Math Olympiad | A Very Nice Geometry Problem
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I solved like this:
From point B we draw a line parallel to line AB until it reaches line DC and mark point E. Square ABDE is formed on the left and triangle BCE on the right.
The question ask the Trapezoid ABCD area = ?
AB = AD = DE = BE = a
DC = 5
∆BCD is a right triangle.
Angle C = 30°
Angle E = 90°
Angle B = 60°
BE = a
CE = 5 - a
BC = 2a (in right triangle: hypotenuse is twice the shorter side)
Applying Pythagoras:
(2a)^2 = a^2 + (5 - a)^2
4a^2 = a^2 + 25 - 10a + a^2
2a^2 + 10a - 25 = 0
a =(-10+-√(100 - 4*2*(-25))/4
a = (-10+-√300)/4
a = (-10 +-10√3)/4
a1 = (-10 - 10√3)/4 is negative number => rejected
The other
a = (-10 + 10√3)/4
a = 10(√3 - 1)/4
a = 5(√3 - 1)/2
a^2 = [5(√3 - 1)/2]^2
a^2 = (25(2 - √3))/2
5a = 5*5(√3 -1)/2
5a = 25(√3 - 1)/2
Trapezoid area = (1/2)*(B+b)*h =
(1/2)* (5+a)*a
(1/2)* (a^2 + 5a) = (1/2)*(25(2-√3)/2) + (25(√3-1)/2) =
Área = 25*(2 - √3 + √3 - 1)/4
Área = 25*(1)/4
Area Trapezoid = 25/4
Area Trapezoid = 6,25 unit^2
A=(5+a)a/2..(...a=(5-a)tg30...,a=5(√3-1)/2...)...A=25/4
Draw perpendicular from the vertex B to DC, Which will intersect DC at E
Let AD=AB =BE =X
So, CE =5-X
In triangle BEC
tan30=BE/EC=1/*3(*=read as root over )
1/*3=x/(5-x)
*3.x=5-x
*3.x+x=5
X(*3+1)=5
X=5/(*3+1)
The area of traizium =
1/2. (h). (AB+CD)
=1/2.{(5/(*3+1)}.{5+.5/(*3+1)}
=1/2.{5.(*3-1)/2}.{10+5(*3-1)}/2
=1/8.{5.(*3-1)}.{5(2+*3-1)}
=1/8.25.{(*3+1)(*3-1)}
=(1/8).25.2
=25/4=6.25
φ = 30° → sin(3φ) = 1; ∎ADEB → AB = BE = DE = AD = a
∆ BEC → CE = 5 - a; BE = a
ECB = φ → sin(φ) = 1/2 → cos(φ) = √3/2 → tan(φ) = sin(φ)/cos(φ) = √3/3 = a/(5 - a) →
a = (5/2)(√3 - 1) → area ABCD = a^2 + (a/2)(5 - a) = 25/4
Equal sides =x , making square of area x.x , base of triangle remaining= 5-x , then the hypotenuse = cos(30) /(5-x) and x will be equal to half hypotenuse because sin 30 degrees = 1/2
So there is an equation in x
x = cos 30 /(5-x) x= root(3) /(2/2)(5-x)) , because cos 30 = (root 3)/2 squaring both sides x.x = 3/((5-x)(5-x))
x.x( (25-10x+x.x) =3 x.x.x.x-10x.x.x.+25x.x -3 =0 is not nice.
Trying extending by reflecting in the line which is 5 across:
Now twice the area is a rectangle of 2x.x and an equilateral triangle. If you do not like where your calculations lead then check for error or try a dfferent approach.
We use the symmetry to deduce the equilateral triangle . The length of each side is BC, which will be 2x
That area will be x times sqrt(3) We get back by halving back to ABCD and I am just at 7 minutes into the lesson. so I am replaying the seventh , eighth and ninth minutes, to get to
x = (5)/ ( root(3)+1) and understand the solution.
Admission that the lesson is very necessary. Thank you.
25/4 or 6.25 Answer
another approach
Let the side of the trapezoid = n, then the height also =n
Hence, the area of the trapezoid in terms of n is (n+5)n/2 = n^2+ 5n)/2
Draw a perpendicular line to form a square and a 30-60- 90 right triangle,
then n + sqrt 3 n = 5 since 60 degrees corresponds to sqrt 3 n
sqrt 3n = 5-n
3n^2 = 25 + n^2 -10n
2n^2 + 10n = 25
n^2 + 5n = 12.5
Let's make the above equation similar to the area of the trapezoid in terms of n, which is (n^2 + 5n)/2 (see above)
by dividing both sides by 2
Hence3, n^2 + 5n =12.5 becomes
(n^2 + 5n)/2 = 12.5/2
6.25 Answer
EからBCに下ろした垂線の足をFとする。
BE=a →BF=a/2
EF=b →EC=2b
△BEC = 4×△BEF
△BEFを□ADEBの周りに1つずつ付けると、1つの大きな正方形になり、その1辺の長さは
a/2 +b
すなわち、求める面積S=(a/2+b)^2
ここでDC=a,EC=2b なので a+2b=5 → a/2+b=5/2
よって S=(5/2)^2=25/4
I enjoy this way of seeing the situation. I suppose BF is 1/3 of CF , to provide material for 4. times BEF to equal BEC and reshape to build the large square.
But it is raw in my mind.
It is easier to complete the trapezoid to the triangle with angles "30-6--90", and then consider the similar triangles. From this it is easy to find the vertical side of the triangle and then the "x".
Let DA = AB = s. Drop a perpendicular from B to E on CD. As ∠ADE = ∠BAD = 90°, and by construction so do ∠DEB and ∠EBA, and as DA = AB = s, then DE and EB also equal s and ADEB is a square.
As ∠ECB = 30° and ∆BEC is a right triangle, then BE/EC = tan(30°). Let EC = x.
BE/EC = tan(30°)
s/x = 1/√3
x = √3s
DC = DE + EC
5 = s + x = s + √3s
s = 5/(√3+1)
s = 5(√3-1)/(√3+1)(√3-1)
s = 5(√3-1)/(3-1) = 5(√3-1)/2
Trapezoid ABCD:
Aᴛ = h(a+b)/2
Aᴛ = s(s+5)/2
Aᴛ = (5(√3-1)/2)(5(√3-1)/2+5)/2
Aᴛ = (5(√3-1)/2)((5√3-5+10)/2)/2
Aᴛ = (5(√3-1)(5√3+5)/4)/2
Aᴛ = 25(√3-1)(√3+1)/8
Aᴛ = 25(3-1)/8 = 50/8 = 25/4 sq units
The simpler way to solve for X is to in 30/60/90 triangle, base leg is X√3. So X + X√3 = 5.
BE=1/2 of BC hypotinuse so BC is double of opposite perpendicular side so BC=2x. Then PYthagorean Theorem is giving (2x)^2= x^2+(5-x)^2 so 4x^2=x^2+5^2+x^2-2*5*x so 2x^2+10x-25=0 10x=0 so x= (-10+ -sqrt300)/4= (-10+ -10*sqrt3)/4 = (-5+ -5*sqrt3)/2 x=(-5-5*sqrt3)/2 < 0 REJECTED or x= (5*sqrt3-5)/2 >0 ACCEPTED. Finally we have the Trapezioum(Trapezoid) ABCD with Big base B.B.=DC=5., Small Base b=AB=x=5*(sqrt3-1)/2 and Height h=AD=x= 5*(sqrt3-1)/2 The type of Area of Trapezoids is (1/2)*(B.B.+b)]*h = (1/2)*[(5+5*(sqrt3-1)/2]* [5*(sqrt3-1)/2] = (1/2)*5*(sqrt3+1)/2* [5*(sqrt3-1)/2] =( 25/8)*[(sqrt3)^2-1^2]= (25/8)*(3-1)=(25/8)*2=25/4=6.25 square units.
(AB+CD)/2× perpendicularAD=area
We know that ∆BCD is a right triangle.
Angle C = 30°
ABDE is a square.
tan30℃ = 1 / √3 → BE / EC = x / (5 - x) = 1 / √3 → x = 5 / (√3 + 1)
The answer is 25/4. And also I think that you should have entitled this video as another, "You should be able to do this". And I will practice this AGAIN because this involves one of the easiest geometric construction: the 30-60-90 triangle. I also guessed *almost* every step that you have shown. And yet I feel like an idiot. I shall connect that to other problems on your channel ans other channels!!!
Be grateful for the videos posted on this channel! Stop criticizing! If you want difficult questions, download the tests from the International Mathematical Olympiads and see if you can understand at least one statement
X = 5/(1+3^1/2)
= 25 / (4 + 2*3^1/2)
= 6.25*(4-2*3^1/2)
X=5/2(√3-1),y=5√3/2(√3-1),
Area=x^2+xy/2
We can use a calculator to save some work with square roots
(5)^2=25 180°ABCD/25 =7.5 (ABCD ➖ 7ABCD+5).
Area= x*(5+x)/2，that would be easier to calculate.
shortest path [(5+x)x]/2: trapezoid area
=5(x/(5-x)=1/3^1/2=>3^1/2 x=5-x=>5/(1+3^1/2).ar=(5+×)x/2.
cảm ơn : Madame Teaches đã cho tôi biết cách giải bài toán quá sức đối với tôi!!!
Area=1/2(5√3-5)/2+5)(5√3-5)=25/4.
OK! I have the same answer, only the earlier one has been reduced.
BC * sin 30 = x BC * cos 30 = 5 - x >> tan 30 = x/(5-x) x = 1.829 ...
30/60/90. 1, 2, sq.rt. 3.
(25√3 + 25) / 4
No entiendo que dice: pero tengo un terreno así está para sacar el metro cuadrado no se cómo se hacer
13.63
No ví cuál es el resultado o no dió resultados
Itiseasy to compute the are a of the trape ziod parallel side s
There was an error in the calculation.
I thought so, too. But I thought certain color text I can't see well explained it.
You lost me with purple writing. I don't see purple very well.
NÃO ACREDITO!!!!! O CARA FEZ DA MANEIRA MAIS DIFÍCIL DO MUNDO!!!!!! EU RESOLVI EM TRINTA SEGUNDOS PELO LADO DO QUADRADO!!!!!!!!!!!!!!!!!!!!!!. OLHE QUE É UMA OLIMPÍADA!!!!!!!!!!!!!!!!!!!!
TAN= O/A....TAN 30= x÷5
x/(5-x)=1/√3 2x^2+10x-25=0 x^2+5x=25/2 Area=x(5+x)/2=25/4----over
Answer is not 6.25 your calculation is totally wrong the answer is 9.37