Once I knew that PB = 4 I tried to calculate the height of the triangle PBT by drawing the perpendicular to BE from the point T, which I called TH. Since the angle TBE is 60° the right triangle BHT is of the type 30°,60°,90°, then if BH = X (height of PBT) TH = X√ 3 I then calculated the value of X by noting that the triangles PBE and HET are similar. In fact, they have a common angle and are right angles. So PB : TH = BE : HE 4 : X√ 3 = 8 : (8 - X) X = (16√ 3 - 8)/11 (PBT triangle height) Area = 1/2* 4 * (16√ 3 - 8)/11 = (32√ 3 - 16)/11
This is another problem that yields easily to coordinate geometry. Set B=(0,0), E=(8,0) and P=(0,4). Because FBE is 60 degrees, F=(4,4*sqrt(3)). The lines containing PT and BF have equations x+2y=8 and y=x*sqrt(3). Solving this system for x, we see their intersection point T has an x-coordinate value of 8/(2*sqrt(3)+1), which is the length of the altitude from T to PB. Double this to get the area of the green triangle, since base PB=4.
Yes, the final discrepancy in my answer is from triangle PTB not being a right angled triangle. I assumed it was. In your solution you formed a right angled triangle to calculate the height using Side PB as the base. This explains the more complicated workings required.
5:00 At this pointcan't you show that △PBT is similar to △DEA? Then DE is easily derived, and you know PB. So you could find PT and TB via ratio, right?
Looking at this Triangles PBE and DAE are right angled and similar with P as diagnal mid point of the rectangle. Triangle FBE is equilateral with angles 60 degrees. Hence PB is 4 units and angle PBT is 30 degrees . sin(30) =0.5 .PT = 2 and BT = sqrt(16 -4) = 3 * sqrt(3). Triangle area is 0.5 base* height = 0.5 * 3 * sqrt(3) * 2 = 3 sqrt(3).
1. Note that AE = 16 and AD = 8. Therefor DE must bisect CB in half at P, since CB is perpendicular to AE and hits AE in the middle at B AB = BE = 8. Thus CP = PB = 4 2. Angle PBT is 30 degrees as explained in the video. 3. Since the angle in C is 90 degrees and DC = 8 and CP = 4 you can solve for the angle DPC = 63,435 degrees since BPT and DCP are opposing angles they must be of the same Value. 4. Now you have 3 measurements of triangle PTB (Angle - Side - Angle) => 63,434 degrees, PB = 4 and 30 degrees. You can calculate all the remaining values using trigonometric formulae.
Dear Rachid, all three sides are equal! Sum of angles in a triangle is 180 degrees. Therefore, each angles is 180/3 degrees => 60 degrees Thanks for asking. Cheers
Game plan: • Find distance from BC to T. This will be the height h in the calculation of the area of triangle ∆TPB. • Find length of PB. This will be the base b in the calculation of the area of triangle ∆TPB. • Alternately, finding the areas of ∆PCD and ∆FTE will let us determine the area of ∆TPB from the areas of the three major polygons in the construction. By observation, ∆DAE and ∆PBE share an angle at E, have 90° angles along AE, and have parallel sides in DA and PB. ∆DAE and ∆PBE are similar. Triangle ∆PBE: DA/AE = PB/BE 8/16 = b/8 b = 8(8)/16 = 4 Drop a perpendicular from T to BE at G. By observation ∆TGE is similar to ∆PBE and, as a right triangle constructed from an equilateral triangle, ∆BGT is a 30-60-90 special triangle. As BG = h, GE = 8-h. As ∆BGT is a 30-60-90 special triangle, TB = 2h and GT = h√3. Triangle ∆TGE: TG/GE = PB/BE h√3/(8-h) = 4/8 = 1/2 2h√3 = 8-h h(2√3 +1) = 8 h = 8/(2√3 +1) = 8(2√3 -1)/(12-1) h = (8/11)(2√3 -1) Triangle ∆TPB: A = bh/2 = 4(8/11)(2√3 -1)/2 A = (16/11)(2√3 -1) ≈ 3.58
Drop perpendicular from T to BE and call the intersection H. Let distance HE be x. Note that right ΔDAE and ΔTHE are similar and the ratio of the short side to the long side is 1/2, so TH has length x/2. ΔBTH is a special 30°-60°-90° right triangle, its long side is √3 times as long as its short side, so length BH = x/(2√3). However, length BH is also 8-x, so 8-x=x/(2√3). Doing the algebra, x=(96-(16√3))/11. Length BH=8-x=((16√3)-8)/11. We note that, if we treat PB as the base of ΔPTB, BH is equal to its height. So, area of ΔPTB=(0.5)(4)((16√3)-8)/11=16(2√3-1)/11, as PreMath also found.
I think this may be incorrect. Angle AED is not 30° but rather arctan 8/(8+8) which is approximately 26.565°. Thus triangles PTB and BTE are not similar and angle PTB is not a right angle. Sorry.
Todella mielenkiintoinen tehtävä ja otti aikaa löytää "yksinkertainen" ratkaisu. DE alenee .5 yksikköä edetessään oikealle. kolmion kylki BF nousee sqrt3 yksikköä adetessään oikealle. Yhteensä lähenevät toisiaan .5+sqrt3. Leikkauspisteen etäisyys viivasta CB on 4/(.5+sqrt3) Kolmion ala A = .5*4*4/(.5+sqrt3)
Thank you for your sharing 👍
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Kind regards
Once I knew that PB = 4 I tried to calculate the height of the triangle PBT by drawing the perpendicular to BE from the point T, which I called TH. Since the angle TBE is 60° the right triangle BHT is of the type 30°,60°,90°, then
if BH = X (height of PBT)
TH = X√ 3
I then calculated the value of X by noting that the triangles PBE and HET are similar. In fact, they have a common angle and are right angles. So
PB : TH = BE : HE
4 : X√ 3 = 8 : (8 - X)
X = (16√ 3 - 8)/11 (PBT triangle height)
Area = 1/2* 4 * (16√ 3 - 8)/11 = (32√ 3 - 16)/11
This is another problem that yields easily to coordinate geometry. Set B=(0,0), E=(8,0) and P=(0,4). Because FBE is 60 degrees, F=(4,4*sqrt(3)). The lines containing PT and BF have equations x+2y=8 and y=x*sqrt(3). Solving this system for x, we see their intersection point T has an x-coordinate value of 8/(2*sqrt(3)+1), which is the length of the altitude from T to PB. Double this to get the area of the green triangle, since base PB=4.
Simply great!
i used coord geometry too.
Me too
I love rationalizing radicals in the denominator by multiplying by the conjugate . One of the best concepts of mathematics in my opinion. 🙂
Yes, the final discrepancy in my answer is from triangle PTB not being a right angled triangle. I assumed it was. In your solution you formed a right angled triangle to calculate the height using Side PB as the base. This explains the more complicated workings required.
Thank you for doing these with all the accuracy and explanation.
5:00 At this pointcan't you show that △PBT is similar to △DEA?
Then DE is easily derived, and you know PB. So you could find PT and TB via ratio, right?
This is not the case. δ ≠ φ → AED = δ ≈ 26,565°; PBT = φ = 30°→
EDA ≈ 63,435° = 3φ - δ ≠ 2φ = 60° = TBE 🙂
Where do we know that its an equilateral triangle and not an isosceles traingle?
Sir, another lengthy problem. Where do you source them?
Another great piece of work, thanks again 👍🏻
Great puzzle. Thanks PreMath 😊
Great explanation👍
Thanks for sharing😊
Looking at this Triangles PBE and DAE are right angled and similar with P as diagnal mid point of the rectangle. Triangle FBE is equilateral with angles 60 degrees. Hence PB is 4 units and angle PBT is 30 degrees . sin(30) =0.5 .PT = 2 and BT = sqrt(16 -4) = 3 * sqrt(3). Triangle area is 0.5 base* height = 0.5 * 3 * sqrt(3) * 2 = 3 sqrt(3).
1. Note that AE = 16 and AD = 8. Therefor DE must bisect CB in half at P, since CB is perpendicular to AE and hits AE in the middle at B AB = BE = 8. Thus CP = PB = 4
2. Angle PBT is 30 degrees as explained in the video.
3. Since the angle in C is 90 degrees and DC = 8 and CP = 4 you can solve for the angle DPC = 63,435 degrees since BPT and DCP are opposing angles they must be of the same Value.
4. Now you have 3 measurements of triangle PTB (Angle - Side - Angle) => 63,434 degrees, PB = 4 and 30 degrees. You can calculate all the remaining values using trigonometric formulae.
Thanks for video.Good luck sir!!!!!!!!!
Plese how do you see that the angle is 60degres
Dear Rachid, all three sides are equal! Sum of angles in a triangle is 180 degrees. Therefore, each angles is 180/3 degrees => 60 degrees
Thanks for asking. Cheers
We don’t have this information. I see the lenght of two sides.
This playlist shows 3 different solutions to this problem:
ua-cam.com/video/NDSDhzi3hHs/v-deo.html
Is that correct? ((4*cos(30))*(4*sin(30)))/2
Game plan:
• Find distance from BC to T. This will be the height h in the calculation of the area of triangle ∆TPB.
• Find length of PB. This will be the base b in the calculation of the area of triangle ∆TPB.
• Alternately, finding the areas of ∆PCD and ∆FTE will let us determine the area of ∆TPB from the areas of the three major polygons in the construction.
By observation, ∆DAE and ∆PBE share an angle at E, have 90° angles along AE, and have parallel sides in DA and PB. ∆DAE and ∆PBE are similar.
Triangle ∆PBE:
DA/AE = PB/BE
8/16 = b/8
b = 8(8)/16 = 4
Drop a perpendicular from T to BE at G. By observation ∆TGE is similar to ∆PBE and, as a right triangle constructed from an equilateral triangle, ∆BGT is a 30-60-90 special triangle. As BG = h, GE = 8-h. As ∆BGT is a 30-60-90 special triangle, TB = 2h and GT = h√3.
Triangle ∆TGE:
TG/GE = PB/BE
h√3/(8-h) = 4/8 = 1/2
2h√3 = 8-h
h(2√3 +1) = 8
h = 8/(2√3 +1) = 8(2√3 -1)/(12-1)
h = (8/11)(2√3 -1)
Triangle ∆TPB:
A = bh/2 = 4(8/11)(2√3 -1)/2
A = (16/11)(2√3 -1) ≈ 3.58
Drop perpendicular from T to BE and call the intersection H. Let distance HE be x. Note that right ΔDAE and ΔTHE are similar and the ratio of the short side to the long side is 1/2, so TH has length x/2. ΔBTH is a special 30°-60°-90° right triangle, its long side is √3 times as long as its short side, so length BH = x/(2√3). However, length BH is also 8-x, so 8-x=x/(2√3). Doing the algebra, x=(96-(16√3))/11. Length BH=8-x=((16√3)-8)/11. We note that, if we treat PB as the base of ΔPTB, BH is equal to its height. So, area of ΔPTB=(0.5)(4)((16√3)-8)/11=16(2√3-1)/11, as PreMath also found.
alternate solution (long but gets it done) :
from congruence triangle DCP=triangle EBP
CP = PB = 8/2 = 4
Agreen = 1/2 * 4 * sin(30) * BT = BT (1)
let a = angle PEB
Agreen = 1/2 * 8 * sin(a) * PT (2)
from (1)&(2):
BT = 4 * sin(a) * PT (3)
sin(a) = 4 / (4^2 + 8^2) ^.5 = 1/(5)^.5 (4)
cos(a)= (1-sin(a)^2)^.5 = 2/(5)^.5
let c= 1/(5^.5)
let d=3^.5
from (3)&(4):
BT = 4 * sin(a) * PT = 4c * PT
let b = angle PTB b = a+60
cos(b) = cos(60) * cos (a) - sin(a) * sin(60)
= (1/2) * (2 c) - c * d/2 = [c-cd/2]
from law of cosines:
BT^2 + PT^2 - 2 * BT * PT * cos(b) = 4^2
(4c * PT)^2 + PT^2 - 2 * (4c * PT) * PT * [c-cd/2] = 4^2
[16 * c^2 + 1 - 8 * c^2 + 4 * c^2 * d] * PT^2 = 16
[8 * c^2 + 4 * c^2 * d +1] * PT^2 =16
[13/5 + 4 * 3^.5/5] * PT^2 = 16
PT=(80 / (13+4*3^.5))^.5
A = 4 * sin(a) * PT = 4 * (1/(5)^.5) * [(80 / (13+4*3^.5))^.5]
= 16/(13 + 4 * 3^.5)^.5
= 3.5841
Yes, the sqrt(16-4) = sqrt(4*3) = 2sqrt(3). I now have 3.4641
When the length of PB was briefly inserted under PB in equation 1, we could already conclude that the area of △ PBT = TB...
Very cool
So nice of you.
Thank you! Cheers! 😀
@@PreMath 👏👏👏
angle BPT = angle ADE
= arc tan (1/2)
= arc sin ( 1/√5)
sin (PTB)
= sin ( 30° + angle TPB)
= (1*2 + √3)/ ( 2 √5)
TB /PB
= (1/√5)/[ (1*2 + √3)/ ( 2 √5)]
= 2 / ( 2 + √3) = (4 - 2 √3)
Hereby TB = AB (2 -√3)
[ ∆ PBT ]
= AB^2 ( 2 - √3) sin(30°) /2
= AB^2 ( 2 - √3) /4
angle BPT = atan(2/1)
What is AB? ,Area?
@@nehronghamil4352 AB is the side of the square. For present problem put 8 unit for AB to get the area
Ab equilateral traingle should have been specified from the very beginning by hatching each side.
similar triangle, 8*8/2*sqrt(5)
Area DCP = Area PBE=16
TE=SQRT(8^2 - 4^2)
TE= 4V3
Area TBE = 1/2 *4*4V3
=8V3
Area of the green=16-8V3
= 16 - 13,85 = 2,15
caution 2nd line false TB not = 4 it's PB
why FBE triangle is equilateral???!!!!!!!!!!!
It was mentioned verbally only.
It’s in the title of the problem, under the problem/drawing.
I think this may be incorrect. Angle AED is not 30° but rather arctan 8/(8+8) which is approximately 26.565°. Thus triangles PTB and BTE are not similar and angle PTB is not a right angle. Sorry.
TB is 4, not PB
Ah, got mixed up as to which triangles were similar!
My answer done quickly is higher so i'll check it over and reconfirm.
Only problem is angle PBT is 26.57 degrees not 30.
i used coord geometry
Agreen=16-(4/11)(48-8rad3)=3,584.…
Great!
Thank you! Cheers! 😀
Başka yol yok mu
FB = 8 ??
Second comment
So nice of you.
Thank you! Cheers! 😀
4
Green area =5.5cm
why FE=BE=FB????
This is incorrect. The triangle is not equilateral.
Todella mielenkiintoinen tehtävä ja otti aikaa löytää "yksinkertainen" ratkaisu.
DE alenee .5 yksikköä edetessään oikealle. kolmion kylki BF nousee sqrt3 yksikköä adetessään oikealle.
Yhteensä lähenevät toisiaan .5+sqrt3.
Leikkauspisteen etäisyys viivasta CB on 4/(.5+sqrt3)
Kolmion ala A = .5*4*4/(.5+sqrt3)