Це відео не доступне.
Перепрошуємо.
Can you find area of the Green shaded Triangle? | (Square) |
Вставка
- Опубліковано 10 лип 2024
- Learn how to find the area of the Green shaded Triangle. Important Geometry and Algebra skills are also explained: Congruent Triangles; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find area of t...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you find area of the Green shaded Triangle? | (Square) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindGreenTriangleArea #Square #PythagoreanTheorem #CongruentTriangles #Geometry #SimilarTriangles #IsoscelesTriangles #RightTriangles #Triangle #AreaOfSquare #AreaOfTriangle #CircleTheorem #GeometryMath #EquilateralTriangle #PerpendicularBisectorTheorem
#MathOlympiad #ThalesTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the Radius
Equilateral Triangle
Pythagorean Theorem
Area of a circle
Area of the sector
Right triangles
Radius
Circle
Quarter circle
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.
BF=a, BC=5√2. FC=√(a^2-50). DF=5√2-√(a^2-50). ΔEDF is isosceles. By the Pythagorean theorem a^2=2*(5√2-√( a^2-50))^2. We square both sides of the equation and solve the biquadrate equation. Choose a^2=400-200√3. S(ΔEBF)=0.5*(400-200√3)*√3/2=23.205. Approximate values are always difficult. It is better to round up at the end of the calculations.
@@rabotaakk-nw9nm СПАСИБО. ИСПРАВИЛА.
@@user-sk9oi9jl2g 👍🥰, умница!
Draw a second diagonal line in the square
Let a = side length of equilateral triangle
Use the side proportionalities of 30-60-90 triangles (1:√3:2) and 45-45-90 triangles (1:1:√2)
a(√3/2) + a/2 = 10
Area of equilateral triangle = a²√3/4
Neat!
That's how I did it! ❤
From your 1st equation, we solve for a:
a(√3+1)/2 = 10
a = 20/(√3+1)
Then we square to get a²:
a² = [20/(√3+1)]²
a² = 400/(4+2√3)
And use the formula for area of equilateral triangle:
A = (√3/4)a²
A = (√3/4)[400/(4+2√3)]
A = (100√3)/(4+2√3)
A = (50√3)/(2+√3)
A = (50√3)(2-√3)
A = 100√3 - 150
A ≈ 23.2
We were taught, and I firmly believe, that the congruence you claim at 2:40 is RHS, for right angle, hypotenuse, side. Technically you do not have an angle other than the Right one to claim SAS, but once you know R, H and S the third sides must be the same also, defining congruence. Just saying.
HL? Hypotenuse-Leg Theorem?
Yes, the initial congruence is by HL theorem, not SAS.
Finding the length of the square's side is easy enough. Then you know that triangle EDF is an isosceles right triangle of 45-90-45, meaning that angle AEB is 75 degrees and angle ABE is 15 degrees. (5*2^.5) divided by cos15 = the hypotenuse which is also the length of each side of the equilateral triangle. Then use formula for area of an equilateral triangle to come up with 23.205...
The is a simple and elegant way to proceed: Once you have noticed that BEF is symetric to AC, then you can mark O the center of EF. Let's consider EO = OF = x. By theorem, BO = x sqrt(3). By construction, EDF is rectangle isocele in O; DO = x.
Therefore, AC = x(sqrt(3) +1) = 10 x = 5(sqrt(3) - 1) and area of EBF = x²sqrt(3) = 50(2sqrt(3)-3).
Another approach at the last part is using the standard formula for the area of the equilateral ∆
A= s²√3/4
1/ Let a be the side of the equilateral triangle. Draw the diagonal DB. Notice that DB is perpendicular bisector of EF and the triangle EDF is a right isosceles triangle.
DB intersects EF at point H
We have DB= DH+HB = a.sqrt3/2 + a/2 =10
-> a =20/(sqrt3+1)
Area of the green triangle= sqa. (sqrt3)/4= 50sqrt3.(2-sqrt3) =23.20 sq units😅
yup, did it same way. it is simpler and doesn't require trigonometry!
It is difficult to prove directly DB is perpendicular bisector of EF. It is easy to prove triangle EDF is a right-angled isosceles triangle (ABE congruent to CBF hence AE = CE hence ED = FD). Diagonal DB bisects triangle EDF into 2 congruent triangles (RHS) hence EF is bisected by DB.
DB = BH + HD(not HB)
@@hongningsuen1348
So sorry about my typo mistakes😂.
1/ The 2 right triangles ABE and BAF are congruent--> AE=CF-> DE=DF --> Triangle EDF is a right isosceles-EF//AC -> the diagonal BD is the perpendicular bisector of EF as well😅
@@phungpham1725 2nd thought. It is actually easy to prove directly diagonal BD is perpendicular bisector of triangle BEF. As triangles ABE and CBF are congruent, angles ABE and CBF = (90 - 60)/2 = 15. Perpendicular bisector of equilateral triangle BEF is also its angle bisector, bisecting 60 into 2 x 30. 15 + 30 = 45, the angle made by diagonal.
2 other methods to compute area after obtaining equilateral triangle side length = 7.32:
1. Heron’s Formula, since we know all 3 side lengths.
2. Drop perpendicular from corner B to bisect EF into 7.32/2 = 3.66. Use Pythagorean theorem to obtain height of triangle = 6.34. Base EF = 7.32, with height = 6.34. So, 1/2 * 6.34 * 7.32 = 23.2, as shown above using 1/2 * EF * BF * sin(60).
Superb fun, Prof. thank you!!
The identity of 45° right triangle can be used to get the side of the square, which is simpler than the Pythagorean formula. 10÷sqrt2=7.0711 next use cosine function to find the side of the equilateral triangle 7.0711÷cos15° = 7.32 once the side of the equilateral triangle is known the Area is determined as [(7.32 ^2)×sine60°]÷2 = 23.21 sq units.
1.△AEB:△CFB:△EDF:△BEF
=1:1:2:(2sqrt3) 😊
2.ABCD=10·10/2=50
△AEB=50/(4+2sqrt3)
=50-25sqrt3
3.△BEF=(50-25sqrt3)(2sqrt3)
=100sqrt3-150😊
10/√2=5√2 EB=BF=FE=√2x ED=DF=x AE=FC=5√2-x
(5√2-x)²+(5√2)²=(√2x)²
50-10√2x+x²+50=2x² x²+10√2x-100=0 (x+5√2)²-150=0 x+5√2=5√6 x=5√6-5√2=5√2(√3-1)
Green Triangle area = √3/4*(√2x)² = √3x²/2 = √3*50*(4-2√3)/2 = 25√3*(4-2√3) = 100√3 - 150
Superb!💡
*Side length of the square: 10/sqrt(2).
*Side length of the equilateral triangle: c.
*DF = c/sqrt(2) in right isosceles triangle EDF, then
FC = DC - DF = (10/sqrt(2)) - (c/sqrt(2)) = (10 -c)/sqrt(2)
*In triangle FCB: FC^2 + CB^2 = FB^2, so we have:
((10 -c)^2)/2 + 50 = c^2, or ((c^2)/2) -10.c +50 + 50 = c^2
or ((c^2)/2) + 10.c -100 = 0, or c^2 + 20.c -200 = 0
Deltaprime = 300, then c = -10 -10.sqrt(3), which is rejected as negative, or c = -10 +10.sqrt(3).
*The area of the equilateral triangle is (sqrt(3)/4).(c^2)
so it is ((sqrt(3)/4).100.(4 -2.sqrt(3))
= 25.sqrt(3).(4 -2sqrt(3)) = 100.sqrt(3) - 150.
(which is about 23.2)
[BEF] - x^2 * sqrt (3)/4 where x is the side of the equilateral triangle. EF parallel to AC -> DEF=DFE = 45 -> BEA = 180-45-60=75 -> ABE=15 -> CBF = 90-15-60=15 -> CBF = 75. Now AC=10 -> AB= 5*sqrt (2) = BC=CD=AD. Now x = AB/cos (15) = 10 * [sqrt (3) -1] --> [BEF] = 10^2 * [sqrt (3) -1] ^2 * SQRT (3)/4 = 100* [4-2*sqrt (3)] * sqrt (3)/4 = [400*sqrt (3) - 600]/4 = 100*sqrt (3) - 150 = 173.2 - 150 = 23.2 sq. units
It can also be solved without resorting to looking up the cosine of 15 degrees. Looking at the angles, it's clear that EDF is an isosceles triangle. If EF is x, then ED = DF = x / sqrt(2). We can use Pythagorus to find AE^2 = x^2 - 50. Also AE = 5 * sqrt(2) - ED. Substituting for ED we find that AE = (10 - x) / sqrt(2).
Plug the AE value into the pythagorean formula for ABE: (10-x)^2/2 + 50 = x^2. Rearranging, we get the quadratic equation x^2 + 20x - 200 = 0. Solving that: x = (-20 +/- sqrt(300))/2. To avoid a negative result, we have to choose "+" from the "+/-". Ultimately, x = 10 * (sqrt(3) - 1) or approximately 7.32.
With side length x, the area of an equilateral triangle is x^2 * sqrt(3) / 4. Plugging in x, I get the triangle area is 50 * (2 * sqrt(3) - 3) which is about 23.2.
From 5:02 : Recalling cos(15º)=(√6+√2)/4, then 1/cos(15º)=4/(√6+√2), rationalizing: 1/cos(15º)=4/(√6+√2)=4(√6-√2)/(6-2)=4(√6-√2)/4=√6-√2=√2(√3-1); then EB=5√2(√2)(√3-1)=10(√3-1). Finally Green Area=(√3/4)EB²=(√3/4)100(4-2√3)=25√3(4-2√3)=50√3(2-√3)=50(2√3-3)=23.2050807568...
This is pretty much similar to an older question on this channel, except we now know the diagonal length instead of the area. Nevertheless, it's still easy.
d = s√2
10 = s√2
s = 10/(√2)
= (10√2)/2
= 5√2
By definition of squares, ∠BAD, ∠BCD, & ∠D are right angles and AB = AC.
So, because BE = EF & △BAE & △BCF are right triangles, △BAE ≅ △BCF by HL.
So, AE = CF by CPCTC. Label AE = CF = x.
Then DE = EF = 5√2 - x.
Use the Pythagorean Theorem on △BAE & △EDF.
a² + b² = c²
(5√2)² + x² = c²
50 + x² = c²
(5√2 - x)² + (5√2 - x)² = c²
(50 - 10x√2 + x²) + (50 - 10x√2 + x²) = c²
100 - 20x√2 + 2x² = c²
2x² - 20x√2 + 100 = x² + 50
x² - 20x√2 + 50 = 0
Use the Quadratic Formula.
a = 1, b = -20√2, c = 50
x = [-b ± √(b² - 4ac)]/2a
{-(-20√2) ± √[(-20√2)² - (4 * 1 * 50)]}/(2 * 1)
[20√2 ± √(800 - 200)]/2
(20√2 ± √600)/2
(20√2 ± 10√6)/2
x = (20√2 + 10√6)/2 or x = (20√2 - 10√6)/2
= 10√2 + 5√6 = 10√2 - 5√6
≈ 26.39 ≈ 1.89
s ≈ 7.07
And x represents the length of a segment shorter than the side of square ABCD. So, x = 10√2 - 5√6.
x² + 50 = c²
(10√2 - 5√6)² + 50 = c²
(200 - 200√3 + 150) + 50 = c²
200 - 200√3 + 200 = c²
400 - 200√3 = c²
The hypotenuses are sides of △BEF, which is equilateral, so we can use a different area formula:
A = (√3)/4 * c²
= [√3 * (400 - 200√3)]/4
= (400√3 - 600)/4
= 100√3 - 150
So, the area of the green triangle is 100√3 - 150 square units, a. w. a. 50(2√3 - 3) square units (exact), or about 23.21 square units (approximation). The video doesn't show the exact answer, so it's right here.
*=read as square root
AB=BC=CD=AD =AC/*2=10/*2=5.*2
The sides of the triangle intersects at the points P, Q, respectively to the sides AD, CD,of the square
Now within triangle ABP & BQC
AB=BC
BP=CQ
Angle BAP=Angle BCQ
SO triangle ABP~BQC
So,
Angle ABP=angle CBQ=15(Angle PBQ=60,given )
Now,
Angle APB=75degree
In triangleAPB
Sin75=AB/BP
(*3+1)/2.*2= 5.*2/BP
BP(*3+1)=5.*2.2.*2
=20
BP=20/(*3+1)=10(*3-1)
Area of triangle =*3/4 ×{10(*3_1)}^2 (^=read as square )
=25×*3(*3_1)^2
=25.*3{4-2.*3)
=100.*3-(50.*3.*3)
=(100×1.732)-150
=173.2-150
=23.2(approximately )
DF is (10-h)*sqrt(2)
EF is 20-2h
Triangle area is h(10-h)
M is the midpoint of EF
EMB is a 30,60,90 triangle with sides (10-h), (10-h)*sqrt(3), and 20-2h
Observe that h = (10-h)*sqrt(3)
h^2 = 3(100 - 20h + h^2)
0 = 300 - 60h + 3h^2 - h^2
2h^2 - 60h + 300 = 0
h^2 - 30h + 150 = 0
(30+or-sqrt(900 - 4*1*150))/2 = h
(30+or-sqrt(300))/2 = h
(30+or-2*sqrt(75))/2 = h
(15+or-sqrt(75))= h
Strip out the positive result as it is >10 (the square's diagonal)
h = 15 - sqrt(75)
Area, as already noted, is h(10-h). so 10h - h^2
150 - 10*sqrt(75) - (15-sqrt(75))^2
---> 150 - 10*sqrt(75) - (225 - 30*sqrt(75) + 75)
---> 150 - 10*sqrt(75) - 225 + 30*sqrt(75) - 75
---> -150 + 20*sqrt(75)
20*sqrt(75) - 150
100*sqrt(3) - 150
173.21 - 150
Area = approx 23.21 un^2
I have now checked your video and see that you followed a different path from me. Although my noting that
h = (10-h)*sqrt(3) led to some moderately fiddly calculations, I found that way a bit easier than yours - but I suppose we are all wired a bit differently.
Thanks once again.
Posto AE=a,risulta a^2+(5√2)^2=l^2..arccos(a/l)+60+arccos((5√2-a)/l)=180...dopo i calcoli risulta a=5√2(2-√3)...l^2=50+50(7-4√3)=400-200√3...Agreen=(1/2)l^2sin60=(200-100√3)√3/2=. =100√3-150=50(2√3-3)
Great solution teacher.
I solved a little different.
AB=BC=CD=AD 😢= a
AC = 10
BE=BF=EF = b
∆ABC applying Pythagoras:
AC^2 = AB^2 + BC^2
10^2 = a^2 + a^2
2a^2 = 100
a^2 = 50
a = √(25*2)
a = 5√2
I did ==>AE = CF = x
∆DEF applying Pythagoras
EF^2 = DE^2 +DF^2
b^2 =(5√2-x)^2 + (5√2-x)^2
∆ABE applying Pythagoras
b^2 = x^2 + (5√2)^2
b^2 = x^2 + 50
Equating the two b^2
we get:
x^2 + 50 = (5√2 - x)^2 + (5√2 - x)^2.
x^2 + 50 = 50 - 10√2x + x^2 + 50 - 10√2x + x^2
x^2 - 20√2x + 50 = 0
x = (20√2 +-√(800- 200))/2
x = (20√2 +- 10√6)/2
x1 = 10√2 + 5√6 =26,38958
Rejected It is > than 5√2 = a
x =10√2 - 5√6 = 1,8946869
Accepted✓
∆ABE applying Pythagoras:
b^2 =
(5√2)^2 + (10√2-5√6)^2
b^2 = 50 + 350 - 200√3
b^2 = (400 - 200√3)
Formula for the area of an equilateral triangle.
A = b^2 * (√3)/4
A = (400 - 200√3)*(√3)/4
A = (400√3 - 600)/4
A = 100√3 - 150
∆BEF area = 23,20508 unit^2
x
a better answer for the lenght of the green triangle is 10*sqrt(3) -10, then the area of the triangle is 100*sqrt(3) -150 square units. Have a good day.
The square ABCD has a diagonal of 10 so each side is 10/sqrt(2). Let the triangle BEF have side lengths of "a" and let AE = x then ED = (10/sqrt(2) - x ) and also by symmetry around the diagonal BD we have DF = ED = (10/sqrt(2) - x ). Applying Pythagoras theorem to triangle ABE and then to triangle EDF we get:
a^2 = x^2 + (10/sqrt(2))^2 and a^2 = (10/sqrt(2) -x)^2 +(10/sqrt(2) -x)^2.
We can solve these equations to get x = 1.895 and a^2 = 53.59. We can find the area of the equilateral triangle BEF using the formula (sqrt(3)/4)*a^2 to be 23.20 square units.
23.21
let the side = a
then a^2 + a^2 =100
2a^2=100
a^2 =50
a= sqrt 50
Since the green triangle is equilateral, then the triangle top is a 75, 15, 90 right triangle ( 90 -60=30/2 =15)
To find the length of green, using the law of sine
Let the length of the green = p,
then p = sqrt 50 * sine 90 degrees/ sine 75 degrees
p = sqrt 50*1.035276
p=7.32050807569
So the length of the equilateral is 7.32051
Hence, its area = 7.32051^2 * sqrt 3/4
= 23.21
The exact value of the area is (25/2)√3 sec²(15°)
Let s be the side length of square ABCD and t be the side length of equilateral triangle ∆EBF.
Triangle ∆ABC:
AB² + BC² = CA²
s² + s² = 10² = 100
s² = 100/2 = 50
s = √50 = 5√2
As EB = BF and AB = BC and ∠EAB = ∠BCF = 90°, then ∆EAB and ∆BCF are congruent right triangles. Let ∠ABE = ∠FBC = θ. As ∠EBF = 60° (being the internal angle of an equilateral triangle) and ∠ABC = 90° (being the internal angle of a square), then:
θ + 60° + θ = 90°
2θ = 90° - 60° = 30°
θ = 15°
cos(15°) = cos(60°-45°)
cos(15°) = cos(60°)cos(45°) + sin(60°)sin(45°)
AB/BE = (1/2)(1/√2) + (√3/2)(1/√2)
s/t = (√3+1)/2√2
t = 2√2s/(√3+1)
t = 2√2s(√3-1)/(√3+1)(√3-1)
t = 2s(√6-√2)/(3-1)
t = 2s(√6-√2)/2 = s(√6-√2)
t = 5√2(√6-√2)
t = 5√2(√2(√3-1)) = 10(√3-1)
Triangle ∆EBF:
Aₜ = t²sin(60°)/2
Aₜ = (10(√3-1))²(√3/4)
Aₜ = (100√3/4)(3-2√3+1)
Aₜ = 25√3(4-2√3) = 25(4√3-6)
Aₜ = 50(2√3-3) ≈ 23.205 units
I got there without using trig, but a whole Lotta algebra:
I call the bisected square side segments "X" and "5Sqrt(2)-X", and Triangle sides "S". This sets up 2 Pythagorean Theorem eqns:
X²+[5•SQRT(2)]²=S²
2•(5•SQRT(2)-X)²=S²
Set equal to each other, solve for X, X=1.89.
Plug back into eqn 1:
(1.89)²+(5•SQRT(2))²=S²
S=7.31
Since triangle is isosceles:
Area=1/2•S•(S/2)SQRT(3)
=S²/4•SQRT(3)
=(7.31)²/4•SQRT(3)
=23.14
Using triangles EDF and BCF, i worked out the side of the green triangle to be Sqrt[200 . (2 - Sqrt[3])] .... (or 7.32)
height of equilateral triangle is easy using pythagorus on half a base and the side.
area works out to be 50 * (2 * sqrt (3) - 3)
You've used SAS to prove triangles ABE & CBF as congruent. Then use congruent triangles to get angles ABE & CBF congruent. These are the included angles in SAS. You can't say the angles are congruent to prove the triangles congruent and then deduce the angles are congruent. Using hypotenuse-leg (HL) resolves this issue.
To be fair, hypotenuse-leg is basically SAS by a different name. The fact that ∆ABE and ∆CBF are both right triangles tells us that the cosine (and thus angle) of ∠ABE and ∠CBF are the same, as we know they're both equal to square side length (leg) over triangle side length (hypotenuse). It's why hypotenuse-leg works (and why it only works on right triangles).
Sir if we agree to ur we may also say it is a case of SSS/SAS congruency.
As using Pythagorean theorem we may say AE = CF etc.
But when we say of SAS congruency we must have to show the angle between the sides are equal. Here it was not done.
@@PrithwirajSen-nj6qq Honestly , I think it s a sss congruency. No need to calculate the angle.
The 2 triangles are right angled ones and they have 2 long legs (= side of the square) and 2 hypothenuses (= side of the equilateral) so, the short legs must be equal. (Using Pythagorean theorem as proof)
Anytime the world looks unfamiliar one can rely on the rigid constructed rules of math that gives potential too solving problems such as this. 🙂
Once you find out that the triangle on the top is 75, 90, 15, after subtracting 60 from 90, then dividing
30 by 2, then the most challenging part is over.
[ABCD]=AB²=½AC² => AB²=100/2=50
BE=AB/cos15°; BE²=AB²/cos²15°=
=AB²/(1+cos30°)/2=50/(1+vʼ3/2)/2=
=200/(2+vʼ3)=200(2-vʼ3)
[BEF]=(vʼ3/4)BE²=(vʼ3/4)200(2-vʼ3)=
=50(2vʼ3-3)≈23.205 😁
ABCD is square
Let side length=a
So AB=BC=CD=DA
AB^2+BC^2=AC^2
a^2+a^2=10^2
So 2a^2=100
So a=5√2
∆ABE ~~∆BCF
So AE=CF=x
So DE=DF=5√2-x
BR=BF=EF=b
In ∆DEF
DE^2+DF^2=EF^2
2(5√2-x)^2=b^2
100-20√2x+x^2=b^2 (1)
In ABE
x^2+(5√2)^2=b^2
x^2+50=b^2 (2)
(1) and (2)
100-20√2x+x^2=x^2+50
So x=1.89
(2) (1.89)^2+50=b^2
So b=7.32
So Green triangle area=1/2(7.32)^2sin(60°)=23.2 square units.❤❤❤
Sir u said triangles ABE and BCF are congruent over SAS theory.
But their two sides are equal but the included angles are not proved equal.
So SAS is not applicable.
We may say these triangles are right angled triangle.
Hence they will be congruent if the hypotenuse equal and any side is equal .
Here it is seen that the hypotenuses are equal and one side of each them is equal. Hence the triangles are congruent over RHS congruent theory
Then angle ABE = angle BCF may be said.
Comment please
S=50(2√3-3)≈23,2
Let's find the area:
.
..
...
....
.....
From the given length of the diagonal we can directly calculate the side length s and the area A of the square:
s = AC/√2
A = s² = AC²/2 = 10²/2 = 50
The triangles ABE and BCF are both right triangles. Additionally we have AB=BC=s and BE=BF=t (with t being the side length of the equilateral triangle). Therefore these two triangles are congruent and we can conclude:
AE = CF
AD − DE = CD − DF
s − DE = s − DF
⇒ DE = DF
Therefore the triangle DEF is an isosceles right triangle. By applying the Pythagorean theorem to the right triangles DEF and ABE we obtain:
EF² = DE² + DF²
t² = 2*DE²
t²/2 = DE²
⇒ DE = t/√2
⇒ AE = AD − DE = s − t/√2
BE² = AB² + AE²
t² = s² + (s − t/√2)²
t² = s² + s² − √2*s*t + t²/2
t²/2 + √2*s*t − 2*s² = 0
t² + (2√2)*s*t − 4*s² = 0
t = −√2*s ± √(2*s² + 4*s²) = −√2*s ± √(6*s²) = −√2*s ± √6*s
Since t>0, the only useful solution is t=(√6−√2)*s. Now we are able to calculate the area of the green equilateral triangle:
A(BEF)
= (√3/4)*t²
= (√3/4)*(√6 − √2)²*s²
= (√3/4)*(6 − 2√12 + 2)*50
= (√3/4)*(8 − 4√3)*50
= √3*(2 − √3)*50
= 100√3 − 150
≈ 23.21
Best regards from Germany
nice
@@devondevon4366 Thanks a lot for your kind feedback and best regards from Germany.
we cant say ABE & CBF are congruent using SAS .
Thank you!
50*(2*sqrt(3)-3)
Exact result for cos 15° is (1 + sqrt(3))/(2 sqrt(2))
Yes
The 15°-75°-90° right triangle appears so frequently in geometry problems that it is worthwhile for all of us to know its properties. The ratio of sides (short):(long):(hypotenuse) is (1 - √3):(1 + √3):(2√2), from which cos(15°) can be readily derived to be (1 + √3)/(2√2), as stvcia points out.
@@jimlocke9320 exactly
Has any one Solved this without using trigonometry ?
Reluctant to do for trigonometric means cannot yield simple answeras usual, but I only can work in this way. First s^2=100/2=50, the answer is s^2-s^2 tan 15°-s^2(1-tan 15°)^2=50(1-tan 15-1-tan^2 15°+2tan 15°)=50(1+tan 15°+tan^2 15)😢😢😢😢😢😢😢😢.
I calculated the area as (25/2)√3 sec²(15°)
@JamesDavy2009 very nice 😁
Why angles 15 deg r same?🧐
First, you can pin it?