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I used Pyphagorean theorem for four triangles: one is buit by the perpendicular on the base line(the same as you did), two are from a perpendicular on AD, and the fourth is APB . I calculated PB and then defined the sin of theta. it is1/2.

Can y'all prove that the desired area is Always (1/3) of the rectangle's area ? OOOPS ! That's exactly what the proposer proved ! I hadn't seen the solution before getting "My Math On !"

There is another way of doing it simplified yielding the same resault.

Vertex angle of shaded triangle is 30° = ⅓ 90° Consequently: Shaded triangle area = ⅓ Rectangle area At = ⅓ Ar At = ⅓ 15 At = 5 cm² ( Solved √ )

Area of rectangle Ar = a. b = 15 cm² However, ratio between 'a' and "b' IS NOT GIVEN. This means we can choose this ratio, and even so we are meeting the original conditions. I chose to have a rectangle with ratio a/b = tan 30° Now, points D and E are unified, and area of right triangle BCE is half of rectangle area, is 15/2=7,5 cm² Area of shaded triangle is 2/3 of area of right triangle At = ⅔ 7,5 At = 5 cm² ( Solved √ ) Note that Ar = 3 . At Everytime we have this configuration, At =⅓.Ar Consequently, the easiest and shortest solution is: A = A / 3 = 15 / 3 A = 5 cm² ( Solved √ )

Area of rectangle Ar = a. b = 15 cm² However, ratio between 'a' and "b' IS NOT GIVEN. This means we can choose this ratio, and even so we are meeting the original conditions. I chose to have a square instead a rectangle Now a = b = √15 Shaded triangle became an isosceles triangle, with side s = a / cos 30° = a . 2/√3 s = √15 .2/√3 = 2√5 cm At = ½ s² sin30° At = 5 cm² ( Solved √ ) Note that Ar = 3 . At Everytime we have this configuration, At =⅓.Ar Consequently, the easiest and shortest solution is: A = A / 3 = 15 / 3 A = 5 cm² ( Solved √ )

So itonic that problems with geometry routinely have figures that are not to scale. Visual learners experience cognitive dissonance.

Hello sir Can yoy tell me why BC = AC x COS 45 Pls reply anybody

Solution by adding the areas of the unshaded triangles and deducting from the area of the rectangle. As in the video, let AB = CD = a and AD = BC = b. Having found that Θ = 30°, ΔABF and ΔBCE can be found to be 30°-60°-90° special right triangles, AF = AB/(√3) = a/(√3), and CE = BC/(√3) = b/(√3). Area ΔABF = (1/2)(AB)(AF) = (1/2)(a)(a/√3) = a²/(2√3). Area ΔBCE = (1/2)(BC)(CE) = (1/2)(b)(b/√3) = b²/(2√3). DE = a - b/√3 and DF = b - a/√3. Area ΔDEF = (1/2)(DE)(DF) = (1/2)(a - b/√3)(b - a/√3) = (1/2)(ab - b²/√3 - a²/√3 +ab/((√3)(√3)) = ab/2 - b²/(2√3) - a²/(2√3) +ab/6. So, sum of areas ΔABF, ΔBCE and ΔDEF = a²/(2√3) + b²/(2√3) + ab/2 - b²/(2√3) - a²/(2√3) +ab/6 = ab/2 + ab/6 = 2ab/3. However, ab = 15 (area of rectangle), so 2ab/3 = (2)(15)/3 = 10. Deduct the sum of areas of unshaded triangles, 10, from the area of the rectangle, 15, leaving area of ΔBEF = 5, as Math Booster also found.

φ = 30°; ∎ABCD → AB = CD = CE + DE = b = 15/a; BC = AF + DF = a; CBE = EBF = FBA = φ → sin⁡(φ) = 1/2 → AF = 5√3/a → BF = 10√3/a; CE = a√3/3 → BE = 2a√3/3 → area ∆ BFE = (1/2)sin⁡(φ)(BE)(BF) = (1/4)(2a√3/3)(10√3/a) = 5

Assign AB = 15/x and BC=x. ABF and BCE are 30-60-90 triangles so respectively BF = 10*sqrt(3)/x and BE=2x*sqrt(3)/3. Now calculate the area directly using Area = 1/2*BF*BE*sin30. Voila! 5. (This method requires that you the generalized formula for area of a triangle = 1/2*side*side*sine of included angle.)

Seus vídeos são ótimos! Parabéns pelo trabalho! 🎉🎉🎉

*Solução Elegante:* No ∆ABF: *AB= cos 30°× BF (1)* No ∆BEC: *BC= cos 30°× BE (2)* multiplicando as equações (1) e (2), obtemos: AB×BC=(cos 30°)^2 × BF×BE Ora, AB×BC=15 e cos 30°=√3/2. Daí, 15= 3/4 × BF×BE, isto é, *BF×BE= 20 (3)* [BEF]= (BF×BE× sin 30°)/2 Substituindo a equação (3), temos: [BEF]= (20 × 1/2)/2 *[BEF]=5.*

Let DC = AB = x and AD = CB = y. The area of the rectangle is thus xy = 15. As ∠CBA = 90° and θ+θ+θ = ∠CBA, then θ = 90°/3 = 30°. cos(θ) = CB/BE √3/2 = y/BE √3BE = 2y BE = 2y/√3 cos(θ) = BA/FB √3/2 = x/FB √3FB = 2x FB = 2x/√3 Triangle ∆EBF: Aᴛ = EB(BF)sin(θ)/2 Aᴛ = (2y/√3)(2x/√3)(1/4) Aᴛ = xy/3 = 15/3 = 5 sq units

Let AB= a BC= b EC = c and AF= d 3@=90° @=30° By soh cah toa 2a/sqrt 3=FB BE = 2b/sqrt 3 Area of red triangle 1/2 absinc 1/2*2a/sqrt 3*2b/sqrt 3*1/2=ab/3=15/3=5

|BE|= 2b oot3 (using angles 30,60 and 90) and |BF|= 2a oot 3(using angles 30,60 and 90). Area of triangle BFE = 1/2 by sine 30 degrees by |BE| by |BF|= 1\4 by 2a/root3 by 2b/ root 3= ab/3=5.

Thank you

If you're gonna get into the weeds with some algebra, you can just chase lengths and sum areas, without the need to draw this extra horizontal line. [ABF]=1/2 x^2/sqrt(3), [BEC]=1/2 y^2/sqrt(3), [FED]=1/2 (x-y/sqrt(3))(y-x/sqrt(3)). Adding those and the desired area up, you get 15, but then any appearances of xy in the expansion are also 15, and the other terms cancel out in the expansion.

BF=x,BE=y AB=xcos30 BC=ycos30 xy(cos30)^2=15 so xy = 20 Shaded area = xysin30/2 = 5

Very nice solution

If I understand this video, the trick is to establish both the relevant chords and the right angles so that the corresponding and congruent angles are established FIRST. After that, then you conclude that the shaded triangle is the summation of TWO triangles bc of the bisected side of the square. Looks like I will have to practice on that problem!!!

You're doing a great work here, man, never stop doing this

No Brasil isto não existe por total falta de conhecimento

If you draw the height H from B, you can compute S = 1/2*H*AC = 64, so H = 8 cm = AC/2. Then, this height is also a mediator, being equal to the radius of the circumscribed circle (the hypotenuse being its diameter). So, the right triangle is also isosceles and θ = 45°. Pure geometric demonstration

Apologize me, I am mistake ,shaded area is 0.464.

2 sqrt3 - 3 = 0.464 .

Shaded area should be 0.478 instead of 0.464.

So,The answer is incorrect.

The area surface of triangle is 12, then height from A is 4,8. If x is a side of the square, then 4,8/5=(4,8-x)/x from similar triangles. x=2,449 and S=5,998

Xcos 4thetha😂

Nice👍

We can solve by using AM-GM inequality.

*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.*

10:55-13:05 Right Triangle Altitude Theorem: CD²=AC•CE => (R-2r)(R+2r)=64 😁

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.

X= 4 sin 54.responsi.

You can also use tan 2Θ and solve for tanΘ then find the ratio of x

*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.*

It is better to let DC be x as area of shaded triangle = (1/2)xDCxsqrt3.

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.

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.

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

Area=2√3-3.❤

❤🎉😊😊😊😊

❤❤

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!

Cok uzun bir çözüm olmuş. Yine de emeğiniz için teşekkürler 🙏

Potencia del punto C respecto a la circunferencia de redio "R" =8² =(R-2r)(R+2r)→ R²=64+4r² → Área sombreada =(πR²/4)-(πr²)=[π(64+4r²)/4]-(πr²) =π(16+r²-r²) =16π. Gracias y un saludo.

We note that length OC is not given, which implies that the solution is the same over a range of values for OC. When presented with this type of problem, I look for special cases. The most obvious special case here is length OC = 0. The small circle's radius becomes 0 and CD and OB become equal in length, so OB = 8. The quarter circle has area (1/4)(π)(8)² = 16π. If this were a multiple choice test, or we were not required to show our work, we could select 16π as the correct answer and move on. However, to check our work, or if the first special case were not apparent, another special case may be considered. Construct OD. As in method #1, we see right triangle ΔOCD. One side, CD, has length 8. We take the familiar Pythagorean triple 3 - 4 - 5 and double each side, so we have a 6 - 8 - 10 right triangle, so assign 8 to CD. OC = 6 and OB = 10. The radius of the quarter circle is OD, or 10, so its area = (1/4)(π)(10)² = 25π. The radius of the circle is OC/2 = 6/2 = 3. Its area is π(3)² = 9π. The difference in area is 25π - 9π = 16π, same answer as for the other special case. The second special case does lead us to the general case solution given in method #1. A more challenging problem: what are the radii for the circle and quarter circle when the circle is tangent to OA, OB and AB? (These should be the maximum radii.)

A = π (8+8) A = 16π cm² ( Solved √ )