Can you find area of the Green shaded Triangle? | (Two Methods) |

[@vara1499]

Both methods are wonderful.

[@PreMath]

Glad you think so! Thanks for the feedback ❤️

[@davidellis1929]

Here's a different kind of solution. The locus of the right angle of the yellow triangle is a semicircle inside the square whose diameter is the left side of the square. If we assume that the area of the green triangle is determined uniquely by the available data, then we can safely place the right angle at the center of the square, making the green triangle an isosceles right triangle with legs 2 and area (1/2)(2)(2), which is 2. The solutions given in the video confirm the above assumption.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@tasfiandnafi6359]

You are a best teacher I ever seen😊😊

[@redfinance3403]

I like this problem very much! Here’s my solution : let x be the side length of the square. draw a perpendicular to AD from E, let the intersection at AD be labelled as G. Let GE = h. draw another perpendicular to AB from E and label its intersection at AB as H. HE = sqrt(2^2 - h^2) = sqrt(4 - h^2) by Pythagoras. The area of triangle AED is either (AD)(GE)/2 or (AE)(ED)/2 so (AD)(GE)/2 = (AE)(ED)/2, so ED = xh/2. By Pythagoras, ED^2 + AE^2 = AD^2. So (xh/2)^2 + 2^2 = x^2. Solving for h^2 gives h^2 = 4 - 16/x^2. Now, HE = sqrt(4 - h^2) = sqrt(16/x^2) = 4/x. Finally, area of green triangle equal to (AB)(HE)/2 = 1/2 * x * 4/x = 2 units squared! When you can’t solve with similar triangles, look for Pythagoras. When you can’t solve by Pythagoras, solve by trigonometry. When you can’t solve by trigonometry, solve by coordinate geometry 😂

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@yatesfletcher1424]

If F is the projection of E onto AB, let y be the length of EF. The area G of the green triangle is 1/2*x*y. Triangle AEF is similar to triangle DAE. Thus 2/x=y/2 => y=4/x so G=1/2*x*(4/x)=2.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@tellerhwang364]

1.extend AE→AH丄BH △ADE,△BAH congruent(AAS) →AE=BH=2 2.△AEB=AE·BH/2=2😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@murdock5537]

Nice! ∎ABCD → AB = AF + BF = BC = CD = AM + DM = r + r; AE = 2; sin⁡(EFA) = 1; EF = h; ADE = FAE = δ → sin⁡(δ) = 2/2r = 1/r = h/2 → hr = 2 → h = 2/r → rh = 2 = area ∆ AEB

[@PreMath]

Excellent! Thanks for sharing ❤️

[@andrejradovic2124]

There is even a simpler way to obtain the area of triangle ABE, and it is as follows: Project the point B orthogonally on the line through A and E to obtain point F. Then the resulting triangle ABF is side-angle-angle-congruent to triangle ADE since the hypotenuse of both of them is x, the angles ADE and BAF are the same, and they both have a right angle in E and F, respectively. Now triangle ABE has base AE = 2 and height BF = 2 and therefore an area equal to 2. No need for analytical geometry, trigonometry and not even Pythagoras - humble congruence does the trick!

[@montynorth3009]

DE can also be 2 units which fully satisfies the 90 degree angle requirement, and then each yellow corner angle will be 45 degrees. Yellow and green will be congruent with areas 1/2 x 2 x 2. 2 units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Let's use an orthonormal center A and first axis (AD) We have A(0; 0) D(c; 0) and E(2.cos(t); 2.sin(t)) with c the side length of the square and t = angleEAD Then VectorAE(2.cos(t); 2.sin(t)) and Vector(ED(2.cos(t) -c; 2.sin(t) These vectors are orthogonal, so we have: (2.cos(t)).(2.cos(t) -c) +(2.sin(t)).(2.sin(t)) = 0, or 4 -2.c.cos(t) = 0 or c.cos(t) = 2 The basis of the green triangle is AB = c, the corresponding height is the abscissa of E: 2.cos(t), so its area is (1/2).(c).(2.cos(t)= = c.cos(t) = 2

[@PreMath]

Excellent! Thanks for sharing ❤️

[@CloudBushyMath]

Super Duper👍

[@PreMath]

Many many thanks❤️

[@reynaldowify]

2 = x * sen alfa. Therefore x = 2 / sen alfa. h = 2 * sen alfa So x * h / 2 = ( ( 2 / sen alfa ) * (2 * sen alfa) ) / 2 = 2;

[@timmcguire2869]

Kinda solved in my head in 10 seconds: Let E be the center point of the square making the diagonals 4, the sides 2sqrt2, and each 1/4 of the square, one of which is our triangle, 2 square units..

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You're welcome!🌹 Thanks a lot❤️

[@phungpham1725]

1/Drop the height BH to AE (extended) We have: angle AEE = angle BAH ( having sides perpendicular) so the two triangles ADE and HAB are congruent so, AE=BH=2 --> Area of the green triangle= 1/2 x2x2= 2 sq units😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@pralhadraochavan5179]

Good morning sir

[@PreMath]

Same to you, dear 😀🌹

[@wackojacko3962]

Trig identities are more fun! 🙂

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@marcelowanderleycorreia8876]

Holly shit! Beautiful question!!!

[@PreMath]

Thanks for the feedback ❤️

[@jimlocke9320]

DE is not given so the solution must be the same for all valid values of DE and we can solve special cases. monthnorth3009 chose DE = 2 and found the area of the green triangle to be 2. A simpler special case is DE = 0. ΔADE is now a line segment equal to the side of the square, so AD = AB = 2 and green triangle has area 2. However, we no longer really have a triangle. To make our case more legitimate, let DE be extremely small but >0. As we make DE smaller, we can make AD as close as we want to AE but never make them equal as long as DE > 0. Eventually, we can make DE small enough to treat the difference between AD and AE as negligible. In mathematics, AD = AE is called the limit as DE approaches 0.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@AmirgabYT2185]

S=4

[@LuisdeBritoCamacho]

I am going to be very quick 01) Draw a Vertical Line from E. Define Point F between Point A and Point B. 02) Triangle ADE is congruent with Triangle AEF 03) 2 / AD = EF / 2 04) AD * EF = 4 05) AD = AB 06) Green Area = 4 / 2 = 2 ANSWER : Area of Green Triangle equal to 2 Square Units.

[@PreMath]

Super!🌹 Thanks for sharing ❤️ I always appreciate your feedback ❤️

[@prossvay8744]

Green area =2 square units.❤❤❤

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergeyvinns931]

Так это тот же болт. только вид с боку! h/2 = 2/x=sina, тоже тригонометрия.

[@PreMath]

Спасибо за отзыв ❤️

[@duantran-bk4mv]

S(green)= 2(dvdt)

[@sergeyvinns931]

Интересно. какой второй метод. Первый решается устно!

[@PreMath]

Спасибо за отзыв ❤️

[@ericwickeywoodworkersurfbo6135]

Pretty killer

[@PreMath]

Thanks for the feedback ❤️

[@scottwebster695]

Don't put any advanced math, physics, engineering degrees on job applications. Sometimes you just need a job to pay the bills and they won't hire you as a truck driver if you are being honest..

[@misterenter-iz7rz]

Usual trick, we suppose the diagonal is 4, side is 4/sqrt(2), area is 1/4×(4/sqrt(2))^2=2.😂😂😂

[@PreMath]

Excellent! Thanks for sharing ❤️