Find area of Green shaded Triangle | Square and Equilateral triangle | Important skills explained

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Find area of Green shaded Triangle | Square and Equilateral triangle | Important skills explained

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[solimana-soli]

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

[misterenter-izrz]

Inspired by your beautiful solution, I attempt to offer a shorter solution, let A be the area of the area of the green triangle, so the area of the adjacent yellow triangle is 2 sqrt(3)A, thus 16=(1+2sqrt(3))A, or A=16/(1+2sqrt(3)).😅

[tombufford]

Thank you for doing these with all the accuracy and explanation.

[davidellis]

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.

[wackojacko]

I love rationalizing radicals in the denominator by multiplying by the conjugate . One of the best concepts of mathematics in my opinion. 🙂

[tombufford]

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.

[theoyanto]

Another great piece of work, thanks again 👍🏻

[misterenter-izrz]

Lovely mathematical figure, it is not too difficult, but I am afraid the computation is a bit cumbersome, as we have the exact figures of the green triangle, one side is 4, two angles are 30, arctan 2=63.435, so the last angle is 86.565, so opposite side is 4xsin(63.435)/(sin 86.565)=3.58415, thus the area is 4x3.58415/4=3.58415.😅

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing😊

[murj]

Where do we know that its an equilateral triangle and not an isosceles traingle?

[tombufford]

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).

[flesby]

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.

[jonathancapps]

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?

[vara]

Sir, another lengthy problem. Where do you source them?

[tombufford]

Yes, the sqrt(16-4) = sqrt(4*3) = 2sqrt(3). I now have 3.4641

[rachidmeknassi]

Plese how do you see that the angle is 60degres

[jimlocke]

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.

[ybodoN]

When the length of PB was briefly inserted under PB in equation 1, we could already conclude that the area of △ PBT = TB...

[nehronghamil]

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

[quigonkenny]

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