Can you find area of the Green shaded region? | (Justify) | #math #maths | #geometry

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Can you find area of the Green shaded region? | (Justify) | #math #maths | #geometry

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[alexundre]

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

[georgebliss]

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.

[santiagoarosam]

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.

[ناصريناصر-سب]

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.

[marioalb]

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

[PrithwirajSen-njqq]

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
=

[ClassicalMusicFan]

"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.

[Emerson_Brasil]

*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.*

[yalchingedikgedik]

Thanks Sir
Very nice
Thanks for PreMath
❤❤❤❤

[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!

[pwmiles]

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

[jimlocke]

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.

[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

[joeschmo]

✨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.

[himo]

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)

[phungpham]

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😅😅😅

[adept]

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

[misterenter-izrz]

，Let A be the total area, 23=(3/4+1/4×1/3)A=10/12 A=5/6A, A=23×6/5=138/5, therefore the answer is (1/4×2/3)A=1/6 ×138/5=23/5.😊

[SkinnerRobot]

G/(G + 23) =
G/(G + 23) = 1/5
5G = G + 23
4G = 23
G = 23/4 cm²

[davidellis]

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.