Can you find area of the Pink Shaded Trapezoid? | Trapezoid | (Trapezium) |

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Learn how to find the area of the Pink Trapezoid. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; Trapezoid area formula; Trigonometry; Square. Step-by-step tutorial by PreMath.com
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Комментарии : 74

@inyomansetiasa Месяц назад

First comment and first like, you can pin it?

@PreMath Месяц назад

Done!

@inyomansetiasa Месяц назад

Thank you sir

@montynorth3009 Месяц назад

Tan 30 = a /(100 - a ). a = tan 30 x (100 - a ). a = 100 tan 30 - a tan 30. a + a tan 30 = 100 tan 30. a (1 + tan 30 ) =100 tan 30. a = 100 tan 30 /1 + tan 30. a = 36.60254. Area = 1/2 ( 36.60254 +100 ) x 36.60254. 2500.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@michaelstahl1515 Месяц назад

You did it exatly like me !

@user-nr9cs8fd5q Месяц назад

1. Connect A&C. Diagonal AC is the hypotenuse of the isosceles triangle ABC. Angle CAB=45 deg., angle ACB=180-45-30=105 deg. From sine teorem for triangle ABC: 100/sin 105 deg=AC/sin 30 deg.

@vincentmudimeli4430 Месяц назад

Can you do more of this sir they are helpful

@123rockstar2010 Месяц назад

Via Sine Rule: AC = 100sin(30)/sin (105) AC = 51.7638 By Pythagorean Theorem: a² + a² = AC² a = sqrt (AC²/2) a = 36.6025 Area = .5(a + 100)(a) Area = 2500 sq. units

@calvinmasters6159 Месяц назад

Good puzzle. Fun to do.

@alster724 Месяц назад

Yesssss!!!!! I got it! Cheers!

@user-re8yc5gc8w Месяц назад

bc^2 = a^2 + (100-a)^2 -> (2a)^2 = a^2 + 100^2 -200a + a^2 -> 4a^2 - 2a^2 +200a = 10000 -> 2a^2 + 200a = 10000 -> a^2 + 100a = 5000 abcd = 1/2(100+a)*a = 1/2( a^2 + 100a) -> 1/2(5000) = 2500

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@remziturhan4367 Месяц назад

Nice problem, nice solving...

@raya.pawley3563 Месяц назад

Thank you

@ChuzzleFriends Месяц назад

Draw a segment thru C and a point E on segment AB, such that segment CE is perpendicular to the bases of trapezoid ABCD. This forms a square ADCE because AD = CD. Label the side length of square ADCE as s. Segment CE also forms △BEC, a special 30°-60°-90° right triangle. BE = (CE)√3 = s√3 So, AB = AE + BE = s + s√3 = 100. s + s√3 = 100 s(√3 + 1) = 100 s = 100/(√3 + 1) = 100/(√3 + 1) * (√3 - 1)/(√3 - 1) = (100√3 - 100)/2 = 50√3 - 50 A = h[(a + b)/2] = (50√3 - 50)[(50√3 - 50 + 100)/2] = (50√3 - 50) * (50√3 + 50) * 1/2 = (7500 - 2500) * 1/2 = 1/2 * 5000 = 2500 So, the area of the pink trapezoid is 2,500 square units.

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@Martin-hi7wx Месяц назад

I used trig. Tan and cosine formula to calculate the length of the diameter. It is a shorter and easier altertive. I believe

@thephilosophyofhorror 19 дней назад

Very positive: that you managed to get rid of the sqr3 without using a calculator or relying on memorizing its value. A little negative: that you did not take the time to present why area of trapezoid = 1/2height(large base-small base) (ie you could have said that area of trapezoid clearly is smallbase(height)+(largebase-smallbase)height/2= (2smallbase+largebase-smallbase)(height)/2=(largebase+smallbase)(height)/2.

@quigonkenny Месяц назад

Drop a perpendicular from C to E on AB. As CD = DA and all internal angles are 90°, AECD is a square. Let CD = DA = AE = CE = x. As ∠CEB = 90°, ∆CEB is a right triangle, and as ∠EBC = 30°, that makes it a 30-60-90 special right triangle, where ∠BCE = 60°, BC = 2CE and EB = √3CE. As CE = AE, EB = √3AE, thus: AB = AE + EB 100 = AE + √3AE 100 = x + √3x = x(1+√3) x = 100/(1+√3) x = 100(1-√3)/(1+√3)(1-√3) x = 100(1-√3)/(1-3) x = 100(√3-1)/2 = 50(√3-1) Pink Trapezoid ABCD: A = h(a+b)/2 = x(x+100)/2 A = ((50(√3-1))² + 100(50(√3-1)))/2 A = (2500(3-2√3+1) + 5000(√3-1))/2 A = 1250(4-2√3) + 2500(√3-1) A = 5000 - 2500√3 + 2500√3 - 2500 A = 2500 sq units

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@benjisurya Месяц назад

tan(30)=x/(100-x)=0,5773 x=36,602 Area=x2/2+50x After substitution and calculation Area=2500s.u

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@giuseppemalaguti435 Месяц назад

a/(100-a)=tg30=1/√3...100/a=√3+1..a=100/(√3+1)..Apink=(100+a)a/2=

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@user-nm3et4wr8f Месяц назад

Solution is okay. Thanks!

@PreMath Месяц назад

You are very welcome! Thanks for the feedback ❤️

@michaelstahl1515 Месяц назад

Great video again Sire ! I used tan (30) = x / ( 100 - x ) when x = CE and EB = 100 - x . I got for x nearly 36,6 and for the pink area A = 2499,78 square units. But using the third binomal formula was more elegant than my solution. You got exactly 2500 sqare units.

@yalchingedikgedik8007 Месяц назад

Thanks Sir That’s wonderful method . I am answer if it is possible find the area by plusing the area of triangle and square area. With glades

@philipkudrna5643 Месяц назад

Before watching: I arrive at 2.500 square units. Approach: the left part is a square, the right part is a 30-60-90 triangle. Side relations: 1-sqrt(3)-2. to arrive at the side of the square, divide 100 by (1+sqrt(3)). The side of the square is approx 36,6. From here calculate the areas of the square and the triangle, which totals to 2500. After watching: exactly what I did, though I didn‘t have pen and paper, so I couldn‘t do the pretty steps with rationalizing the denominator etc, but simply used the calculator and my iPhone…)

@harrymatabal8448 Месяц назад

Mr Philip excellent work

@PrithwirajSen-nj6qq Месяц назад

Thanks. To make more clear sol, we may draw EF so that F is on BC and angle BEF is 30 degrees. Then angle CEF will be 60 degrees. 🔺 CEF will be an equilateral 🔺 .This means CE =CF = EF 🔺 BEF is an isosceles 🔺 and EF=BF Then CE=CF=EF=BF If CE=a Then BC =CF+EF=CE+CE=2a

@PreMath Месяц назад

Thanks for sharing ❤️

@wackojacko3962 Месяц назад

Excellent! 🙂

@PreMath Месяц назад

Thank you! Cheers!❤️

@procash1968 Месяц назад

Amazing !!

@PreMath Месяц назад

Thank you! Cheers!🌹❤️

@mehdismaeili3743 Месяц назад

Excellent '

@PreMath Месяц назад

Many thanks dear!❤️

@MrPaulc222 Месяц назад

I missed a few tricks there, including the initial factorisation. However, even using decimal approximations and a calculator I got to 2499.9ish.

@Nothingx303 Месяц назад

Thanks I solved it too 😊😊😊❤❤❤ It was tasty 😋

@misterenter-iz7rz Месяц назад

100=(sqrt(3)+1)a, a=100/(sqrt(3)+1)=50(sqrt(3)-1), therefore the area is (1/2)×(100+50(sqrt(3)-1))×50(sqrt(3)-1)=(2500/2)(sqrt(3)+1)(sqrt(3)-1)=2500.🎉

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@GillesF31 Месяц назад

I got the right answer with a different way: I M P O R T A N T : • my calculations are carried out with 6 digits after the decimal point because I use a trigonometric function (tangent) • the point "E" is on the line AB, at the opposite of C get x = CE tan(30) = x/(100 - x) tan(30)·(100 - x) = x 0.577350·(100 - x) = x 57.7350 - 0.577350x = x x + 0.577350x = 57.7350 1.577350x = 57.7350 x = 57.7350/1.577350 x = 36.602529 trapezoid area: area = x² + (x·(100 - x))/2 area = 36.602529² + (36.602529·(100 - 36.602529))/2 ----------------------------- | area = 2499.99 | ----------------------------- 🙂

@prossvay8744 Месяц назад

Connect C to E AD=DC=CE=EA=x In ∆BEC BE=100-x Tan(30°)=x/100-x x=50√3-50 So area trapezoid=1/2(100+50√3-50)(50√3-50)=2500 square units.❤❤❤ Thanks sir.

@PreMath Месяц назад

Excellent! You are very welcome! Thanks for sharing ❤️

@violetadeliu734 Месяц назад

Anxhela besoj se e ke me te lehte tani?

@papilgee4evaeva Месяц назад

Maybe it's a cultural difference, but I was taught never to refer to the hypotenuse of a right triangle as a "leg." Good solution though.

@PreMath Месяц назад

Thanks for the feedback ❤️ Kind regards🌹

@andryvokubadra2644 15 дней назад

I thought value of angle 30° has wrong. AD = DC AD = CE AD = DC = CE AB = 100 = AE + BE BE = 100-AE We can write : Tan CBE = CE/BE Then : Tan CBE = CE/(100-AE) Tan CBE = CE/[100-(100-CE)] Tan CBE = CE/CE Tan CBE = 1 CBE = arc Tan 1 CBE = 45° But. . . it in this video 30° not 45°.

@rachidmsmdi6433 Месяц назад

x+x√3=100. X=100/(1+√3) X=50(-1+√3) 2❤️=(X+100)X ❤️=50² ❤️=2500 💙💚💛💜❤️🖤. Hamas=Résistance ❤️🖤

@PreMath Месяц назад

Thanks for sharing ❤️

@gokudragonball2605 Месяц назад

so great

@PreMath Месяц назад

Glad to hear that! Thanks for the feedback ❤️

@marcgriselhubert3915 Месяц назад

Much simple.

@nilsalmgren4492 Месяц назад

Easy once you see the square and the 30-60-90 triangle

@PreMath Месяц назад

Thanks for the feedback ❤️

@Waldlaeufer70 Месяц назад

Let's give this one a try: 100 = s(square) + x x = s√3 (since there is a 30-60-90° triangle) 100 = s√3 + s s(√3 + 1) = 100 s = 100/(√3 + 1) = 100 (√3 - 1) / ((√3 + 1)(√3 - 1)) = 100 (√3 - 1) / (3 - 1) = 100 (√3 - 1) / 2 = 50 (√3 - 1) A = A(square) + A(triangle) = s² + 1/2 * s√3 * s = s² + 1/2 * s² * √3 = s² (1 + 1/2 √3) A = (50 (√3 - 1))² (1 + 1/2 √3) = (2500 (3 - 2√3 + 1)) (1/2 (2 + √3)) = (2500 (4 - 2√3)) (1/2 (2 + √3)) = 1250 (4 - 2√3) (2 + √3) = 1250 (8 + 4√3 - 4√3 - 6) = 1250 * 2 = 2500 square units

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@AmirgabYT2185 Месяц назад

S=2500

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@stevegardner9910 Месяц назад

Tan 30 = a/(100-a). Go from there. Easy stuff.

@PreMath Месяц назад

Thanks for the feedback ❤️

@LuisdeBritoCamacho Месяц назад

A little bit late but here I am!! 1) tan(30º) = sqrt(3) / 3 2) AD = CD = X 3) So, CC' = X 4) BC' = (100 - X) 5) CC' / BC' = tan(30º) 6) X / (100 - X) = tan(30º) 7) X / (100 - X) = sqrt(3) / 3 8) X = 50 * (sqrt(3) - 1) ~ 36,6 lin un 9) T = B + b * (h/2) 10) T = (100 + 36,6) * 36,6 / 2 12) T = 136,6 * 18,3 13) T = 2.499,78 sq un 14) My Best Answer is : The Pink Trapezoid Area is approx. equal to 2.500 Square Units.

@kennethstevenson976 Месяц назад

Would 2499.78 square units be acceptable? That is what I got using fewer steps.

@PreMath Месяц назад

Yes! Close enough. Thanks for sharing ❤️

@rajendraameta7993 Месяц назад

I think without using trapezium area formula , area of trapezium can be found in this problem

@PreMath Месяц назад

Thanks for the feedback ❤️

@kalavenkataraman4445 Месяц назад

2500

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️

@violetadeliu734 Месяц назад

Shoh vetem figuren. Jepen te 4 kendet e nje trapezi qe jane 90, 30, 150 dhe 90 grade. Jepet baza e madhe 100. Zbato formuat dhe gjindet sip. e trapezit. Po si mbajte mend, i nxjerr duke formuar trek. Dhe nje katror me brinje ma bazen e vogel te trapezir. Diferenca ndermjet tyre edhte kateti i trk. kendrejte me kende 30 dhe 60 grade. Tani vetem .....

@unknownidentity2846 Месяц назад

Let's find the area: . .. ... .... ..... First of all we add point E on AB such that AE=CD. Then ADCE is a square and since ∠AEC=∠BEC=90°, the triangle BCE is a 30°-60°-90° triangle. Therefore we can conclude immediately: BE = √3*CE So from the known length AB we obtain: AB = AE + BE = CE + BE = CE + √3*CE = CE*(1 + √3) ⇒ CE = AB/(1 + √3) = (1 − √3)*AB/[(1 + √3)(1 − √3)] = (√3 − 1)*AB/2 Now we are able to calculate the area of the trapezoid: A(ABCD) = (1/2)*(AB + CD)*AD = (1/2)*(AB + CE)*CE = (1/2)*[AB + (√3 − 1)*AB/2]*(√3 − 1)*AB/2 = [1 + (√3 − 1)/2]*(√3 − 1)*AB²/4 = [(√3 + 1)/2]*(√3 − 1)*AB²/4 = AB²/4 = 100²/4 = 2500 Best regards from Germany

@PreMath Месяц назад

Excellent! Thanks for sharing ❤️