7-3√2)(7-3√2)/2 A₁ = (49-42√2+18)/2 = 67/2 - 21√2 A₂ = bh/2 = (2)(2)/2 = 2 Aᴘ = Aᴛ - A₁ - A₂ = 107/2 - 21√2 - (67/2-21√2) - 2 Aᴘ = 107/2 - 67/2 - 2 = 40/2 - 2 = 18 sq units

@santiagoarosam430 Ай бұрын

DC tiene una pendiente del 100%→ Las alineaciones DC, AB y AE se cortan y definen un triángulo rectángulo de catetos de longitud 5+2=7 → Área amarilla =(7*7/2)-(3*3/2)-(2*2/2)=18 ud². Gracias y un saludo cordial.

@marcgriselhubert3915 Ай бұрын

Let's use an orthonormal, center A, first axis (AB) We have A(0; 0) B(5; 0) C(5; 2) E(0; a) with a = AE As angleAED = 135° = 90° + 45°, (ED) is parallel to "y = x", and so VectorED(3/sqrt(2); 3/sqrt(2)) and then we have D(3/sqrt(2); a + 3/sqrt(2)) and VectorDC(5 -3/sqrt(2); 2 -a -3/sqrt(2)) This vector is orthogonal to VectorED or VectorU(1; 1) ten 5 - 3/sqrt(2) +2 -a -3/sqrt(2) = 0 and a = 7 -6/sqrt(2) or a = 7 -3.sqrt(2) Finally A(0; 0) B(5; 0) C( 5; 2) E(0; 7- 3.sqrt(2) and D((3/2).sqrt(2); 7- (3/2).sqrt(2)) , VectorDC(5 -(3/2).sqrt(2); (-5 +(3/2).sqrt(2)) = (5 -(3/2).sqrt(2)).VectorV, with VectorV(1; -1) Area of trapezoïd EABC: ((EA +BC)/2). AB = ((9 -3.sqrt(2))/2).5 = (45 -15.sqrt(2))/2 Area of triangle DEC: (1/2).ED.DC = (1/2).(3). DC with DC = (5- (3/2).sqrt(2)) .sqrt(2) = 5.sqrt(2) -3 Then area of triangle DEC = (15.sqrt(2) -9)/2 The yellow area is the sum of these two areas, it is then (45 -9)/2 = 18

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@aljawad Ай бұрын

Nice and simple “trigonometry-less” solution. I solved this problem by assuming (A) is the origin, and proceeded to find the X,Y coordinates of (D) & (E) to find the total area.

@toninhorosa4849 Ай бұрын

I like your solution teacher, but I find another way this solution. From point E, I drew a parallel to straight AB until I found straight CD and marked point F. The triangle DEF was formed and the angle Ê, which in the polygon was 135° - 90°, became 45° in the triangle. And this triangle will be isosceles, as angle F will also be 45° to complete 180° the sum of the 3 internal angles of triangle DEF. From point F, I drew a line parallel to BC until I reached straight AB where I marked point H. At the bottom, the rectangle AHFE and the trapezoid BCFH were formed. Yellow area = ∆DEF area + Rectangle AHFE area + trapezoid BCFH area. ∆DEF area = (3*3)/2 = 9/2 EF^2 = 3^2 + 3^2 EF^2 = 2*3^2 EF = 3√2 ==>AH= 3√2 BH = 5 - 3√2 Drawing from point C a line parallel to straight EF until reaching straight AE, I meet point G. And straight CG when meeting straight FH forms point P. And we will have segments FP and CP equal, as angles F and C will have 45 °. FP=CP=BH = 5 - 3√2 FH = 2 + (5 - 3√2) FH = 7 + 3√2 Rectangle AEFH area = 3√2(7 + 3√2) = (21√2 - 18) Trapezoid BCFH area = (B+b)*h=(7-3√2+2)(5-3√2)/2 = ((9-3√2)(5-√3))/2 Trapezoid area=(63-42√2)/2 Yellow Area = 9/2+(42√2-36)/2 +(63-42√2)/2 = (9 + 42√2 -36 +63 -42√2)/2 Yellow area =36/2= 18 unit^2

@user-yx9kr8ur5q Ай бұрын

Extend AE upwards to F and extend BC upward to G such that FDG is a straight line and AFGB is a rectangle. Then EFD is a 45-45-90 triangle with hypotenuse 3 and sides x = 3/sqrt(2) and triangle CDG is also a 45-45-90 triangle with sides: y = 5 - x = (5 - (3/sqrt(2))). The area of the polygon is the area of the rectangle AFGB minus the areas of the two 45-45-90 triangles EFD and DGC and is just 5(7 - (3/sqrt(2)))- 1/2( (3/sqrt(2))^2) - 1/2( (5 - (3/sqrt(2))) ^2) = 18 square units.

@phungpham1725 Ай бұрын

I solved it the same as yours😅

@PreMath Ай бұрын

Thanks for sharing ❤️

@murdock5537 Ай бұрын

Nice! φ = 30° → sin⁡(3φ) = 1; ABCDE → AB = AN + BN = (3√2)/2 + (5 - 3√2/2); BC = AK = 2 CD = b; DE = 3; sin⁡(CDE) = 1; DEA = 9φ/2; CE = k; AE = AK + EK = 2 + a → CK = CM + KM = (5 - 3√2/2) + 3√2/2 DM = a + 3√2/2 → k^2 = 9 + b^2 = 25 + a^2 → b^2 = 16 + a^2 = (5 - 3√2/2)^2 + (a + 3√2/2)^2 → a = 5 - 3√2 → b = √(59 - 30√2) → area ABCDE = 10 + 5a/2 + 3b/2 = 10 + (1/2)(5a + 3b) = 18

@bobbyheffley4955 Ай бұрын

Angle C=135° (the sum of the interior angles of a pentagon is 540°).

@yalchingedikgedik8007 Ай бұрын

Very nice and useful Thanks Sir Thanks PreMath ❤❤❤❤❤❤ With my respects

@wackojacko3962 Ай бұрын

Drawing that big triangle reveals the most important facts of the problem and the effects of that triangle on the isoceles triangles it created...you know, gotta see the big picture so too speak to readily solve. 🙂

@ashutoshkumardalei3264 Ай бұрын

Excellent 👌🏻 👌🏻 👌🏻 observation for construction.

@PreMath Ай бұрын

Glad to hear that! Thanks for the feedback ❤️

@prossvay8744 Ай бұрын

Yellow Polygon area=1/2(7)(7)-1/2(2)(2)-1/2(3)(3)=18 square units.❤❤❤

@PreMath Ай бұрын

Excellent! Thanks for sharing ❤️

@user-tr7ol9us1w Ай бұрын

My teacher, I think there is an easier way. The opposite side of the triangle at the top is equal to 5 squared by 25, then we subtract 9 from it because it is the square of 3, and the root of the result, which is 16, equals 4. Then we multiply 4 by 3 by dividing 2, and the result is 6. Then we multiply 2 by 5, and the result becomes 10 . and 10Plus 6 equals 16

@PreMath Ай бұрын

Thanks for the feedback ❤️

@user-tr7ol9us1w Ай бұрын

@@PreMath Thanks for your lessons

@johnbrennan3372 Ай бұрын

I did it by extending ae and bc and drawing a line thro’ d parallel to ab. It worked out but not as nice as the method on the video.

@jamestalbott4499 Ай бұрын

Thank you!

@pralhadraochavan5179 Ай бұрын

Good morning sir

@misterenter-iz7rz Ай бұрын

Not difficult but clumsy in computation 😮. ((5-3/sqrt(2))+2)×5-1/2((3/sqrt(2))^2+(5-3/sqrt(3))^2).😅

@PreMath Ай бұрын

Thanks for the feedback ❤️

@nenetstree914 Ай бұрын

@PreMath Ай бұрын

Thanks for sharing ❤️

@unknownidentity2846 Ай бұрын

Let's find the area: . .. ... .... ..... Let's assume that A is the center of the coordinate system and that AB is located on the x-axis. In this case we obtain the following coordinates: A: ( 0 ; 0 ) B: ( 5 ; 0 ) C: ( 5 ; 2 ) D: ( xD ; yD ) E: ( 0 ; yE ) The slope of the line DE is tan(135°−90°)=tan(45°)=1. So we can conclude: yD − yE = xD − xE From the given length of DE we obtain: DE² = (xD − xE)² + (yD − yE)² DE² = 2*(xD − xE)² 3² = 2*(xD − xE)² 3²/2 = (xD − xE)² ⇒ yD − yE = xD − xE = 3/√2 The lines DE and CD are perpendicular to each other. Therefore the product of their slopes is −1 and we can conclude: yC − yD = −(xC − xD) Now we can calculate the missing coordinates: xD − xE = 3/√2 xD − 0 = 3/√2 ⇒ xD = 3/√2 yC − yD = −(xC − xD) 2 − yD = −(5 − 3/√2) yD − 2 = 5 − 3/√2 ⇒ yD = 7 − 3/√2 yD − yE = 3/√2 7 − 3/√2 − yE = 3/√2 ⇒ yE = 7 − 6/√2 Now we are able to calculate the area of the yellow region: A(ABCDE) = (1/2)*(yE + yD)*(xD − xE) + (1/2)*(yD + yC)*(xC − xD) = (1/2)*(7 − 6/√2 + 7 − 3/√2)*(3/√2 − 0) + (1/2)*(7 − 3/√2 + 2)*(5 − 3/√2) = (1/2)*(14 − 9/√2)*(3/√2) + (1/2)*(9 − 3/√2)*(5 − 3/√2) = (1/2)*(42/√2 − 27/2) + (1/2)*(45 − 27/√2 − 15/√2 + 9/2) = (1/2)*(42/√2 − 27/2) + (1/2)*(45 − 42/√2 + 9/2) = (1/2)*(42/√2 − 27/2 + 45 − 42/√2 + 9/2) = (1/2)*36 = 18 Best regards from Germany