Perpendicular height of the green area is (3*sqrt(3))/2, due to the 30,60,90 on the left. The width of its base is 1/3 of the distance between 4 and 13, so the green base is 7. (7 + 4)/2 * (3*sqrt(3))/2 (11/2) * 3*sqrt(3)/2 (33/2) * sqrt(3)/2 (33*sqrt(3))/4 un^2 Decimal approximates to 14.29 un^2 I have now checked the video. There were a few ways to tackle this one. Thank you.
CD = EF + (EC/EA).(AB-EF) = 4 +(3/9).9 = 7. The green area is then ((EF+CD)/2).(EC.sin(60°)) = ((4+7)/2).(3.(sqrt(3)/2) = (33/4).sqrt(3). (Very quick!)
Slightly different method, same answer : Complete external left triangle. From memory, sin(30)= 0.5 External triangle width is 9 x 0.5 = 4.5 By subtraction, the right hand external triangle is also 4.5 wide. The trapezoid is symmetrical. Green rectangle is (1.5 + 4) * 2.6 = 14.29
Drop perpendiculars from E and F to AB at M and N respectively. EM and FN will intersect CD at J and K respectively. As ∠CJE = ∠AME = 90° and ∠ECJ = ∠EAM = 60°, being corresponding angles, ∆CJE and ∆AME are similar triangles. Similarly, ∠FKD = ∠FNB = 90° and ∠KDF = ∠NBF = θ, so ∆FKD and ∆FNB are similar triangles. As ∠EAM = 60°, ∆AME and ∆CJE are 30-60-90 special right triangles, so AM = AE/2 = (6+3)/2 = 9/2 and EM = FN = AM√3 = (9/2)√3 = 9√3/2. As EM and FN are parallel to each other and perpendicular to MN, JK, and EF, MN = JK = EF = 4. NB = AB-AM-MN = 13-9/2-4 = 26/2-17/2 = 9/2. As AM = NB = 9/2, EM = FN = 9√3/2, and ∠AME = ∠FNB = 90°, ∆AME and ∆FNB are congruent by SAS, and as EJ = FK, by similarity, ∆CJE and ∆FKD are congruent. KD = CJ = CE/2 = 3/2. FK = EJ = CJ√3 = (3/2)√3 = 3√3/2. CD = CJ+JK+KD = 3/2+4+3/2 = 7. Green Trapezoid CDFE: Aɢ = h(a+b)/2 = (3√3/2)(4+7)/2 Aɢ = 3√3(11)/4 = 33√3/4 ≈ 14.29 sq units
Your solution is so impressive! Mine is a litte bit different😅 1/ EA and FB intersect at point G. Label GE= a By triangle similarity, we have: GE/GA=EF/AB= 4/13 -> a/(a+9) =4/13 -> a= 4 -> the triangle GEF is an equilateral one and so is the triangle GDC ( and GAB) Area of the green trapezoid= Area of GDC- Area of GEF = sq7( sqrt3)/4 - sq4(sqrt3)/4 = sqrt3/4 x (sq7-sq4) = 33sqrt3/4 sq units 😅😅😅
Just another way to show variety: the big trapezoid area = the sum of the other smaller ones and by setting side CD= x and substituting on the above equation you get x=7 and then you can find the area. Thank you for the video ❤
H=9 Sin 60=7.7942 Base= 9 cos60=3 Total base=3+4+6=13 Base of green area=1.5+4+2=7.5 h for green area=3sin60=2.5981 Area of green =(4+7)/2*2.5981 =14.94 square units
The area is 1/4[33*sqrt(3)]. At the 2:10 mark, I think that I now know that drawing auxiliary lines is useful due to the arrows. While it is not drawn to scale, it helps to know how to draw the auxiliary lines *to scale* and this is probably why we got a AA similarity relation. I am wondering if a playlist of trapezium could be made or a playlist involving AA similarity relation could be made. Or "Think outside of the box". Just asking.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) h = 3 * sin(60º) ; h = 3sqrt(3) / 2 02) B = 7 03) b = 4 04) A = 11 * 3sqrt(3) / 4 05) A = 33sqrt(3) / 4 06) A = 8,25sqrt(3) 07) A = 14,289 sq un 08) NOTE : From AB to EF it follows : 13 ; 10 ; 7 ; 4 Therefore, OUR ANSWER : The Area of Green Region is approx. 14,3 Square Units.
My way of solution ▶ Let's consider point G between A and B G ∈ [AB] and [AG] ⊥ [EG] cos(60°)= [AG]/[EA ] [EA]= 9 lenght units ⇒ 1/2= [AG]/9 [AG]= 9/2 sin(60°)= [EG]/[EA ] [EA]= 9 lenght units ⇒ √3/2= [EG]/9 [EG]= 9√3/2 By considering point H between A and B H ∈ [AB] and [HB] ⊥ [FH] [HB]= [AB] - [AG] - [GH] [HB]= 13 - (9/2) - 4 [HB]= 9/2 ⇒ [HB]= [AG]= 9/2 [EG]= [FH]= 9√3/2 ⇒ ΔEAG = ΔFHB ⇒ ∠HBF= 60° By considering point P between C and D P ∈ [CD] and [CP] ⊥ [EP] we also see that ∠ECP= 60° ⇒ cos(∠ECP)= [CP]/[EC] 1/2= [CP]/3 [CP]= 3/2 ⇒ [KD]= 3/2 K ∈ [CD] and [KD] ⊥ [FK] It is seen that: ΔEAG ~ ΔECP [CP]/[AG]= [EP]/[EG] (3/2)/(9/2)= [EP]/9√3/2 [EP]= 3√3/2 ⇒ A(ECDF)= ([EF]+[CD])*EP/2 [EF]= 4 [CD]= (3/2)+4+(3/2) [CD]= 7 [EP]= 3√3/2 ⇒ A(ECDF)= (4+7)* 3√3/2 /2 A(ECDF)= 33√3/4 A(ECDF) ≈ 14,29 square units
Let's find the area: . .. ... .... ..... First of all we extend the lines AE and BF such that they intersect. May G be the point of intersection. According to the intercept theorem we can conclude: AG/EG = AB/EF (AC + CE + EG)/EG = AB/EF (6 + 3 + EG)/EG = 13/4 (9 + EG)/EG = 13/4 4*(9 + EG) = 13*EG 36 + 4*EG = 13*EG 36 = 9*EG ⇒ EG = 4 CG/EG = CD/EF (CE + EG)/EG = CD/EF (3 + 4)/4 = CD/4 ⇒ CD = 3 + 4 = 7 Since AB and CD are parallel to each other, we obtain ∠DCE=∠BAC=60°. Therefore the height h of the green trapezoid turns out to be: h/CE = sin(∠DCE) h/3 = sin(60°) h/3 = √3/2 ⇒ h = 3√3/2 Now we are able to calculate the area of the green trapezoid: A(CDEF) = (1/2)*(CD + EF)*h = (1/2)*(7 + 4)*3√3/2 = 33√3/4 ≈ 14.29 Best regards from Germany
Before my lawn goes dormant for the winter I gotta paint it Green . It's Trapezoidal shaped so I have too calculate the area so I know how much paint I need. Perfect timing for this video too come up. 🙂
Over-complicated. As soon as you establish your new parallelogram divides the base into 9 and 4, you have immediately established you are dealing with 2 equilateral triangles: 60˚ between 2 sides of equal length.Therefore CD is EC + 4 = 7
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Perpendicular height of the green area is (3*sqrt(3))/2, due to the 30,60,90 on the left.
The width of its base is 1/3 of the distance between 4 and 13, so the green base is 7.
(7 + 4)/2 * (3*sqrt(3))/2
(11/2) * 3*sqrt(3)/2
(33/2) * sqrt(3)/2
(33*sqrt(3))/4 un^2
Decimal approximates to 14.29 un^2
I have now checked the video. There were a few ways to tackle this one.
Thank you.
You are very welcome!
Thanks for sharing ❤️
CD = EF + (EC/EA).(AB-EF) = 4 +(3/9).9 = 7. The green area is then ((EF+CD)/2).(EC.sin(60°)) = ((4+7)/2).(3.(sqrt(3)/2) = (33/4).sqrt(3). (Very quick!)
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Slightly different method, same answer :
Complete external left triangle. From memory, sin(30)= 0.5
External triangle width is 9 x 0.5 = 4.5
By subtraction, the right hand external triangle is also 4.5 wide. The trapezoid is symmetrical.
Green rectangle is (1.5 + 4) * 2.6 = 14.29
Thank you!
AB=13=26/2=[(6+3)/2]+(8/2)+[(26-9-8)/2]=(9/2)+(8/2)+(9/2)---> Ángulos A=60º=B---> Área sombreada =[(3/2)+(8/2)]*(3√3/2)=33√3/4 ud².
Gracias y saludos.
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Drop perpendiculars from E and F to AB at M and N respectively. EM and FN will intersect CD at J and K respectively.
As ∠CJE = ∠AME = 90° and ∠ECJ = ∠EAM = 60°, being corresponding angles, ∆CJE and ∆AME are similar triangles. Similarly, ∠FKD = ∠FNB = 90° and ∠KDF = ∠NBF = θ, so ∆FKD and ∆FNB are similar triangles.
As ∠EAM = 60°, ∆AME and ∆CJE are 30-60-90 special right triangles, so AM = AE/2 = (6+3)/2 = 9/2 and EM = FN = AM√3 = (9/2)√3 = 9√3/2. As EM and FN are parallel to each other and perpendicular to MN, JK, and EF, MN = JK = EF = 4. NB = AB-AM-MN = 13-9/2-4 = 26/2-17/2 = 9/2.
As AM = NB = 9/2, EM = FN = 9√3/2, and ∠AME = ∠FNB = 90°, ∆AME and ∆FNB are congruent by SAS, and as EJ = FK, by similarity, ∆CJE and ∆FKD are congruent.
KD = CJ = CE/2 = 3/2. FK = EJ = CJ√3 = (3/2)√3 = 3√3/2. CD = CJ+JK+KD = 3/2+4+3/2 = 7.
Green Trapezoid CDFE:
Aɢ = h(a+b)/2 = (3√3/2)(4+7)/2
Aɢ = 3√3(11)/4 = 33√3/4 ≈ 14.29 sq units
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Thanks for sharing ❤️
Your solution is so impressive!
Mine is a litte bit different😅
1/ EA and FB intersect at point G.
Label GE= a
By triangle similarity, we have:
GE/GA=EF/AB= 4/13
-> a/(a+9) =4/13
-> a= 4 -> the triangle GEF is an equilateral one and so is the triangle GDC ( and GAB)
Area of the green trapezoid= Area of GDC- Area of GEF
= sq7( sqrt3)/4 - sq4(sqrt3)/4
= sqrt3/4 x (sq7-sq4)
= 33sqrt3/4 sq units 😅😅😅
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φ = 30° → EAB = ABF = 2φ → CP = DQ = 3/2 → green area = 9√3/4 + 6√3 = 33√3/4
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❤
base of the green = x + 4
x = (13 - 4)/3 = 3
height of the green = 3sin60° = (3✓3)/2
green area = (4 + 4 + x)(3✓3)/4 = (33✓3)/4
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@@PreMath Thank YOU for the math problem 🙏
3:45 Since AE=AK=9 the AEK is isosceles.
Angle KAE=60°, and angles AEK=EKA=(180°-60°)/2=60°.
Triangle AEK is thus equilateral.
So CT=CE=3...
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Thanks for the feedback ❤️
Where do you find all these math problems. Very enjoyable.
From from the similarity of triangles, it is much easier to find the area
Just another way to show variety: the big trapezoid area = the sum of the other smaller ones and by setting side CD= x and substituting on the above equation you get x=7 and then you can find the area.
Thank you for the video ❤
You are very welcome!
Thanks for the feedback ❤️
By my calculations, it's isosceles trapezoid but it doesn't look like it
H=9 Sin 60=7.7942
Base= 9 cos60=3
Total base=3+4+6=13
Base of green area=1.5+4+2=7.5
h for green area=3sin60=2.5981
Area of green =(4+7)/2*2.5981
=14.94 square units
Ans 14.29 square units
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S=33√3/4≈14,289≈14,29 square units
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3×3-1.5×1.5=6.75squroot=2.598×5.5=14.2894
14.2894
The area is 1/4[33*sqrt(3)]. At the 2:10 mark, I think that I now know that drawing auxiliary lines is useful due to the arrows. While it is not drawn to scale, it helps to know how to draw the auxiliary lines *to scale* and this is probably why we got a AA similarity relation. I am wondering if a playlist of trapezium could be made or a playlist involving AA similarity relation could be made. Or "Think outside of the box". Just asking.
Excellent!
Thanks for the feedback ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) h = 3 * sin(60º) ; h = 3sqrt(3) / 2
02) B = 7
03) b = 4
04) A = 11 * 3sqrt(3) / 4
05) A = 33sqrt(3) / 4
06) A = 8,25sqrt(3)
07) A = 14,289 sq un
08) NOTE : From AB to EF it follows : 13 ; 10 ; 7 ; 4
Therefore,
OUR ANSWER :
The Area of Green Region is approx. 14,3 Square Units.
Excellent!
Thanks for sharing ❤️
My way of solution ▶
Let's consider point G between A and B
G ∈ [AB] and [AG] ⊥ [EG]
cos(60°)= [AG]/[EA ]
[EA]= 9 lenght units
⇒
1/2= [AG]/9
[AG]= 9/2
sin(60°)= [EG]/[EA ]
[EA]= 9 lenght units
⇒
√3/2= [EG]/9
[EG]= 9√3/2
By considering point H between A and B
H ∈ [AB] and [HB] ⊥ [FH]
[HB]= [AB] - [AG] - [GH]
[HB]= 13 - (9/2) - 4
[HB]= 9/2
⇒
[HB]= [AG]= 9/2
[EG]= [FH]= 9√3/2
⇒
ΔEAG = ΔFHB
⇒
∠HBF= 60°
By considering point P between C and D
P ∈ [CD] and [CP] ⊥ [EP]
we also see that ∠ECP= 60°
⇒
cos(∠ECP)= [CP]/[EC]
1/2= [CP]/3
[CP]= 3/2
⇒
[KD]= 3/2
K ∈ [CD] and [KD] ⊥ [FK]
It is seen that: ΔEAG ~ ΔECP
[CP]/[AG]= [EP]/[EG]
(3/2)/(9/2)= [EP]/9√3/2
[EP]= 3√3/2
⇒
A(ECDF)= ([EF]+[CD])*EP/2
[EF]= 4
[CD]= (3/2)+4+(3/2)
[CD]= 7
[EP]= 3√3/2
⇒
A(ECDF)= (4+7)* 3√3/2 /2
A(ECDF)= 33√3/4
A(ECDF) ≈ 14,29 square units
I would say the middle horizontal line has length 7....(1.5+4+1.5) and that yields and area of 33sqrt(3)/4
Thanks for the feedback ❤️
А=33\/3/4.
Let's find the area:
.
..
...
....
.....
First of all we extend the lines AE and BF such that they intersect. May G be the point of intersection. According to the intercept theorem we can conclude:
AG/EG = AB/EF
(AC + CE + EG)/EG = AB/EF
(6 + 3 + EG)/EG = 13/4
(9 + EG)/EG = 13/4
4*(9 + EG) = 13*EG
36 + 4*EG = 13*EG
36 = 9*EG
⇒ EG = 4
CG/EG = CD/EF
(CE + EG)/EG = CD/EF
(3 + 4)/4 = CD/4
⇒ CD = 3 + 4 = 7
Since AB and CD are parallel to each other, we obtain ∠DCE=∠BAC=60°. Therefore the height h of the green trapezoid turns out to be:
h/CE = sin(∠DCE)
h/3 = sin(60°)
h/3 = √3/2
⇒ h = 3√3/2
Now we are able to calculate the area of the green trapezoid:
A(CDEF) = (1/2)*(CD + EF)*h = (1/2)*(7 + 4)*3√3/2 = 33√3/4 ≈ 14.29
Best regards from Germany
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Very Simple Sum
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a/4=(a+9)/13, 13a=4a+36, a=4, then CD=7, the area is 1/2(49-16)sin 60=33sqrt(3)/4.😊
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Before my lawn goes dormant for the winter I gotta paint it Green . It's Trapezoidal shaped so I have too calculate the area so I know how much paint I need. Perfect timing for this video too come up. 🙂
Sounds great!😀
Thanks for the feedback ❤️
😊
33r3÷4 😊
9 square unit ?
1st comment😁
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Thanks ❤️
Over-complicated. As soon as you establish your new parallelogram divides the base into 9 and 4, you have immediately established you are dealing with 2 equilateral triangles: 60˚ between 2 sides of equal length.Therefore CD is EC + 4 = 7