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

[@SAHIRVLOGCLP]

Thank you for your sharing 👍

[@PreMath]

Thanks for visiting Kind regards

[@soli9mana-soli4953]

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

[@davidellis1929]

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.

[@murdock5537]

Simply great!

[@spiderjump]

i used coord geometry too.

[@shafikbarah9273]

Me too

[@tombufford136]

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

[@wackojacko3962]

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

[@tombufford136]

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.

[@misterenter-iz7rz]

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)).😅

[@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.

[@theoyanto]

Another great piece of work, thanks again 👍🏻

[@misterenter-iz7rz]

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.😅

[@michaelkouzmin281]

Did it the same way.

[@PreMath]

Thank you! Cheers! 😀

[@waheisel]

Great puzzle. Thanks PreMath 😊

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!

[@tombufford136]

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).

[@nehronghamil4352]

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

[@murj617]

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

[@HappyFamilyOnline]

Great explanation👍 Thanks for sharing😊

[@vara1499]

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

[@jimlocke9320]

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.

[@tombufford136]

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

[@rachidmeknassi3709]

Plese how do you see that the angle is 60degres

[@PreMath]

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

[@Brunayres1]

We don’t have this information. I see the lenght of two sides.

[@jonathancapps1103]

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?

[@murdock5537]

This is not the case. δ ≠ φ → AED = δ ≈ 26,565°; PBT = φ = 30°→ EDA ≈ 63,435° = 3φ - δ ≠ 2φ = 60° = TBE 🙂

[@honestadministrator]

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

[@nehronghamil4352]

angle BPT = atan(2/1) What is AB? ,Area?

[@honestadministrator]

@@nehronghamil4352 AB is the side of the square. For present problem put 8 unit for AB to get the area

[@ybodoN]

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

[@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

[@Su4ji]

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

[@acte1313]

caution 2nd line false TB not = 4 it's PB

[@romilsonmagalhaes6487]

Very cool

[@PreMath]

So nice of you. Thank you! Cheers! 😀

[@romilsonmagalhaes6487]

@@PreMath 👏👏👏

[@lk-wr2yn]

Is that correct? ((4*cos(30))*(4*sin(30)))/2

[@misterenter-iz7rz]

First to add like 😮.

[@PreMath]

So nice of you. Thank you! Cheers! 😀

[@rc_youtubeaccount1331]

similar triangle, 8*8/2*sqrt(5)

[@prof.isaquecosta]

This playlist shows 3 different solutions to this problem: kzbin.info/www/bejne/hHW2dZuwnpibftU

[@spiderjump]

i used coord geometry

[@29brendus]

Ab equilateral traingle should have been specified from the very beginning by hatching each side.

[@eliasgiannakopoulos4005]

4

[@giuseppemalaguti435]

Agreen=16-(4/11)(48-8rad3)=3,584.…

[@PreMath]

Great! Thank you! Cheers! 😀

[@aimilios321]

why FBE triangle is equilateral???!!!!!!!!!!!

[@JasperJones-jq7yh]

It was mentioned verbally only.

[@1ciricola]

It’s in the title of the problem, under the problem/drawing.

[@halitiskender1324]

Başka yol yok mu

[@tombufford136]

My answer done quickly is higher so i'll check it over and reconfirm.

[@CharlesWeiss-e4c]

Only problem is angle PBT is 26.57 degrees not 30.

[@clivemitchell3229]

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.

[@iberg]

TB is 4, not PB

[@clivemitchell3229]

Ah, got mixed up as to which triangles were similar!

[@barouchkrakauer7815]

FB = 8 ??

[@فراسمعابره-ج5خ]

Green area =5.5cm

[@A_MIGHTY]

Second comment

[@PreMath]

So nice of you. Thank you! Cheers! 😀

[@aimilios321]

why FE=BE=FB????

[@iodian]

This is incorrect. The triangle is not equilateral.

[@vierinkivi]

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)