First comment and first like, you can pin it?

@PreMath

13 күн бұрын

Done!

@inyomansetiasa

13 күн бұрын

Thank you sir

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

@michaelstahl1515

13 күн бұрын

You did it exatly like me !

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.

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

Nice problem, nice solving...

Can you do more of this sir they are helpful

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

13 күн бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Solution is okay. Thanks!

@PreMath

13 күн бұрын

You are very welcome! Thanks for the feedback ❤️

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

Good puzzle. Fun to do.

Thank you

Amazing !!

@PreMath

13 күн бұрын

Thank you! Cheers!🌹❤️

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.

Excellent! 🙂

@PreMath

13 күн бұрын

Thank you! Cheers!❤️

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

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

10 күн бұрын

Mr Philip excellent work

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

Excellent '

@PreMath

13 күн бұрын

Many thanks dear!❤️

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

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

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

13 күн бұрын

Thanks for sharing ❤️

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

so great

@PreMath

13 күн бұрын

Glad to hear that! Thanks for the feedback ❤️

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

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

13 күн бұрын

Thanks for the feedback ❤️ Kind regards🌹

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

Much simple.

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

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

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 | ----------------------------- 🙂

Anxhela besoj se e ke me te lehte tani?

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

@PreMath

12 күн бұрын

Thanks for sharing ❤️

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

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

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.

S=2500

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

@PreMath

13 күн бұрын

Yes! Close enough. Thanks for sharing ❤️

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

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

2500

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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 .....

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

13 күн бұрын

Excellent! Thanks for sharing ❤️

Ur explanation confusing bad, u let it looks very difficult, it’s too long 🤦‍♀️