Find area of the Purple triangle | (Important math skills explained) |

[@ybodoN]

Thank you for this very beautiful solution! 🙏

[@PreMath]

Thanks dear ❤️🌹

[@troyj3292]

Did I miss how we know that CD is co-linear to DE? Am I correct that without that constraint, there is not a unique answer?

[@ausautarch]

Why have you assumed that ED || DC? There is no indication and EC is a linear segment.

[@gervaischouinard9809]

Yes, it would need an indication the angle ADC is right angle

[@marioalb9726]

Side of blue square: s = √49 = 7 cm Base of green right triangle A = ½ b.h b = 2A / h = 2 . 73,5 / 7 b = 21 cm Hypotenuse of green right triangle: c²= b²+h² = 21²+7² = 490 c = 22,136 cm Similarity of right triangles: b'/ ½c = c / b b' = c²/2b = 490/(2.21) b' = 35/3 = 11,667 cm Area of purple isosceles triangle A = ½ b' h = ½. 35/3 . 7 A = 40,83 cm² ( Solved.√ )

[@robertbourke7935]

Got it. Thanks again.

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!

[@PreMath]

Thank you too, dear ❤️🌹

[@MoBade123]

Thanks teacher ❤❤❤❤

[@PreMath]

You're welcome 😊 ❤️

[@wcdaniel7]

The area of the isosceles triangle can be determined by calculating the area of triangle BPC (32.66), then adding that result to triangle ABD (73.5), then subtracting that result from the area of rectangle ABPD (147).

[@gervaischouinard9809]

I divided the purple triangle with a perpendicular to have two triangles similar to the green triangle. I calculate that sides of each purple triangle are quare root of 10 divided by 6 less than the green ones. I assumed that if sides are x times less, then the area would be x square times less. So 73.5 x 10 / 36 = 20,41666666. The total area is 2 times that so 40,8333333!

[@Copernicusfreud]

Yay! I solved the problem. The video was much nicer than what I did.

[@wcdaniel7]

Q: When calculating the area of the isosceles triangle, why was the value for the height (h) 7 used? Why wasn’t the actual ‘h’ (3.71) not used?

[@cyruschang1904]

If we extend the top of the triangle by drawing a line to the right by 7 cm, then a vertical line of 7 cm down, we have a small triangle the size of half of the blue square, the red triangle plus the small triangle is the same size as the green triangle: green triangle - small triangle = red triangle green triangle - half the blue square = red triangle 73.5 - (49 ÷ 2) = 73.5 - 24.5 = 49

[@xyz9250]

Could also just drop a perpendicular line from C on AB

[@yalchingedikgedik8007]

Thanks Sir for the beautiful exercise Thanks PreMath. If we can solve the area by accounting the area of rectangular and subtract the areas of triangles ( after account the area of white triangle ) . Thanks

[@PreMath]

Yes, definitely Thanks dear ❤️🌹

[@yalchingedikgedik8007]

@@PreMath Thanks sir Good luck with my respects 🌹🌹🌹🌹🌹🌹🌹

[@mike_nelsonTM]

I don't understand... Isn't side BP a height to triangle BPD??? Please help me understand

[@mrmoon6540]

cool task thank you!

[@PreMath]

Glad you liked it! ❤️

[@MrPaulc222]

AB = 21, so DB = sqrt(490) = sqrt(49)*sqrt(10) = 7*sqrt(10). Split DB in two for two right triangles with M as BD's midpoint. CDM is similar to ABD As the side lengths are 3:1 within a triangle, I have calculated base DM as (7*sqrt(10))/2, so height must be (7*sqrt(10))/6. Multiply them out to 490/12 un^2 This is 40 plus 10/12 so 40.8333...un^2 (rounded) Haha. We went totally different ways about it, but ended up at the same place.

[@hguy4100]

How does 7 is the height of the triangle DCB please reply because it is not logical at all

[@phungpham1725]

1/ AD =7, AB=21, DB =7 sqrt 10 and we have the angle DBA=alpha ------> tan alpha= 1/3. 2/Drop the height CH=h to DB, because the angle CDB= alpha so CH/DH=1/3 ------> DH = 3h and DB=6h Area of the purple triangle = 1/2 .h.6h = 3 sq h = 3. sq(7.sqrt10/6) =3. (490/36)= 40.83 sq cm

[@quigonkenny]

Square ADEF: A = s² 49 = s² s = √49 = 7 Triangle ∆DAB: A = bh/2 73.5 = AB(7)/2 AB = 73.5/3.5 = 21 BD² = DA² + AB² BD² = 7² + 21² = 49 + 441 BD = √490 = 7√10 Draw CG, where G is the midpoint of DB. As ∆BCD is isosceles, ∠CGB is 90°. If ∠BDA = α, and ∠ABD = β, where β = 90°- α, then as ∠CDA = 90°, ∠CDB = β as well. As ∠CDB = β and ∠DGC = 90°, ∠GCD = α, and ∆DGC is similar to ∆DAB. By symmetry, ∆CGB is congruent to ∆DGC, and thus also similar to ∆DAB. Triangle ∆DGC: GC/DG = DA/AB GC/(7√10/2) = 7/21 = 1/3 GC = (7√10/2)(1/3) = 7√10/6 Purple Triangle ∆BCD: A = bh/2 = 7√10(7√10/6)/2 A = 490/12 = 245/6 ≈ 40.83cm²

[@terakhanthis]

It feels like you added a step. You had the area of BPD, as 73.5. From there, subtract the area of BPC, or 7×14/3, or 98/3, or 32.66.

[@ybodoN]

In both cases, I count two steps: subtract then multiply or multiply then subtract 🤔

[@prayasmadhurgogoi4028]

Sorry for being out of context ... But .. Assertion (A): - If a1,a2,a3......,an is an AP such that a1+a4+a7+...a16 = 147 , then a1+a6+a11+a16=98 Reason (R): - In an AP, the sum of the terms from the beginning and the end is always same and equal to the sum of first and last term. Can anyone explain how *Reason* is the correct explanation for *Assertion* ???

[@corentinlesaux]

Merci pour ce joli problème que j'ai su résoudre 50 ans après avoir quitté l'école.

[@rollbruv]

Did I miss something? It appears that you've missed an important detail while sharing the problem statement that E-D-C are co-linear. I assumed that they are and solved it by drawing a perpendicular bisector of BD from point C(let's say CO). Then, Triangle COD is similar to Triangle DAB(Angle CDO = Angle DBA as these are alternate angles between parallel line segments CD & AB, and Angle COD = Angle DAB = 90 degrees ), then we get the value of CO as 7/6sqrt(10) cm. Ofcourse, the steps to calculate EF = AD = 7cm, AB = 21cm & BD = 7sqrt(10) remain the same. Thus, we get Area of Triangle CDB as 245/6 sqcm (1/2 x CO x BD, where CO is the height & BD is the base of the triangle).

[@NahidMiah-e6m]

Would you like to say something about how to love Math , i like math But durian doing math i can't. 😢

[@PreMath]

Thank you ❤️

[@Mascig83]

Good evening, responsible for the Pre Math profile. Your classes are very useful because I am studying to enter a public career in my country, Brazil. Please, clarify two doubts for me: 1) Why does the formula for calculating the area of ​​a triangle have the fraction 1/2? 2) By what numbers did you simplify the fraction 392/42?

[@calototube]

1) The area of a triangle is half the area of a rectangle. 2) If you divide both by 2 you get 196/21.

[@Mascig83]

@@calototube Thanks🙂

[@devondevon4366]

40.8333 Answer Length of Square = 7 Length of AB: 73.5 * 2 = 147/7 = 21 Extend point B and C to form a right triangle BCP and a retangle ABDP with area 147 (7 * 21) Let CD= n, then CB=n, then CP= 21-n , then (21-n)^2 + 7^2 = n^2 n^2 -41n + 441 + 49 =n^2 490 = 41n 11.6667 = n Hence, CP = 9.333 (21 - 11.6667) Hence, area of BCP = 9.3333 * 7 * 1/2 =32.67 Hence, area of BCP + green triangle = 106.1666 (32.67 + 73.5) Hence area of purple region = ABDP - 106.1666 = 147 - 106.1666 = 40.83333 Answer

[@PreMath]

Super! Thanks ❤️🌹

[@angusmcpherson]

I checked your answer by determining the green triangle hypotenuse (22.1359436212), and with the CD value of 35/3 (11.66667) I determined the height of the purple triangle-- 3.689. When I multiplied 1/2 bh I got the same answer: 40.8331

[@devondevon4366]

@@angusmcpherson Nice job

[@devondevon4366]

40.8333

[@prossvay8744]

area of purple region=40.83cm^2

[@PreMath]

Thanks ❤️

[@prossvay8744]

@@PreMath thanks teacher

[@lk-wr2yn]

((10,5^2+3,5^2)^0,5)*(((3,5^2)+((3,5/3)^2))^(0,5))=245/6=40,83333

[@JSSTyger]

40.84

[@laxmikatta1774]

Day 2 Of asking premath sir to reveal his face

[@giuseppemalaguti435]

AV=490/12

[@billycox475]

Correct me if I'm wrong, but the height of BCD is a line drawn from POINT C to LINE DB.

[@ybodoN]

Another one is the perpendicular dropped from B on DC (extended). The last one is the perpendicular dropped from D on BC (extended). Since BCD is an isosceles triangle, these two have the same length.

[@angusmcpherson]

The height from C to DB is 3.689

[@andirijal9033]

Sir where is your face ?

[@mfayyazpardasi2352]

Mujhe bahut Khushi Hui aur bahut Achcha laga hai aapki Mehfil Mein Aakar apni Ammi Ka Vasta Hai Hamara bhi Jarur Sath Dena Mujhe Bhi AAP Sab Ka Intezar rahata hai

[@PreMath]

Thanks dear ❤️🌹 Ji zaroor...

[@wackojacko3962]

Well, isn't that special! 🙂

[@PreMath]

Thank you ❤️