Find area of the Yellow region in the Square | (Important skills explained) |

[@HappyFamilyOnline]

Very well explained 👍 Thanks for sharing 😊

[@PreMath]

Thanks for liking ❤️

[@phungpham1725]

1/ Calculating the sides of the right triangle ABC: (x+5).(2x-5)=2 . 54-------> 2.sqx +5x - 133 =0 -----> x= 7-----> AB=9 and BC= 12 2/ Calculating the side a of the square: BCF and ABE are similar triangles so: CF/12 =BE/9 = (CF-BE)/(12-9) = FB/3------> CF/12= FB/3-----> CF/FB =12/3=4 -----> FB= a/4 -----> sq a+ sq(a/4) =sq12----> sq a =(144x16)/17 3/The area of the yellow region= (144x16)/17 -54 = 81.53 sq cm

[@PreMath]

Thanks ❤️

[@yalchingedikgedik8007]

Thats very nice Thanks Sir Very easy explained.

[@Abby-hi4sf]

Fantastic approach. Amazing how triangle similarity cut short the process! Thank you again

[@PreMath]

You are so welcome! ❤️🌹

[@bigm383]

Lovely work, Professor!❤

[@PreMath]

Glad you like it! Thank you ❤️

[@johnwindisch1956]

I did it by adding all the areas of the triangles and solving for X. And it worked!

[@amarendrasingh7327]

Upload some more hard questions videos please 🙏🙏

[@PreMath]

Definitely! Thanks ❤️🌹

[@santiagoarosam430]

(X+5)(2X-5)=2*54→ X=7 → CB=12 ; AB=9 → Razón de semejanza entre AEB y BFC, s=9/12=3/4→ CF=b→ BE=3b/4 ; FB=b/4 ; AE=3b/16 ; DA=13b/16 → (b*b/4*2)+(3b*3b/4*16*2)+(13b*b/16*2)+54=b²→ b²=48²/17→ Área amarilla =(2304/17)-54=81.5294 Gracias y un saludo cordial.

[@PreMath]

Thanks ❤️🌹

[@Stanislaw2344]

Moc pěkný příklad.

[@Copernicusfreud]

Yay! I solved the problem. I used the quadratic formula to find the value of x.

[@PreMath]

Fantastic ❤️

[@wackojacko3962]

Can the unknown side length of the Blue Triangle be found using Heron's Formula? Just curious...🙂

[@giuseppemalaguti435]

(x+5)(2x-5)=108...x=7.....Ay=l^2-54=16*144/17-54,con Pitagora e i triangoli simili

[@PreMath]

Thanks ❤️🌹

[@shadrana1]

IABCI=54 cm^2 2*IABCI=108 cm^2= 9*12=AB*BC=(2x-5)(x+5) (2x-5)=9 >>>>x=7cm.............(1) (x+5)=12>>>>x=7cm.............(2) x=7 satisfies the area equation of ABC. Triangle CFB is similar to triangle BAE (angles are the same in both triangles) Hence, CF/BE=12/9=4/3. Therefore, CF=4/3 BE..................(3) Let CF=FE=ED=DC= 4t say, 4t= 4/3 BE EB=3/4*4t= 3t..............................(4) BF=FE-BE= (4t-3t)=t.....................(5) Consider triangle CFB, CB^2=CF^2+FB^2 Pythagoras, 12^2=(4t)^2+t^2=17t^2 t^2=144/17..................................(6) IDEFCI=16t^2 =(16*144)/17 ICBAI=(54*17)/17 IYellow shaded regionI=IDEFCI-ICBAI=[ (16*144)-(54*17)]/17=(2304-918)/17 =1386/17 is approx. equal to 81.53 cm^2 Thanks for the puzzle PreMath and have a good Festive Season.

[@ChandanRoy-sr6yu]

Sir's voice is very heavy😅😅

[@RaymineRaymine]

Agree

[@PreMath]

Thanks ❤️

[@unknownidentity2846]

Because of a lack of time this time only a short comment: Again a very nice problem and I solved it essentially the same way you showed in the video. Just a short remark: When units are used for the area, then units should also be used for the lengths. In this case it would mean x+5cm and 2x−5cm. I think it is easier to skip the units completely as long as the focus lies on math. Best regards from Germany

[@PreMath]

Thanks for sharing 🌹

[@prossvay8744]

Area of the yellow shaded region=2304/27-54=1386/17=81.53 cm^2

[@PreMath]

Thanks ❤️🌹

[@himo3485]

(x+5)(2x-5)/2=54 2x^2+5x-133=0 (2x+19)(x-7)=0 x>0 , x=7 CB=12 BA=9 CFB♾️BEA 12 : 9 = 4 : 3 CF=4a FB=4b BE=3a EA=3b CF=FB+BE 4a=4b+3a 4b=a b=a/4 FB=4＊a/4=a (4a)^2+a^2=12^2 17a^2=144 a^2=144/17 DEFC=4a＊4a=16a^2=2304/17 area of the Yellow region : 2304/17 - 54 = 2304/17 - 918/17 = 1386/17cm^2

[@PreMath]

Thanks ❤️🌹

[@MegaSuperEnrique]

X is not a length, it could be negative, however x+5 is a length, so x can't be that far negative.

[@Mycroft616]

It cannot be negative at all due to the other side length being 2x - 5. 2x - 5 > 0 2x > 5 x > 5/2 (or 2.5) Premath should have included this tidbit, but we do know x cannot be negative.

[@johnwindisch1956]

Your way was much easier!😂😂