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".
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.
An easier way to calculate would be: Let x be the equal sides. Drop a vertical line from the top intersection of slanted line and horizontal line. Length of this line is x. If y is the slanted line, then y=2x (as sin(30°)=0.5) (5-x^2)+x^2=y^2 (Pythagoras theorem) Or (5-x^2)+x^2=4x^2 x^2+5x=25/2 Area = x^2 + ½x.(5-x) = (x^2 + 5x)/2 =(25/2)/2=25/4
Right side length is x. sin(30°)=1/2 cos(30°)=sqrt(3)/2 Height is x/2. S = (x/2)^2 + (x/2)*x*sqrt(3)/2/2 S = x^2/4*(1+sqrt(3)/2) 5 = x/2 + x*sqrt(3)/2 x = 10/(1+sqrt(3)) S = 100/4/(1+sqrt(3))^2*(1+sqrt(3)/2) S = 25*(1+sqrt(3)/2)/(4+2*sqrt(3)) S = 25/4*(1+sqrt(3)/2)/(1+sqrt(3)/2) S = 25/4 = 6.25 That's all. I solved before listening.
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, the area of the trapezoid, is the height times the average width. That is, add the top and bottom segments, divide by 2, multiply by height. You use x as the common segment length that matches the height and the top and left segments. Thus, A = x*(x+5)/2 . There's no need to find the area of each part (square and triangle) to add later. There's no need to find all the segment lengths, as only x is needed. To find x, use the given angle, trigonometry, and the labelled sides of the triangle. tan(30⁰) = x/(5-x) = 1/√3 √3*x = 5-x (√3+1)*x = 5 x = 5/(√3+1) = 5*(√3-1)/2 A = [5*(√3-1)/2]*[5*(√3-1)/2 + 5]/2 = (5/2)*[√3-1]*(5/2)*[√3-1 + 2]/2 = (25/8)*[√3-1]*[√3+1] = (25/8)*2 = 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
You cannot write 6.25 because you use the "." to multiply. That means you're saying the answer is 150. Consider using "*" for multiply, and leave "." as the radix mark. Then, use "√" or "^.5" to indicate the principal square root.
6.25 is the correct answer but I solved it in a simpler way using the tangent function instead of building squares. I constructed a big rectangle triangle by prolonging the opposite side and the hypotenuse, and then found the area of the small triangle that will be reduced from the area of the big triangle. Very simple.
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
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.
There's no need to go through the trouble of finding the hypotenuse, by dealing with the squares and square root for the Pythagorean Theorem. Just use the tangent function. tan(30⁰) = 1/√3 = x/(5-x) 5-x = √3*x 5 = (√3+1)*x x = 5/(√3+1) = 5*(√3-1)/2 The area of the trapezoid is x*(x+5)/2 . Plug in x and simplify to get 25/4 .
FYI, your formula for the hypotenuse is wrong. NB: Try to use standard symbols and characters in your work. Choose the ones accepted by most software, including programming languages and math tools. For multiplication, use "*". Avoid implied multiplication, as variable names can be longer than 1 character. For example, "GET" may mean "General Excise Tax" and not "G*E*T". For square root, be consistent and avoid mixing "root" and "sqrt". It's fine to use prefix "√" or infix "^.5" or "**(1/2)" , but add parentheses when precedence rules require them: x = (-b ± √(b² - 4*a*c))/(2*a) -1 = e^(i*π*(2*k+1)) , i = √-1 , k ∈ ℤ Also, use the "⁰" symbol for angles in degrees. You can use Copy & Paste to save the useful symbols in a text file, which you can later reference in forming your new comments. You can get the symbols from here, or even do web searches for math fonts.
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.
@@oahuhawaii2141 Of Course when you are in Math Olympiads time is valuable.Also many times Trigonometry is the short way or the only way. But i'm Geometry lover and i prefer Geometrical Methods.
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
This violates precedence rules; it's bad form: s = 5(√3-1)/(√3+1)(√3-1) It should be fixed as either of the next 2 lines: s = 5*(√3-1)/(√3+1)/(√3-1) s = 5*(√3-1)/((√3+1)*(√3-1)) Avoid implied multiplication.
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
You can use the shortcut for the area of the trapezoid, rather than finding the areas of the square and triangle, and summing them. This is something you should know.
@michaeldoerr5810: The area of any trapezoid is its height times the average of its widths. That is, add the top and bottom segments, divide by 2, and multiply by the height. It's easy to prove for a general trapezoid. In this case, the left side is vertical, and not sloping, so it's an even simpler case to prove. Duplicate the shape, flip it upside-down, put it on the right side of the original, and slide it over so both slopes meet. You've got a rectangle of height x and width (x+5). Its area is twice the area of the original shape, making A = x*(x+5)/2 .
Thank you very much. We know that ∆BCD is a right triangle. Angle C = 30° ABED is a square. tan30° = 1 / √3 → BE / EC = x / (5 - x) = 1 / √3 → x = 5 / (√3 + 1)
There's no need for approximate values, when the expression can evaluated symbolically to resolve into a simple fraction. You don't need logarithms to compute 25/4 .
Angles listed without "°" is in radians, not degrees. 30⁰ = π/6 radians ≠ 30 radians Use "=" for equality, and "≈" for approximations: x = 5/(√3+1) = 5*(√3-1)/2 x ≈ 1.8301270...
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You've got the order of the ratios wrong. Angles 30⁰ : 60⁰ : 90⁰ are related to the sides with relative lengths of 1 : √3 : 2 because the sine of those angles are 1/2 : √3/2 : 1 and the triangle can be scaled by a value, such as 2 . For the trapezoid here, the scale is 2*x , so the sides are x : √3*x : 2*x .
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".
A=(5+a)a/2..(...a=(5-a)tg30...,a=5(√3-1)/2...)...A=25/4
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.
An easier way to calculate would be:
Let x be the equal sides. Drop a vertical line from the top intersection of slanted line and horizontal line. Length of this line is x.
If y is the slanted line, then y=2x (as sin(30°)=0.5)
(5-x^2)+x^2=y^2 (Pythagoras theorem)
Or
(5-x^2)+x^2=4x^2
x^2+5x=25/2
Area = x^2 + ½x.(5-x)
= (x^2 + 5x)/2
=(25/2)/2=25/4
Right side length is x.
sin(30°)=1/2
cos(30°)=sqrt(3)/2
Height is x/2.
S = (x/2)^2 + (x/2)*x*sqrt(3)/2/2
S = x^2/4*(1+sqrt(3)/2)
5 = x/2 + x*sqrt(3)/2
x = 10/(1+sqrt(3))
S = 100/4/(1+sqrt(3))^2*(1+sqrt(3)/2)
S = 25*(1+sqrt(3)/2)/(4+2*sqrt(3))
S = 25/4*(1+sqrt(3)/2)/(1+sqrt(3)/2)
S = 25/4 = 6.25
That's all. I solved before listening.
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
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, the area of the trapezoid, is the height times the average width. That is, add the top and bottom segments, divide by 2, multiply by height. You use x as the common segment length that matches the height and the top and left segments. Thus, A = x*(x+5)/2 . There's no need to find the area of each part (square and triangle) to add later. There's no need to find all the segment lengths, as only x is needed.
To find x, use the given angle, trigonometry, and the labelled sides of the triangle.
tan(30⁰) = x/(5-x) = 1/√3
√3*x = 5-x
(√3+1)*x = 5
x = 5/(√3+1) = 5*(√3-1)/2
A = [5*(√3-1)/2]*[5*(√3-1)/2 + 5]/2
= (5/2)*[√3-1]*(5/2)*[√3-1 + 2]/2
= (25/8)*[√3-1]*[√3+1]
= (25/8)*2
= 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
You cannot write 6.25 because you use the "." to multiply. That means you're saying the answer is 150. Consider using "*" for multiply, and leave "." as the radix mark. Then, use "√" or "^.5" to indicate the principal square root.
You also have "+.5" , but intended "+5" .
6.25 is the correct answer but I solved it in a simpler way using the tangent function instead of building squares.
I constructed a big rectangle triangle by prolonging the opposite side and the hypotenuse, and then found the area of the small triangle that will be reduced from the area of the big triangle. Very simple.
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
You have a typo:
(n+5)n/2 = n^2+ 5n)/2
@@oahuhawaii2141 It appears I have
(n^2 + 5n)/2 without the left bracket. Thank you
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.
There's no need to go through the trouble of finding the hypotenuse, by dealing with the squares and square root for the Pythagorean Theorem. Just use the tangent function.
tan(30⁰) = 1/√3 = x/(5-x)
5-x = √3*x
5 = (√3+1)*x
x = 5/(√3+1) = 5*(√3-1)/2
The area of the trapezoid is x*(x+5)/2 . Plug in x and simplify to get 25/4 .
FYI, your formula for the hypotenuse is wrong.
NB: Try to use standard symbols and characters in your work. Choose the ones accepted by most software, including programming languages and math tools. For multiplication, use "*". Avoid implied multiplication, as variable names can be longer than 1 character. For example, "GET" may mean "General Excise Tax" and not "G*E*T". For square root, be consistent and avoid mixing "root" and "sqrt". It's fine to use prefix "√" or infix "^.5" or "**(1/2)" , but add parentheses when precedence rules require them:
x = (-b ± √(b² - 4*a*c))/(2*a)
-1 = e^(i*π*(2*k+1)) , i = √-1 , k ∈ ℤ
Also, use the "⁰" symbol for angles in degrees. You can use Copy & Paste to save the useful symbols in a text file, which you can later reference in forming your new comments. You can get the symbols from here, or even do web searches for math fonts.
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.
Skip all the messy work for squares and square root of the Pythagorean Theorem. Use tan(30⁰) = 1/√3 = opposite/adjacent .
@@oahuhawaii2141 Of Course when you are in Math Olympiads time is valuable.Also many times Trigonometry is the short way or the only way. But i'm Geometry lover and i prefer Geometrical Methods.
shortest path [(5+x)x]/2: trapezoid area
The simpler way to solve for X is to in 30/60/90 triangle, base leg is X√3. So X + X√3 = 5.
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
This violates precedence rules; it's bad form:
s = 5(√3-1)/(√3+1)(√3-1)
It should be fixed as either of the next 2 lines:
s = 5*(√3-1)/(√3+1)/(√3-1)
s = 5*(√3-1)/((√3+1)*(√3-1))
Avoid implied multiplication.
1/3^0.5=x/5-x=>5-x=3^0.5x=>x=5/3^0.5 +1=5×0.732=3.66.area=4.33×3.66sq unit.
If you know BE:EC:BE=1:2:√3 and trapezoid area formula, you can totally solve the quiz in mind.
Если вы не знаете что
S(ABDC) =(AB+DC)*BE/2
Лучше не надо решать если вы не знаете простых действий
(5)^2=25 180°ABCD/25 =7.5 (ABCD ➖ 7ABCD+5).
OK! I have the same answer, only the earlier one has been reduced.
We can use a calculator to save some work with square roots
Yes, but that's only approximate values. The final answer has no square root, and is in fact a fraction, 25/4 .
(AB+CD)/2× perpendicularAD=area
X=5/2(√3-1),y=5√3/2(√3-1),
Area=x^2+xy/2
But you still haven't found the area.
Area= x*(5+x)/2，that would be easier to calculate.
[5/(1 + ✓3) + 5] [5/(1 + ✓3)] ÷ 2 = 25/(4 + 2✓3) + 25/(1 + ✓3) ÷ 2 = (50 + 25✓3)/(4 + 2✓3) ÷ 2 = 25/4
Area=1/2(5√3-5)/2+5)(5√3-5)=25/4.
cảm ơn : Madame Teaches đã cho tôi biết cách giải bài toán quá sức đối với tôi!!!
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
You can use the shortcut for the area of the trapezoid, rather than finding the areas of the square and triangle, and summing them. This is something you should know.
@@oahuhawaii2141 I did not know. I guess I stand corrected.
@michaeldoerr5810: The area of any trapezoid is its height times the average of its widths. That is, add the top and bottom segments, divide by 2, and multiply by the height. It's easy to prove for a general trapezoid. In this case, the left side is vertical, and not sloping, so it's an even simpler case to prove. Duplicate the shape, flip it upside-down, put it on the right side of the original, and slide it over so both slopes meet. You've got a rectangle of height x and width (x+5). Its area is twice the area of the original shape, making A = x*(x+5)/2 .
@@oahuhawaii2141 I thought Math Booster tried a more subtle method than that. That was why I sounded obtuse.
1.5×2=3)(3×1.5=4.5÷2=2.25+3=5.25
X = 5/(1+3^1/2)
= 25 / (4 + 2*3^1/2)
= 6.25*(4-2*3^1/2)
Nope, 3^1/2 = (3^1)/2 = 3/2 = 1.5 .
You should use 3^(1/2) or 3^.5 or √3 or sqrt(3).
Why did you write the following form?
X = k
= k²
If X = k = k² , then X = k = 0, 1 .
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)
You used the wrong letter or order:
ABE is the right triangle.
ABED is the square.
"tan30⁰C" doesn't make sense mathematically.
We can state ∠C = 30⁰ and write tan(30⁰) = 1/√3 .
tan(θ) uses θ that's an angle; 30⁰C is a temperature.
Thank you very much.
We know that ∆BCD is a right triangle. Angle C = 30°
ABED is a square.
tan30° = 1 / √3 → BE / EC = x / (5 - x) = 1 / √3 → x = 5 / (√3 + 1)
@@clementchiu1315: You may be able to edit your original comment.
Why not use logarithmic table for easy calculation
There's no need for approximate values, when the expression can evaluated symbolically to resolve into a simple fraction. You don't need logarithms to compute 25/4 .
=5(x/(5-x)=1/3^1/2=>3^1/2 x=5-x=>5/(1+3^1/2).ar=(5+×)x/2.
Se resuelve de un modo más fácil y rápido con trigonometría.
BC * sin 30 = x BC * cos 30 = 5 - x >> tan 30 = x/(5-x) x = 1.829 ...
Angles listed without "°" is in radians, not degrees.
30⁰ = π/6 radians ≠ 30 radians
Use "=" for equality, and "≈" for approximations:
x = 5/(√3+1) = 5*(√3-1)/2
x ≈ 1.8301270...
(25√3 + 25) / 4
Divide by (√3+1).
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5÷3=1.66666×1.666666×2=5.55555
No entiendo que dice: pero tengo un terreno así está para sacar el metro cuadrado no se cómo se hacer
30/60/90. 1, 2, sq.rt. 3.
You've got the order of the ratios wrong.
Angles 30⁰ : 60⁰ : 90⁰ are related to the sides with relative lengths of 1 : √3 : 2 because the sine of those angles are 1/2 : √3/2 : 1 and the triangle can be scaled by a value, such as 2 . For the trapezoid here, the scale is 2*x , so the sides are x : √3*x : 2*x .
There was an error in the calculation.
I thought so, too. But I thought certain color text I can't see well explained it.
Final result: 750/98
You lost me with purple writing. I don't see purple very well.
65
3.33333......
13.63
6.25
No ví cuál es el resultado o no dió resultados
Area = 25/4 = 6.25 .
This can not be an olympiad question.Not very hard.simple question
Itiseasy to compute the are a of the trape ziod parallel side s
TAN= O/A....TAN 30= x÷5
tan(30⁰) = 1/√3 = x/(5-x)
You are taking too long steps .
Cut it short
Very easy , typical , I am not impressed
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
You're off by 50%. You have an extra factor of 3/2 in your calculations.