Can you find area of the Green shaded triangle? | (3 Methods) |

[@wackojacko3962]

This one is fun and I like 2nd method mostly. 🙂

[@PreMath]

Glad you liked the second method! 🙂

[@soli9mana-soli4953]

The fastest way is tracing a perpendicular from D to AC we get 3 congruent right triangles whose area is 3/2sqrt3

[@marcgriselhubert3915]

Very good.

[@ritanjanpaul]

You solved it in a complicated way. Simple and easy methods are there.

[@EliteMathsAU]

Great job.

[@billylowgroundvytaszukas5797]

thank You. You help me to " back to reality". Thank you for your videos.

[@ناصريناصر-س4ب]

The triangle ADC is isosceles, so

[@sergeyvinns931]

RUSSIA! Есть три метода решения задачи, самый простой это геометрический, следующий тригонометрический, за тем, алгебраический. Решение самое простое, это разбить треугольник ACD на два, опустив перпендикуляр из угла ADC гипотенузу АС, получится 3 равных треугольника, площадь одного из которых, окрашенного в зелёный цвет, нужно найти, для этого достаточно поделить площадь 3\/3 на 2, получим А=3\/3/2. Два других способа происходят от первого.

[@marioalb9726]

α = 60°-30° = 30° (Angle BAD) sin α = h₁ / h₂ = 1/2 Two triangles, same base (AD=DC), and its heights with a ratio 1/2 A₁ = ½ A₂ = 1,5√3 cm² ( Solved √)

[@alexundre8745]

Good Morning Master Parabéns pelas aulas Forte Abraço do Rio de Janeiro

[@shrijjithnatakala]

4:10 here we can simply prove that triangle abd is congruent to triangle aed therefore the area is 3√3/2

[@TurquoizeGoldscraper]

U got the angles of the green triangle and found it was a 30-60-90 triangle, therefore BD = 1/2 AD = 1/2 DC. Triangle ABD and ADC share the same height, therefore cutting the base in half, cuts the area in half. Area(ABD) = 1/2 Area(ADC) = 1/2 * 3 sqrt(3) = 3/2 sqrt(3)

[@georgebliss964]

Triangle ABD. Sin BAD = sin 30 = 1 / 2. Therefore BD = 1/2 AD. Since AD = DC, BD = 1/2 DC. Since triangles ABD & ADC have the same height, Areas are proportionate to base lengths. Thus area of ABD = 1/2 area of ADC. 1/2 x 3 root 3.

[@laurenamyhanks]

I found the side lengths of the isosceles triangle to be √12 using the area of a triangle (1/2absinC) equal to 3√2. Then used right angled trig to work out the adjacent and opposite of the RAT. I got 3 and √3. Then used area of a triangle (1/2 base X height) to get the final answer. That was fun!

[@MrPaulc222]

Make the midpoint of AC, M, and draw a line between them. This splits the large triangle into 3 by 30,60,90 triangles. Because AD is the hypotenuse of both AMD and ABD, those two triangles are of equal area, so ABD (the green triangle), has an area of (3*sqrt(3))/2. In decimal, this approximates to 2.598 un^2

[@himo3485]

30° : 60° : 90° = 1 : √3 : 2 BD = DC/2 area of the Green triangle : 3√3/2 cm^2

[@Z-eng0]

Method is more efficient step-wise, since there's no need for the complex calculations in method 2, all calculations in method 1 are direct and quicker, BUT method 1 is less obvious for anyone to go to (I personally wouldn't go dividing a triangle of known are into 2 halves if I don't know yet whether or not that would be more useful than using the area to find sides in the future calculations) so if one can go for method 1 without any prior knowledge good for them, but I still prefer method 1 if a person has no prior knowledge of this kind of problem and has a limited time with a calculator at hand

[@devondevon4366]

Angle A = 30 degrees, isosceles, hence angle D =120 degrees Draw a perpendicular line from D to form two congruent- 30-60-90 right triangles. The area of each, hence, is 3 sqrt 3/2. Hence, the product of the bases is n * n sqrt 3 (let n= the base) Hence, n* n sqrt 3 = 3 sqrt 3 n^2 = 3 Hence, n = sqrt 3 Hence, the sides are sqrt 3 and 3 Hence, the hypotenuse of triangle ABD= 2 sqrt 3 Hence, BD= sqrt 3 and AB = sqrt 3 * sqrt 3 =3 Hence, the area of ABD= 3 sqrt 3 * 3* 1/2 = 3 sqrt 3/2 Answer

[@quigonkenny]

Drop a perpendicular from D to E on CA. As AD = DC, ∆ADC is an isosceles triangle, so DE is a perpendicular bisector and forms two congruent right triangles ∆CED and ∆DEA. As ∠CAD = ∠DCA = 30°, ∆CED and ∆DEA are 30-60-90 special right triangles, and ∠ADE = ∠EDC = 60°. As ∠BDA is an exterior angle to ∆ADC at D, then ∠BDA = ∠DCA+∠CAD = 30°+30° = 60°. As ∠ADE = ∠BDA = 60°, ∠ABD = ∠DEA = 90°, and DA is common, ∆ABD and ∆DEA are congruent, so the area of ∆ABD is the same as ∆DEA, which is 1/2 that of ∆ADC, or 3√3/2 cm².

[@giuseppemalaguti435]

AD=a...3√3=(1/2)a^2sin120=√3a^2/4..a=√12...h=√12sin60=3...Agreen{(1/2)3*√12sin30=(3/2)√3

[@michaeldoerr5810]

The answer is x = sqrt(3). I have noticed that all three methods are basically explanations for why the x value is the way it is. And the first method definitely requires the most reasoning. I shall use that for practice!!!

[@alster724]

Method 2 is more convenient

[@cyruschang1904]

Sin30° = 1/2 Green triangle is similar to the big triangle Green triangle area to the big triangle area = (✓3)^2 : 3^2 Green triangle area to the white triangle area = (✓3)^2 : 3^2 - (✓3)^2 = 3 : 6 = 1 : 2 Green triangle area = (3✓3) cm^2 ÷ 2 = [(3✓3)/2] cm^2

[@AmirgabYT2185]

S=3√3/2≈2,6

[@waheisel]

I was taught that the "legs" of triangle are the sides that form a right angle?

[@petohunor6737]

Könnyű. 3*sqrt(3)/2. Az egyenlőszárú háromszöget megfelezve kapunk 3 egybevágó háromszöget. Vagyis a zöld háromszög területe fele a megadott háromszög területének. (A 120°-os szöget megfelezve 60°-os szögeket kapunk . így a harmadik szög 90°-os lesz ...

[@santiagoarosam430]

Si E es proyección ortogonal de D sobre AC---> DEA; DEC y DBA son triángulos congruentes---> Área DBA =ACD/2=3√3/2 cm². Gracias y saludos.

[@sergioaiex3966]

Solution: A = 3√3 cm² 3√3 = ½ x x sin 120° 3√3 = ½ x x √3/2 3√3 = √3/4 x² 3√3 . 4/√3 = x² x² = 12 x = 2√3 A = ½ base height 3√3 = ½ 2√3 h h = 3 Tan 60° = h/a √3 = 3/a a = 3/√3 a = 3√3/3 a = √3 A = ½ √3 . 3 A = 3√3/2 cm² A = 2,598 cm²

[@phungpham1725]

1/The triangle ABD is a special 30-90-60 Label BD= a --> AD=CD= 2a Focus on the green and the white isosceles triangle, they have the same height and the base BD=1/2CD So, area of the green triangle = 1/2 3sqrt3 sq cm😅😅😅

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Since ACD is an isosceles triangle (AD=CD), we can conclude: ∠CAD = ∠ACD = 30° ⇒ ∠ADC = 180° − ∠CAD − ∠ACD = 180° − 30° − 30° = 120° From the known area of the triangle ACD we obtain: A = (1/2)*AD*CD*sin(∠ADC) (3√3)cm² = (1/2)*AD²*sin(120°) (3√3)cm² = (1/2)*AD²*(√3/2) 12cm² = AD² ⇒ AD = √(12cm²) = (2√3)cm Since ∠ABD=90° and ∠ADB=180°−∠ADC=180°−120°=60°, we know that ABD is a 30°-60°-90° triangle. Therefore we can conclude: BD:AB:AD = 1:√3:2 BD/AD = 1/2 ⇒ BD = AD/2 = (2√3)cm/2 = (√3)cm AB/AD = √3/2 ⇒ AB = √3*AD/2 = √3*(2√3)cm/2 = 3cm Now we are able to calculate the area of the green triangle: A(ABD) = (1/2)*AB*BD = (3√3/2)cm² Best regards from Germany

[@alexniklas8777]

3√3/2

[@PrithwirajSen-nj6qq]

Fourth method ABC is 30-60-90 triangle Hence if AB is x then AC is 2x Now AD is angle bisector of 🔺 ABC Hence AD /CD =x/2x=1/2 Then we notice the height of 🔺 s ABD & ADC is AB Area of ABD/area of 🔺 ADC =AD/CD=1/2 Area of 🔺 ABD= Area of 🔺 ADC/2=3√3/2 sq units [ in this case we need not find out the length of any side]

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Let AD = DC = X cm 02) Parallelogram Area (P) Formula : P = X * X * sin(60º) ; P = X^2 * sin(60º) 03) P = 6sqrt(3) 04) X^2 * sqrt(3)/2 = 6sqrt(3) ; X^2 = 12sqrt(3) / sqrt(3) ; X^2 = 12 ; X = sqrt(12) ; X = 2sqrt(3) cm 05) sin(60º) = AB / X ; AB = 3 cm 06) cos(60º) = BD / X ; BD = sqrt(3) 07) Green Triangle Area (GTA) = BD * AD / 2 08) GTA = (sqrt(3) * 3) / 2 09) GTA = [3sqrt(3)] / 2 sq cm Thus, OUR BEST ANSWER : Green Triangle Area equal to (3sqrt(3)/2) Square Centimeters.

[@Birol731]

My way of solution ▶ For the given white area of the triangle ΔADC A(ΔADC)= [DC]*[AB]/2 [DC]= x [AB]= h A(ΔADC)= 3√3 cm² ⇒ 3√3= hx/2 hx= 6√3........Eq-1 for the triangle ΔADC ∠DCA= ∠CAD= 30° ⇒ ∠CAD= 180°- 2*30° ∠CAD= 120° ⇒ ∠BDA= 60° sin(∠BDA)= √3/2 h/x= √3/2 2h= √3x h= √3x/2.......Eq-2 If we pu the value h in equation-1 we get: hx= 6√3 √3x/2*x= 6√3 x²= 12 x= 2√3 cm ⇒ h= √3/2*2√3 h= 3 cm tan(∠BDA)= √3 tan(60°)= √3 ⇒ h/[BD]= √3 [BD]= 3/√3 [BD]= √3 cm A(ΔBDA)= h*[BD]/2 A(ΔBDA)= 3*√3/2 ⇒ Agreen= 3√3/2 cm²

[@devondevon4366]

3 sqrt 3/2

[@wasimahmad-t6c]

1.5625

[@EPaozi]

Ne pas se lancer tête baissée. Et ... voir que ABD est la moitié de ACD. !

[@wasimahmad-t6c]

3×4=12÷2=6

[@wasimahmad-t6c]

4×8=32÷2=16full area 3×4=12÷2=6 )(6/16