Can you find area of the Green shaded region? | (Justify) |

[@alexundre8745]

Bom dia Mestre Esse foi difícil Mas com as explicações do Sr ficou fácil Grato por mais essa aula

[@PreMath]

Olá querido❤️ De nada! Obrigado pelo feedback ❤️

[@alexundre8745]

@@PreMath O Sr é gente boa. Grato

[@georgebliss964]

In short:- Area of ABC = 1/2 * 5x * 2y * sin theta. Area of AED = 1/2 * 2x * y * sin theta. So area ratio ABC / AED = 10 / 2 = 5. Then (A + 23) / A = 5. 5A = A + 23. 4A = 23.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@EliteMathsAU]

Love it

[@santiagoarosam430]

Trazamos BD---> BED=a ; EAD=2a/3 ; ADB=DCB---> (2a/3)+a=23-a---> a=69/8---> EAD =2*69/3*8 =23/4 =5,75 cm² = Área sombreada verde. Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@joeschmo622]

✨Magic!✨ I never heard of that ratio-of-areas rule... nifty! I was going to go all trig on that problem, but that's so much easier and direct.

[@PreMath]

Thanks for your appreciation of the ratio-of-areas rule! 😊❤️

[@ناصريناصر-س4ب]

Suppose the area of triangle AED is K, then the area of triangle BED is 1.5K, and the area of triangle BCD is 1.5K + K = 2,5K, so Area of BCDE=2,5K+1,5K=4K = 23, so K = 5.75.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@timeonly1401]

Appreciate the VERY simple, direct answer!! A things that could've been clearer (to me) was (1) HOW you got Area(∆BED) = 1.5 Area(∆AED) and (2) how you got Area(∆BCD). I had to fill in the parts that weren't said... (1) Consider ∆BED & ∆AED. Drop a perpendicular from point D onto line segment AB, which becomes the altitude of BOTH triangles. Since the area of a triangle is jointly proportional to base length and altitude length, and since the altitudes are the same for both ∆BED & ∆AED, we have Area(∆BED)/A(∆AED) = 3x/2x = 1.5 . So Area(∆BED) = 1.5 A(∆AED) = 1.5K. (2) Consider ∆ABD & ∆BCD. Drop a perpendicular from point B onto line segment AC, which becomes the altitude of BOTH triangles. Since the area of a triangle is jointly proportional to base length and altitude length, and since their altitudes AND their bases are the same for both ∆ABD & ∆BCD, we have Area(∆BCD) = Area(∆ABD) = Area(∆AED) + Area(∆BED) = K + 1.5K = 2.5K

[@ناصريناصر-س4ب]

Thank you I just wanted to be brief in the answer.​@@timeonly1401

[@waheisel]

Triangle DCE can quickly be shown to equal in area to the green triangle. And Triangle EBC can be quickly shown to be 3x the area of the green triangle. Ergo the green triangle is one fourth the area of the yellow quadrilateral. Thanks PreMath for the fun daily puzzle!

[@yalchingedikgedik8007]

Thanks Sir Very nice Thanks for PreMath ❤❤❤❤

[@PreMath]

You are very welcome! Thanks for the feedback ❤️🙏

[@marioalb9726]

(A₁+A₂)/A₁ = 5x.2y / 2x.y 1 + A₂/A₁ = 5 A₂/A₁ = 4 A₁ = A₂/4 = 5,75 cm² ( Solved √)

[@pwmiles56]

Alternatively, let the green area be g. Make line BD. Triangle BDC is 1/2 the height of ABC, on the same base, so BDC = (g + 23)/2 Triangle BED is the same height as AED, on 3/2 times the base, so BED = (3/2) g The yellow area is the sum of BDC and BED (g + 23)/2 + (3/2) g = 23 2g + 23/2 = 23 2g = 23/2 g = 23/4

[@PreMath]

Excellent! Thanks for sharing ❤️

[@adept7474]

▲ACE = 2/5, ▲ADE = 1/5, BEDC = 4/5 = 23, ADE = 23/4 = 5,75.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@PrithwirajSen-nj6qq]

We may use trigonometry Area of green triangle =1/2 *2x*y *sin theta =xy sin theta Area of big triangle = 1/2*5x*2y*sin theta =5xy sin theta Area of yellow quadrilateral =area of big triangle - area of green triangle = 4xy sin theta Hence Area of green triangle = xy sin theta =4xy sin theta/4 =23/4 square units =

[@PreMath]

Excellent! Thanks for sharing ❤️

[@himo3485]

AD=DC=y Green shaded region AED : S 5x*2y=10xy=S+23 2xy=S xy=S/2 5S=S+23 4S=23 S=23/4=5.75(cm²)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@1ClassicalMusicFan]

"If two triangles share a height, then the ratio of their areas is equal to the ratio of their bases." Draw the segment EC. △EDC and the green triangle share a height and their bases are equal, so they have the same area, denoted by a. --> area of △EAC = 2a and △EBC = 23 - a. Similarly, area of △EAC/area of △EBC = 2x/3x = 2/3. --> area of △EBC = 3a. From these two results, we get a = 4.75.

[@jimlocke9320]

Construct EF parallel to BC, where F lies on AC. Construct DG parallel to BC, where G lies on AB. Let BC = b be the base of ΔABC and designate the height as h. ΔAEF, ΔADG and ΔABC are similar. ΔABC has area (1/2)bh. The dimensions of ΔAEF are 2/5 of those of ΔABC. So, its base EF is 2b/5 and height 2h/5. ΔAEF has area (1/2)(2b/5)(2h/5) = (2/25)bh. The dimensions of ΔADG are 1/2 of those of ΔABC, so its height is h/2. If EF = 2b/5 is considered the base of ΔDEF, its height is h/2 - 2h/5 = h/10, area of ΔDEF = (1/2)(2b/5)(h/10) = (1/50)bh. Green area = ΔAEF + ΔDEF = (2/25)bh + (1/50)bh = (1/10)bh. Ratio of areas of green area to ΔABC = (1/10)bh/((1/2)bh) = 1/5. So, green area is 1/5 the area of ΔABC. If green area = A, then area of ΔABC = A + 23. So, 5A = A + 23, 4A = 23, A = 5.75 cm², as PreMath also found.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@davidellis1929]

It's easier if you draw the median BD, so the area of ABD is half that of ABC. Then DE divides the area of ABD in a 2:3 ratio, so the green area is 2/5 that of ABD, or 1/5 that of ABC. That's 1/4 the yellow area, so the green area is 23/4.

[@PreMath]

Thanks for the feedback ❤️

[@quigonkenny]

Long method: Let M be the midpoint of BC and N be the midpoint of AB. Draw MD, DN, and NM. As DA = CA/2, AN = AB/2 = 5x/2, and ∠A is common, then ∆AND and ∆ABC are similar triangles with side length ratio 1:2. As ∠AND = ∠ABC and ∠NDA = ∠BCA, then these pairs of angles are corresponding angles, and thus ND and BC are parallel. Similarly, as NB = AB/2 and BM = BC/2 and ∠B is common, then ∆NBM is also similar to ∆ABC with a 1:2 ratio, and thus congruent to ∆AND. Same for ∆DMC. By similar deduction as above for ND and BC, NM is parallel to CA and MD is parallel to AB. As ND and BC are parallel and MD and AB are parallel, ∠MDN and ∠DMC are alternate interior angles and thus congruent. As NM is parallel to CA, ∠DNM and ∠BMN are similarly congruent. As ∠MDN = ∠DMC = ∠AND, ∠DNM = ∠BMN = ∠NDA, and ND is common, then ∆MDN and ∆AND (as well as ∆NBM and ∆DMC by extension) are congruent. As each of ∆AND, ∆NBM, ∆DMC, and ∆MDN are congruent with the others and all make up the larger triangle ∆ABC, then if the area of each is U, then the area of ∆ABC is 4U. Draw BD. As AE = 2x and EB = 3x, then the area of ∆AED is 2/3 that of ∆EBD, as their bases are in a 2:3 ratio but their heights are identical. Similarly, the area of ∆AED is in a 2:2.5 ratio with that of ∆AND, or is 4/5 of the area or 4U/5. From this we can determine the area of the yellow quadrilateral in terms of U, and thus determine the value of U. 4U - 4U/5 = 23 16U/5 = 23 U = (5/16)23 = 115/16 = 7.1875 As the green triangle area is 4U/5: 4U/5 = 4(115/16)/5 = 23/4 = 5.75 cm² Short method: The area of a triangle, given two side lengths a and b and the angle between them C is absin(C)/2. Let AD = DC = y and let the sine of ∠CAB be k. Triangle ∆ABC: Aᴛ = absin(C)/2 = 5x(2y)k/2 Aᴛ = 5xyk Green Triangle ∆AED: Aɢ = absin(C)/2 = 2x(y)k/2 Aɢ = xyk Aᴛ - Aɢ = 23 5xyk - xyk = 23 4xyk = 23 xyk = 23/4 Aɢ = 5.75 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@shrikrishnagokhale3557]

Is it 23/4 cm^2?

[@SkinnerRobot]

G/(G + 23) = [½(2x)(y)sinθ]/[½(5x)(2y)sinθ] G/(G + 23) = 1/5 5G = G + 23 4G = 23 G = 23/4 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@imetroangola17]

*Solução Simples:* Seja AD=DC=y. Além disso, seja a área [ABC] = A. Daí, (AB×AC sen A) /2 = A (5x . 2y sen A) /2 = A (2x . y sen A)/2 = A/5 Ora, área [ADE]=(2x . y sen A)/2 Logo, [ADE] = A/5 → _A = 5[ADE]._ [ABC] - [ADE] = 23 5[ADE] - [ADE] = 23 4[ADE] = 23 → [ADE] = 23/4 *[ADE] = 5,75 cm².*

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@wackojacko3962]

@ 3:38 , who woulda' thunk it? ...a primitive mind? No, I think not! 🙂

[@PreMath]

😀 Thanks for the feedback ❤️

[@phungpham1725]

1/ Label the area of the green triangle= G Focus on the two triangles AED and ECD,their areas are equal ( same base, same height) so the area of the triangle AEC= 2G 2/ Focus on two trisngles AEC and EBC: Area of AEC/ Area of EBC=2/3 -> 2G/area of Area of EBC= 2/3 -> G/area of EBC= 1/3 --> G + 3G = Area of the yellow quadrilateral G=23/4 sq cm😅😅😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@DB-lg5sq]

شكرا لكم على المجهودات S(AED)=a S(CED)=a S(CEB)=23-a ارتفاع CEB هو CH ........... CH=(46-2a)/3x S=23/4

[@DB-lg5sq]

Merci beaucoup pour votre effort On peut procéder de la manière suivante S(AED)=1/2 AE AD sinA S(ABC)=1/2AB AC sinA S(EAD) / S(ABC) =1/5 ...... S(ABC) =23/4

[@marcgriselhubert3915]

The theorem you use (product of the ratios on the two sides of common angle theta )is interesting as it is not often seen. It would benefit of a proof. Here is one: Area of ABC = (1/2). norm(Vectorial product of VectorAB and VectorAC) = (1/2).abs(det(VectorAB, VectorAC)) in any orthonormal Now if E is as VectorAE = k.VectorAB (k>0), and D as VectorAD = k'.VectorAC (k'>0), then the area of AED is (1/2).abs(det(VectorAE, VectorAD)) = (1/2).abs(det(k.VectorAB, k'.VectorAC) = (k.k').area of ABC (by bi-linearity of the determinant). Another proof (more elementary): Area of ABC = (1/2).AB.AC.sin(angleBAC) and Area of AED = (1/2).AE.AD.cos(angleEAD) = (1/2).(k).AB.(k').AC.sin(angleBAC) = (k.k').area of ABC.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@michaeldoerr5810]

The area is 23/4 units square.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wasimahmad-t6c]

23÷3.6=6.38

[@よっこいしょういち-w9m]

AED = 2/5 * 1/2 = 2/10 = 1/5

[@PreMath]

Thanks for the feedback ❤️

[@sergeyvinns931]

RUSSIA! Треугольник АВС равнобедренный, АВ=АС, АЕ+ВЕ=AD+DC; 2х+3х=5х; AD=5x/2=2,5x; площадь треугольника АВС равна АВ*АС*sina=(5x)^2*sina; площадь треугольника EAD равна AE*AD*sina=2x*2,5x*sina; найдём сколько треугольников ЕАD поместится в треугольнике АВС; 25x^2*sina/5x^2*sina=5; обозначим площадь треугольника ЕАD через S, тогда площадь АВС=5S, запишем чему равна площадь 23, через S, 5S-S=23, 4S=23, S=23/4=5,75.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@nenetstree914]

23/4

[@PreMath]

Excellent! Thanks for sharing ❤️