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 √)
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.
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.
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)
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😅😅😅
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]
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.
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
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!!!
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
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.
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This one is fun and I like 2nd method mostly. 🙂
Glad you liked the second method! 🙂
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 √)
AD=a...3√3=(1/2)a^2sin120=√3a^2/4..a=√12...h=√12sin60=3...Agreen{(1/2)3*√12sin30=(3/2)√3
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.
The triangle ADC is isosceles, so
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.
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)
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😅😅😅
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]
The fastest way is tracing a perpendicular from D to AC we get 3 congruent right triangles whose area is 3/2sqrt3
30° : 60° : 90° = 1 : √3 : 2
BD = DC/2
area of the Green triangle : 3√3/2 cm^2
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.
S=3√3/2≈2,6
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
Good Morning Master
Parabéns pelas aulas
Forte Abraço do Rio de Janeiro
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!!!
4:10 here we can simply prove that triangle abd is congruent to triangle aed therefore the area is 3√3/2
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.
Let's find the area:
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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
3√3/2
Ne pas se lancer tête baissée. Et ... voir que ABD est la moitié de ACD. !
3×4=12÷2=6
4×8=32÷2=16full area 3×4=12÷2=6 )(6/16
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.