Area of Red Triangle

[peterdelaney-cntd]

Your channel is strangely addictive. Good work

[zmollon]

Love that you provide the formulae required to solve it for my students who haven't learned those yet.

[danielboone]

Can you make a proof showing that dry humor is associated with math? Just think about how exciting that would be.

[mare]

"How exciting" Said Andy enthusiastically

[guyosborn]

Thank you - this is the first of your puzzles I solved in less than ten seconds (not the answer but the steps). Nice, fast-paced explanation, thank you again

[rwen]

1/2 ab sinc(c) is such a huge skip, I totally forgot about that. I used cosine law instead to find the third side. and then did a really long solution by making that side as the base, and solving for the height using a sort of systems of equation by dividing the area into 2 more triangles with the height

[rkeating]

Missed opportunity to show the included angle side-angle-side formula derivation. It’s not difficult and takes the half-base times height triangle area formula that you use a lot just a step further. That said, this channel is fantastic and I wish it had been around when I was learning geometry and math thinking/problem solving in general. Thank you for your work!

[FlyGuy]

A lot of the trick to understanding math is knowing the fundamental properties of things like the 30/60/90 triangle and the area formula that you showed. Unfortunately, about the only thing I remember from my math days is the pythagorean equation and the soh/cah/toa stuff.
It's still fun watching you go through these, though.

[IgS]

Side of the large square is 6. If the base of the red triangle is then 6, then its height is half of 3 sqrt 3 using similar triangle where hypotenuse is 3. Then it is a simple matter of the area being half the base times the height, which is 9/2 sqrt 3

[macmay]

That's interesting... I did it by finding the "height" of the red triangle with the 6 as the base. I dropped down a line from the upper right corner of the red triangle, and extended the top line of the lower square, to give me a new triangle where the middle side would be perpendicular to the 6 and equal to the height of the triangle.

Since the whole obtuse angle of the red triangle was 120, subtracting the 90 degrees of the right angle I made meant the tip just of my new triangle was 30, so I had a 30-60-90 triangle. The hypotenuse we already knew was three, so the middle length would be half that times root three, so the height was 1.5 root three.

Then I had the base and the height of the red triangle. So multiply those values over 2 was the area, which was the same thing you got, nine times root three all over 2.

[Qermaq]

I noted that 3 is the hypotenuse of a little 30-60-90 where the long leg is the height of the red triangle, and that works out to be 3sqrt(3)/2, and the red triangle's base is 6, so I got 9sqrt(3)/2 that way.

[charlesdusautoy]

Use the 30/60/90 logic to get the sides of the big square = 6. Then extend the right side of the big square vertically and take a horizontal line from the bottom-right corner of the small square to meet the vertical line at a right angle. You get a 30/60/90 triangle with hypotenuse = 3 (side of small square) meaning the vertical (short) side is 3/2 and the horizontal (middle) side is 3/2 x square root of 3. This becomes the height of the red triangle with a base of 6. Area is 1/2 x 6 x 3/2 x square root of 3 = 9/2 x square root of 3.

[sawyergrimm]

Pretty similar approach, I just used Pythagorean theorem to get the 6.

[oxiibwx]

solved with law of sines, law of cosines, and trigonometric area formula! Simple and nice!

[popugai]

I did almost the same thing, but I used the cosine theorem to find the third side of the triangle and after it the Heron formula for finding the area

[chadddisowned]

Using the sin formula was too much of a leap for me, so

1.I formed a quadrilateral by extending the sides of the 1st and 2nd square, finding the bottom angle of the red triangle to be 30 degrees due to sum of angles being 360
and
2. Also splitting the red triangle across from the hypotenuse of the first 30-60-90 triangle.

This formed two distinct red triangles, the 1st of which had only one known 30 degree angle.

3.I again extended this triangle until it became a 30 60 90 and
then
4. used the fact that I had found the bottom red angle to find the angle of the top red angle.

This let me find that the triangle formed by completing the top red triangle was an equilateral one, allowing me to

5.find it's area and subtract it from the complete triangle found in step 2

6. adding this triangle I found, which was 1.5Rad(3), added with 3Rad(3), the which produced 4.5Rad(3), also approximately 7.9, all this you can do without a calculator, very nice :)

Note: the way I used to find the rest of the info is almost the exact same as the video except for how I found the actual area of the red triangle.

[UnbeatenPath]

I went a different way but yours is way easier. I made a right triangle in the red area with the hypotenuse of 3 which split the 120 angle into 90 and 30. Then I used law of sin to get the base. Then I used that as the height of the red triangle and did (bh)/2 to get the area

[someonespadre]

Area≈7.9 on my slide rule

There is an opportunity to produce the top line of the big square to the right until it meets a perpendicular dropped from the rightmost point of the 3x3 square which makes a triangle half the size of the original triangle therefore it can be seen that the base of the little triangle is the same as the height of the red triangle and it is 3*sqrt3/2 therefore the area of the red triangle is 1/2*6*3*sqrt3/2=9*sqrt3/2

[yoloswagbi]

Can you do a video explaining why you can find the area of a triangle using that formula?

[Duck-expz]

First time ever i can do qes like this without hint 🤑