Can you find the area of this triangle??

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[flaviusclaudius]

At first I was like "no", but then you specified the triangle was equilateral and not an arbitrary triangle

[wallywutsizface]

Here's how I did it:

Look just at the 30/60/90 triangle formed on the left. The height is 1, and the base is S-1, where S is the side length of the whole triangle. Using the formula for a 30/60/90 triangle, we know S-1 = 1/sqrt3, so S = 1+sqrt3. Now that we have the side length of the whole triangle, we just plug it in to the formula for the area of an equilateral triangle, S^2 * sqrt3/4. This gives the same answer, 1/2 + sqrt3/3.

[jackhandma]

* Everytime Michael Penn does a simple looking problem*

Me: *My time has come.*

[jessejordache]

You really should specify that it's an equilateral triangle before you "get started". The way you have it set up, the way you introduce that fact makes it clear that the solution is going to involve trig.

[manucitomx]

I’m always happy when I can solve it by myself. I’m positively joyful when my method is the same.
Thank you, professor.

[lovecastle]

Never said it was an equilateral triangle at the start

[georgethomas]

This can be done more simply and quickly. The area is 1 + A where A is the area of the small triangle at the top. The base b of that small triangle can be found to be 1 - 1/sqrt(3), from a simple consideration of tan(60). And then the area of the big triangle is 1 + sqrt(3)*(b^2)/4 = 1. 0774 which is same as the value derived in the video.

[abby]

1:40 Him "Maybe that's the most important one"
Me: "bars"

[s.m.m]

Hi.
We could also say the area equals the area of the square which is 1, plus the area of the smaller equilateral triangle in the picture.
Thanks

[jhjhjh]

When you solve the problem just from thumbnail:
This does put a smile on my face

[waterunderthebridge]

Oh wow, I did it almost exactly like him (although I just plugged 1+sqrt(3) into the area formula for an equilateral triangle), how interesting so many can look at the same problem and think of such similar solutions (albeit in this case there probably aren’t a lot of alternative, very different solutions)

[zarinakhatun]

I used the (√3/4)×a^2 formula for right angle triangle,
where 'a' is the side length of the right angle triangle.
We got the length of BF and the length of FC given 1, the total length of BC would be (√3+(1/√3)) which is our 'a'. After further calculation, we would find our good place to stop.

[garylillich]

Most Geometry students should be able to solve without Trig.
In a 30-60-90∆, the ratio of sides is 1: √3: 2
Which means the sides of the ∆ on the left are √3/3, 1 & 2√3/3
So each side of the EQ∆ is 2√3/3+1
Altitude is (√3+1)/2

[stephenbrand]

Yippee, got it correct. Used a bit of congruent triangles as well as trigonometry. So area of big triangle is 1 plus the area of the little triangle on top,

[sharpfang]

I approached it from another side. Let's also denote point G at intersection of AC and DE. BEF and CDG are identical triangles, so the trapezoid BCEG has the same area as the square CDEF = 1. What's left is to add the area of the tiny triangle, AEG, with side (EG) equal to 1-DG. And DG can be obtained with trigonometry.

[Bodyknock]

Just as an aside an interesting note is that triangle CD_ (the _ is the intersection he didn’t label) is congruent to triangle BEF (they’re both right triangles with equal AAS). So their areas are equal, and thus the area of the whole triangle ABC is just 1 (the area of the square) plus the area of the small equilateral triangle AE_ at the top. You can then solve for the area at the top since you know the side length of the large equilateral triangle and thus its height and area, and just subtract 1 from that area.

[Shadow__X]

Haven't watched yet but one thing that is immediately apparent to me is that I can get the side length of the left part of the triangle, which is a right triangle with one side equal to 1 and angle 60°. I can then get the missing sides and add one to the bottom one, which gives me the side length of the triangle, from that I can calculate the area

[graikoyt]

As Penn said, there's a formula for the area of an equilateral triangle given one of its sides.
I'd just like to point out that the lower part of the equilateral triangle has an area of 1. This is because triangle BEF is congruent to the right triangle formed by the part of the square that's outside the equilateral triangle.

[ziyaaddhorat]

You didn't say it was an equilateral in the thumbnail/description, I was sat for a good minute trying to figure it out

[nasim]

I think easier to use:

area = ½ a b (sin C )
= ½ a a (sin π/3) = ½ a² (√3/2) = (√3/4) a²

where a = 1 + (1/√3)