Calculate area of the Blue shaded Quadrilateral | Learn Geometry and Trigonometry skills fast

[@vidyadharjoshi5714]

Sin 30 = 0.5 = h/28 = H/42. h = 14, H = 21. Area = 0.5*(H*56 - h*21) = 0.5*(1176 - 294) = 441

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@arthurschwieger82]

Nice solution! I used the 30/60/90 triangle approach by drawing a line from E down, perpendicular to AB. Also one from C down, perpendicular to AB. That gives me the height of each of the same triangles you were doing. From there, our numbers are basically the same as the ration of the height of each triangle is 1/2 of the hypotenuse. You used Sin(30) to get 1/2 and I used the 30/60/90 triangle relationship to get the 1/2. And when I did my addition correctly, I ended up with the same value as I should. It seems that 21 + 35 = 56 not 46. ;-)

[@solomou146]

Ακριβώς ίδια λύση.

[@yalchingedikgedik8007]

Thanks PreMath That very useful and nice

[@murphygreen8484]

Yay, I got one on my own first!

[@HappyFamilyOnline]

Very well explained👍 Thanks for sharing😊

[@anthonycheng1765]

Another method: use similar triangle. Triangle ACB ~ Triangle ADE (SAS Similarity). AE/AB = AC/AD = 1/2 Area ratio = 1:4 Area of Triangle: Area of quadrilateral is 1:(4-1) (1:3). Area = 147*3=441

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@soli9mana-soli4953]

Can you say better how did youi find Area of quadrilateral is 1:(4-1) (1:3). Area = 147*3=441 ??

[@alster724]

Paused the video at 2:08 and used the same concept for the big ∆ABC and solved it on my own. Same drill, fast forwarded to the end to double check and got the same answer. Easy! God bless from The Philippines 🇵🇭

[@theoyanto]

Another Gem !! I've started to catalogue the rules and theorems you present in the Google Keep app, so i can quickly reference various key lessons, it would be all to easy to let all those sparkling gems fade with time Thanks again 🤓👍🏻

[@ybodoN]

Alternatively: AD = ½ AC and AE = ½ AB, thus ABC and AED are in the ratio 4:1. Therefore, the area of the blue quadrilateral is 3·½·AE·AD·sin 30°.

[@santiagoarosam430]

∠CAB=30º → La altura de ∆ACB respecto a la base AC es h=(21+35)/2=28 → Área ∆ACB =(28+14)h/2=588 →→→ Valores relativos de las áreas componentes: ∆CED=a → ∆EAD=(28/14)a=2a → ∆DCB=(35/21)3a=5a → ∆ABC=a+2a+5a=8a → Área relativa del cuadrilátero DBCE =a+5a=6a →→→ Área azul =588(6/8) =441 Gracias y un saludo.

[@williamwingo4740]

No peeking, no calculators: It's simpler just to remember that in a 30-degree right triangle, the side opposite the 30-degreee angle is one-half the hypotenuse. Construct two right triangles by dropping perpendiculars from C and E to the bottom line AB. Call them CF and EG. CF is the altitude of the big triangle ABC and equals (1/2)(28 + 14) = (1/2)(42) = 21; so the area of ABC is (1/2)(21)(21 + 35) = (1/2)(21)(56) = (21)(28) = 588 (did use pencil and paper for that last multiplication). EG is the altitude of the small triangle AED and equals (1/2)(28) = 14; so the area of AED is (1/2)(14)(21) = (7)(21) = 147. Finally, the area of the quadrilateral ECBD is the difference; or, 588 -- 147 = 441. And the day is off to a good start. Cheers. 🤠

[@Maaloma125]

How are you how do you write in this wonderful way? I hope you will answer me

[@PreMath]

Hello dear, we use Camtasia Techsmith utility along with Canva/photoshop... Thanks for asking. Take care

[@Chris-gd4gj]

Just use formula 1/2*a*b*sin angle: 1/2(28+14)(21+35)sin30-1/2(28)(21)sin30 Then u can find the answer

[@AmjadAli-lk3pf]

Please explain the quadratic formula briefly....

[@PreMath]

Dear Amjad, I've already made a video on this topic! Please check out the link: kzbin.info/www/bejne/iZyxZ4qFo8llaZo Take care

[@AmjadAli-lk3pf]

Yes sir I already checked this vedio..please explain with a sketch,where is use this formula quadratic

[@AmjadAli-lk3pf]

My concept still not clear on this quadratic formula.

[@giuseppemalaguti435]

56*21/2-14*21/2=21*21=441

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@carlinoiavarone8342]

I find 882 the area of irregulare quadrilater with the formula of bramapughta

[@wackojacko3962]

Circular functions like Sine flourished in the Gupta empire of India. 🙂

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

[@prabhagupta6871]

The word sine was first used by Aryabhatta and then later it was translated to Arabic and then latin and greek which we use today

[@militarymatters685]

Solved it

[@PreMath]

Excellent! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@militarymatters685]

@@PreMath thank you so much sir Looking forward to your next video

[@philippeganty]

Since your data is enough, here is a solution without trigonometry: Let's draw the height of the triangle ADE which passes through E and intersects AD in F, and the height of the triangle ABC which passes through C and intersects AB in G. We then have right triangles of type 30°/60°/90°, so EF=AE/2=14 and CG=AC/2=21. Then Area ABC-Area ADE=(56*21/2)-(21*14/2). By simplification, we get = 3^2*7^2=441.

[@prossvay8744]

Area of blue shaded=441 sqare unit