A Very Nice Geometry Problem
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- Опубліковано 24 чер 2024
- A Very Nice Geometry Problem
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This seems to be the second time that geometry was used to calculate the shaded AND it was done WITHOUT subtracting the whole triangle from the non-shaded triangle. Then again doing that would have been cumbersome, I guess. And why I shall compare that with other similar problems and practice them!
AC*√3/2=2√3/2=√3→ 2θ=30º→ Con eje AD hacemos la simetría de ABD y obtenemos el triángulo DB´C, semejante al ABC→ AC=AB´+B´C=(√3)+(2-√3)→ DC=2*(B´C)=4-2√3→ Área ADC =DC*AB/2=2√3 -3 =0,4641...
Gracias y un saludo cordial.
We notice, from the ratio of side to hypotenuse, that ΔABC, with hypotenuse 2 and one side √3, is a 30°-60°-90° special right triangle, so BC = 1 and 2Θ = 30°. Area of ΔABC = (1/2)(1)(√3) = (√3)/2. Because 2Θ = 30°, Θ = 15°. ΔABD is a 15°-75°-90° right triangle, which, while not considered "special" in geometry, we should consider "familiar" because of its frequent appearance in problems, and we should know its properties. The ratio of sides is (short):(long):hypotenuse is (√3 - 1):(√3 + 1):2√2. So, BD = (√3)(√3 - 1)/(√3 + 1). Area ΔABD = (1/2)(BD)(AB) = (1/2)((√3)(√3 - 1)/(√3 + 1))(√3) = (3/2)(√3 - 1)/(√3 + 1)). To remove the radical from the denominator, multiply by (√3 - 1)/(√3 - 1) and simplify. (√3 + 1)(√3 - 1) = 2 and (√3 - 1)(√3 - 1) = 3 - 2√3 + 1 = 4 - 2√3. So, area = (3/2)(4 - 2√3)/2 = 3(2 - √3)/2. Shaded area = area ΔABC - area ΔABD = (√3)/2 - 3(2 - √3)/2, which simplifies to 2√3 - 3, as Math Booster also found.
После нахождения BD проводим перпендикуляр из D к AC. Перпендикуляр равен BD, он же вьісота треугольника ADC. Площадь ADC= 1/2×AC×BD=BD
Reflect ∆ABD about AD to form ∆AD'D.
∠D'DC = ∠BAC =2θ, hence ∆BAC is Similar to ∆DD'C
D'C=2-√3
DD'/(2-√3)=√3/1
DD'= 2√3-3
Area(ADC)=½*DD'*AC= 2√3-3
That's exactly what I did : )
Yay, I'm thinking like a genius !
I notice that you find quite ingenious solutions to problems. What is your background? Are you a teacher or a math enthusiast? 💯
@@PS-mh8ts Thanks you, I am an artist and love maths.
@@harikatragadda Excellent! 💯🙏
Base of the large triangle
=√[4-3]=1
Area of large 🔺
=1/2*1*√3=√3/2
Area of coloured triangle
= 2/(2+√3)*√3/2
=√3/(2+√3)
=√3(2-√3)=(2√3-3 ) sq unit
Comment please.
It is better to let DC be x as area of shaded triangle = (1/2)xDCxsqrt3.
*Outra solução:*
[ADC]/DC=[ABD]/BD, isto é,
[ADC]/[ABD]=DC/BD. Pelo teorema da bissetriz interna, temos:
DC/BD=AC/AB=2/√3. Assim,
[ABD]/[ADC]=2/√3, daí
*[ABD]/{[ABD]+[ADC]}=2/(2+√3)*
Ora, [ABD]+[ADC]=[ABC].
Por Pitágoras no ∆ABC, temos que BC=1. Logo:
[ABC]=√3/2. Assim,
[ABD]/(√3/2)=2/(2+√3)
[ABD]= √3 × 1/(2+√3)
Como 1/(2+√3)= 2 - √3 então
*[ABD]= 2√3 - 3.*
Shaded area should be 0.478 instead of 0.464.
I get 0.464 when I calculate 2√3 - 3.
Or, total area of triangle ABC = .8660. Going Pythagorean gives a Cos of angle ACB = 1 Now, AB/AC stands in the same ratio as BD/DC. Thus BD = .4638 Area of unshaded triangle ABD is thus .4016. So, subtracting .4016 from total area of ABC gives the shaded area = .4644 Pretty damn close to what the Boosterman came up with. But with less gobbledygook. Don't get me wrong, I support the Boosterman.
Area=2√3-3.❤
❤❤
You can also use tan 2Θ and solve for tanΘ then find the ratio of x
Bydroppinh perpendicular from o to ac we getthe height of the triangle aoc and the height is equal to x.
❤🎉😊😊😊😊
2 sqrt3 - 3 = 0.464 .
Apologize me, I am mistake ,shaded area is 0.464.
*Solução 2:*
Pelo teorema de Stewart, temos:
(AD)^2= AB × AC - BD × CD
(AD)^2= 2√3 - BD × CD.
Como ∆ABD é triângulo retângulo, por Pitágoras:
(AD)^2= 3 + (BD)^2, daí
3 + (BD)^2= 2√3 - BD × CD
(BD)^2 + BD × CD= 2√3 - 3
BD×( BD+ CD)= 2√3 - 3
BD× BC= 2√3 - 3
Por Pitágoras no ∆ABC, temos BC=1.
Logo,
*BD= 2√3 - 3.*
Note que:
[ADC]= (AD×AC×sin θ)/2 e
sin θ = BD/AD, assim
[ADC]=(BD × AC)/2= (BD×2)/2
[ADC]= BD ( De forma numérica)
Portanto,
*[ADC]=2√3 - 3.*
So,The answer is incorrect.