Can you calculate area of the Yellow shaded Quadrilateral? | (Triangles) |

[@zupitoxyt]

Such a great holidays are going on with incoming fantastic questions everyday ❤

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤

[@zupitoxyt]

@PreMath ur welcome

[@harikatragadda]

Draw BE perpendicular to AC. ∆BEC is Congruent to ∆BDC by ASA. Since ∠ABE =α , ∆AEB is Similar to ∆BEC. EC/BC = 2x/3x, EC= 16, and hence BE=8√5 and AB=12√5 AREA = ½AB*BC +½BE*EC= 208√5

[@reynaldowify]

Great, simpliest approach

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Both right triangles are similar. b / 2x = 24 / 3x --> b = 16cm Pytagorean theorem: (2x)² = 24² - b² = 24² - 16² x = √80 = 4√5 cm Yellow area : A = ½bh + ½bh A = ½*16*8√5 + ½*24*12√5 A = 64√5 + 144√5 = 208√5cm²

[@PreMath]

Thanks for the feedback ❤️

[@marioalb9726]

Both right triangles are similar. Labelling 'c'=AC and 'b'=DC c + b = 3/2 *24 + 2/3 *24 = 52 cm cosα = 2x/3x = 2/3 --> α= 48,19° A = ½.a.c.sinα + ½a.b.sinα A = ½.a.(c+b).sinα A = ½.24.52.√5/3 = 208√5 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Both right triangles are similar. Labelling 'c'=AC and 'b'=DC c - b = 3/2 *24 - 2/3 *24 = 20cm Pytagorean theorem: (3x)²=(2x)²+(c-b)² 9x²-4x²= 5x²= 20² --> x= 4√5 cm Yellow area : A = ½b₁h₁ - ½b₂h₂ A = ½(6x)*24 - ½(2x)(c-b) A = 288√5 - 80√5 = 208√5cm²

[@PreMath]

Thanks for sharing ❤️

[@allanflippin2453]

I have no new concepts, but I find the math is simplified by reordering it. 1) Discovering that ABC and BCD are similar, the dimensions in BCD have a 2/3 ratio to those in ABC. 2) Due to the 2/3 ratio, DC = 16 3) Area = (2x * 16 + 3x * 24) / 2 = 52x 4) Use pythagorean theorem on BCD. Notice that all three sides are divisible by 2, so divide them by 2: x^2 + 8^2 = 12^2. 5) x^2 = 144 - 64 = 80. x = 4 * sqrt(5) 6) Multiply by step 3's result: area = 208 * sqrt(5)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Both right triangles are similar b = 2/3 a = 2/3*24 = 16 cm c = 3/2 a = 3/2*24 = 36 cm cosα = 2/3 --> sinα = √5/3 A= ½a.c.sinα + ½a.b.sinα A= ½.24.(36+16).√5/3 = 208√5 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@WernHerr]

My solution: 3x:24 =2x:DC and DC=16. Pythagoras in BDC and x=4*sqrt(5). Area see solution of PreMath!

[@Aligakore]

Did it this way too 😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@RK-tf8pq]

Exactly, I had used the proportionality to the other triangle to get AC = 36. Then Pythagorean theorem to the triangle ABC to get X.

[@giuseppemalaguti435]

3x=24tgα...2x=24sin α...3/2=1/cosα .cosα=2/3..sin a=√5/3..2x=24*√5/3...x=4√5.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! I'm glad you found it helpful. 😀

[@alexundre8745]

Bom dia Sr Muitíssimo obrigado pelas aulas

[@PreMath]

You are very welcome!😀 Thanks for the feedback ❤️

[@PrithwirajSen-nj6qq]

Extend AB to E so that BE =3x Extend CD to E In triangle BDE DE = √(9x ^2 -4x^2)=x√5 tan α=x √5/2x=√5/2 In triangle ABC tan α=3x /24 =√5/2 x = 4√5 3x =12√5 2x =8√5 In triangle BCD tanα= 8√5/CD =√5/2 >CD =16 Area of quadrilateral =1/2*12√5*24 +1/2*8√5*16 =144√5 +64√5 =208√5 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Using a bit more trigonometry: Let's name alpha = t (easier to write). In ABC: tan(t) = (3.x)/24 = x/8. In BDC: sin(t) = (2.x)/24 = x/12. So by division we have: 1/cos(t) = (x/8)/(x/12) = 3/2 As (1 + (tan(t))^2)) = 1/((cos(t)^2) we then have: 1 + ((x^2)/64) = 9/4, and then x^2 = (5/4).64 = 80, and so x = sqrt(80) = 4.sqrt(5) The area of ABC is (1/2).(3.x).24 = (1/2).(12.sqrt(5)).24 = 144.sqrt(5) The area of BDC is (1/2).(2.x).DC = (1/2).(2.x).((2.x)/tan(t)) = (1/2).(2.x).16 = (1/2).(8.sqrt(5).16 = 64.sqrt(5), and finally the yellow area is 144.sqrt(5) + 64.sqrt(5) = 208.sqrt(5).

[@PreMath]

Excellent! Thanks for sharing ❤️

[@AmirgabYT2185]

S=208√5≈465,92≈466

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

We have two similar right triangles whose linear ratio = 3 : 2 The long leg of the larger triangle = the hypotenuse of the smaller triangle = 24 So the three sides of the large triangle = 3x - 24 - 36 And the three sides of the small triangle = 2x - 16 - 24 (2x)^2 + (16)^2 = (24)^2 x^2 = 144 - 64 = 80 Area = (3√80)(24)/2 + (2√80)(16)/2 = (4√80)(9 + 4) = 208√5)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

@ Thank YOU

[@MrPaulc222]

The first part is checking on triangle similarity. As they both have two angles that are identical, the third must be, too., so similar by AAA As the corresponding side lengths are 2/3 or 3/2 (due to 2x, 3x), depending on perspective, AC = 36 and DC = 16. This doesn't seem right, but no matter. 24/3x = 16/2x, which just gives 48x = 48x, so not much help. The total area is 52x un^2, but I need to find a value for x. Ref BCD: 16^2 + (2x)^2 = 24^2 256 + 4x^2 = 576. 4x^2 = 320 x^2 = 80 x = sqrt(80) or 4*sqrt(5). 8*2*sqrt(80) + 12*3*sqrt(80) = 52*sqrt(80) = 208*sqrt(5) approximates to 465.1 un^2. Although I went down a dead end at first, I have left that part up to show my initial error.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@himo3485]

ABC∞BDC 3x/24=2x/DC DC=16 (2x)²+16²=24² 4x²+256=576 4x²=320 x²=80 x=4√5 Yellow shaded area = 12√5*24*1/2 + 8√5*16*1/2 = 208√5

[@PreMath]

Excellent! Thanks for sharing ❤️

[@andrepiotrowski5668]

Why don’t you use proportions? AB/BC = 3x/24 = tan α and BD/BC = 2x/24 = sin α => cos α = 2/3 sqrt(3²-2²) = sqrt(5) plus 24/2 = 12 is the scaling factor for the upper triangle. A_upper = 12 * sqrt(5) * 24 / 2 = 144 * sqrt(5) A_lower = A_upper * (2/3)² = 4/9 * 144 * sqrt(5) = 64 * sqrt(5) A = 208 * sqrt(5)

[@PreMath]

Thanks for the feedback ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Since ABC and BCD are both right triangles, we can conclude: tan(∠ACB) = AB/BC sin(∠BCD) = BD/BC tan(α) = AB/BC sin(α) = BD/BC BC = AB/tan(α) BC = BD/sin(α) AB/tan(α) = BD/sin(α) AB*cos(α)/sin(α) = BD/sin(α) AB*cos(α) = BD (3x)*cos(α) = 2x ⇒ cos(α) = 2x/(3x) = 2/3 For the right triangle BCD we obtain: cos(∠BCD) = CD/BC cos(α) = CD/BC 2/3 = CD/24 ⇒ CD = 24*2/3 = 16 Now we can apply the Pythagorean theorem to the right triangle BCD: CD² + BD² = BC² 16² + (2x)² = 24² 256 + 4x² = 576 4x² = 320 x² = 80 ⇒ x = √80 = 4√5 Now we are able to calculate the area of the yellow region: AB = 3x = 12√5 BD = 2x = 8√5 A(yellow) = A(ABC) + A(BCD) = (1/2)*AB*BC + (1/2)*BD*CD = (1/2)*(12√5)*24 + (1/2)*(8√5)*16 = 144√5 + 64√5 = 208√5 ≈ 465.10 Best regards from Germany

[@zupitoxyt]

Hnn I had a dought on you

[@PreMath]

Excellent! Thanks for sharing ❤️

[@imetroangola17]

Solução: sen α = 2x/24 e tg α = 3x/24 cos α = sen α/tg α =2/3. Como sen² α + cos² α =1, então: (2x/24)² + (2/3)² = 1 (x/12)² = 1 - 4/9 = 5/9 x/12 = √5/3. Portanto, x = 4√5. Assim, cos α = 2/3 = CD/24 → CD = 16. A área S amarela é dada por: S = 3x. 24/2 + 2x. 16/2 S = 36x + 16x = 52x S = 52 × 4√5 *S = 208√5 unidades quadradas.*

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

Both right triangles are similar cosα = b/24= 2x/3x = 2/3 b = 16 cm ; sinα = √5/3 A₁= ½absinα= ½.24.16.√5/3= 64√5 cm² A₂= A₁/cos²α= 64√5/(2/3)²= 144√5 cm² A= A₁+ A₂ = 208√5 cm² ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jonathanburros6762]

Since angle with measure a is in both right triangles sin a = (3X)/[sqrt([ 9 X^2 +24^2)] = (2 X)/24 Solve for X DC =sqrt( 24^2 -24^2) The sum of the areas equals (1/2)(3 X)(24) + (1/2)(2 X)DC

[@PreMath]

Thanks for the feedback ❤️

[@santiagoarosam430]

Razón de semejanza entre ABC y BDC, s=2/3→ s²=4/9→ 24*3x*4/9=DC*2x→ DC=16→ (2x)²=24²-16²→ x=4√5→ Área ABDC =(3*4√5*24/2)*[1+(4/9)] =208√5 u². Gracias y un saludo cordial.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@relens2]

Since they are similar, and the ratio of the smaller to the larger is ⅔, the area of the smaller triangle will be (⅔)^2 or 4/9 of the larger. So the total area will be 13/9 the area of the larger. After solving for x as you did, one need only calc the area of the larger and multiply it by 13/9 - no need to solve for y. Yay!

[@sergioaiex3966]

Solution: Let's label AC = a and CD = b The Triangles ABC and BCD are similar, so we have proportions a/24 = 24/b ab = 576 ... ¹ a/3x = 24/2x a = 3x . 24/2x a = 36 Replacing in Equation ¹ 36 . b = 576 b = 16 Applying Pythagorean Theorem in Triangle BCD to calculate "x" (2x)² + (16)² = (24)² 4x² + 256 = 576 4x² = 320 x² = 80 x = 4√5 AB = 3x = 3 . (4√5) AB = 12√5 BD = 2x = 2 . (4√5) BD = 8√5 Once again, we'll applying Pythagorean Theorem in Triangle BCD to calculate "b" (8√5)² + b² = (24)² 320 + b² = 576 b² = 256 b = 16 Yellow Shaded Area = (½ 12√5 . 24) + (½ 8√5 . 16) Yellow Shaded Area = 144√5 + 64√5 Yellow Shaded Area = 208√5 Square Units ✅ Yellow Shaded Area ≈ 465,1021 Square Units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; BCA = δ = DCB; AB = 3x; BD = 2x; BC = 24 sin⁡(ABC) = sin⁡(BDC) = 1; sin⁡(δ) = 2x/24 = 3x/√(24^2 - 4x^2) → x = 4√5 → tan⁡(δ) = √5/2 → sin⁡(δ) = √5/3 → cos⁡(δ) = 2/3 → CD = 16 → AC = 36 → area ABDC = (1/2)sin⁡(δ)(16(24) + 24(36)) = 12sin⁡(δ)52 = 208√5 or: φ = 30° → sin⁡(3φ) = 1; AC = AE + CE → sin⁡(CEB) = 1 → CE = CD → BE = BD = 2x BCA = δ → BAB = θ = 3φ - δ → ABE = δ → sin⁡(δ) = x/12 cos⁡(δ) = 2/3 → sin⁡(δ) = √5/3 = x/12 → x = 4√5 → 3x = 12√5 → CD = 16 → AC = 36 → AE = 20 → 2x = 8√5 → area ABDC = 32x + 20x = 208√5 btw: AC = AE + CE → sin⁡(CEB) = 1 → ABE = BCE = δ → cos⁡(δ) = 2x/3x = 2/3 → sin⁡(δ) = √5/3 = x/12 → x = 4√5 → tan⁡(δ) = √5/2 → tan^2(δ) = 5/4 = area ∆ABE/∆BCE = P/Q → 5Q = 4P → P = 5Q/4 = Qtan^2(δ) → area ABDC = 2Q + 5Q/4 = 13Q/4 = (13/4)16x = 208√5

[@PreMath]

Excellent! Thanks for sharing ❤️

[@spiderjump]

CD/2x= 24/3x CD= 16 (2x)^2 + 16^2 = 24^2 x= square root of 80 areas of the 2 triangles can calculated .

[@PreMath]

Thanks for the feedback ❤️

[@blogfilmes1134]

Fiz por sen e tg

[@wackojacko3962]

I choke on math exams. ...so for study group before an exam always ask what is gonna be the easiest problem on the exam and what is gonna be the hardest problem on the exam cuz I'm gonna overthink both and get em wrong. ...So if this is misinformation please delete my comment. I'm thinking AC=36. am I close? 🤞

[@PreMath]

😀 Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP-RESOLUTION PROPOSAL : Statement : Triangle [ABC] similar to Triangle [BCD] 01) Let DC = Y 02) 3X / 24 = 2X / Y ; X / 8 = 2X / Y ; Y = 16 lin un 03) DC = 16 lin un 04) 2X / 16 = 3X / 24 ; 2X / 3X = 16 / 24 ; 2 / 3 = 16 / 24 ; (2 * 24) = (3 * 16) ; 48 = 48 - True !! 05) Let AC = Z 06) 24 / 2X = Z / 3X ; 12 / X = Z / 3X ; Z = 36 lin un 07) AC = 36 lin un 08) Linear Ratio (R) = 3 / 2 or R = 1,5 09) Let's find X value!! 10) (24^2 - 16^2) = (2X)^2 ; 320 = 4X^2 ; 320 ; X^2 = 320 / 4 ; X^2 = 80 ; X = sqrt(80) ; X = 4*sqrt(5) lin un Or; 11) (36^2 - 24^2) = (3X)^2 ; 720 = 9X^2 ; X^2 = 80 ; X = sqrt(80) ; X = 4*sqrt(5) lin un 12) 3X = 12*sqrt(5) and 2X = 8*sqrt(5) 13) A(1) = (8*sqrt(5) * 16) / 2 ; A(1) = 64*sqrt(5) sq un ; A(1) ~ 143 sq un 14) A(2) = (12*sqrt(5)) * 24 / 2 ; A(2) = 144*sqrt(5) sq un ; A(2) ~ 322 sq un 15) R = 1.5 => R^2 = 2,25 16) 143 * R^2 ~ 322 ; 143 * 2,25 ~ 322 - True !! 17) Yellow Shade Area (YSA) = A(1) + A(2) ; YSA = (143 + 322) ; YSA = 465 sq un So, after all this Tedius Algebra; OUR BEST ANSWER IS : Yellow Shaded Area is approx. equal to 465 Square Units. Exact Form = (144 + 64)*sqrt(5) Square Units = 208*sqrt(5) Square Units. Best Regards from Al - Andalus; Al - Ushbuna !!

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@nicholastoop7635]

3z/24 = 2x/CD so CD=16 4x² = 24²-16² = 8×40 so x²=80 area of triangles= ½72x + ½32x = 52x area = 52×sqrt(80} =208sqrr(5)

[@LuisdeBritoCamacho]

@@nicholastoop7635 Thanks for sharing.