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
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.
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!
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
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
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.
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.
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.
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
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
hello andy! sorry to bother you, but by looking at the formula -> (1/2)*a*b*sin(θ) shouldn't it be all divided by 2 like -> (a*b*sin(θ))/2 which is ≈5.2255..... the way you used the formula was like -> ((a*b)/2)*sin(θ)...... and if not why isn't it "(a/2)*b*sin(θ)"..... the absence of parentheses changes the final result...... i am just a "little" confused if its something that i dont now.... but let me know! thanks a lot i like your videos very much😁!!
You’re mistaken; division is the same for each of these. If you have 10 sets of 5 apples and split them in half that is the same as splitting the 10 sets in half and saying they have 5 apples in each set. You don’t change the number of apples.
that formula is just a pattern used when given a specific scenario. It is possible to use pythagoras, but in this case it was possible to use prior knowledge of this specific scenario of a 30 60 90 triangle.
I have this little confusion in my head, let's assume that that red triangle is half of a parallelogram in which the area is 18sqm so the area of that said triangle should be 9sqm because it's the half of it can someone please enlighten me 😅😅
Your channel is strangely addictive. Good work
Love that you provide the formulae required to solve it for my students who haven't learned those yet.
Can you make a proof showing that dry humor is associated with math? Just think about how exciting that would be.
"How exciting" Said Andy enthusiastically
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
Heron's formula, at least 😅
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.
You can literally find the area only using trigonometry and pythagorean theorem.
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!
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
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
Love your videos
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.
solved with law of sines, law of cosines, and trigonometric area formula! Simple and nice!
Pretty similar approach, I just used Pythagorean theorem to get the 6.
I solved this using area of parallelogram
Im learning SOLIDWORKS so I sketched this and then used the evaluate feature to find the area. 😂😂
Гарна задача, чудове рішення.
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.
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
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.
Should it be 1/2 × 6 × 3 = 9
Can you do a video explaining why you can find the area of a triangle using that formula?
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
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
hello andy!
sorry to bother you, but by looking at the formula -> (1/2)*a*b*sin(θ) shouldn't it be all divided by 2 like -> (a*b*sin(θ))/2 which is ≈5.2255..... the way you used the formula was like -> ((a*b)/2)*sin(θ)...... and if not why isn't it "(a/2)*b*sin(θ)"..... the absence of parentheses changes the final result......
i am just a "little" confused if its something that i dont now.... but let me know!
thanks a lot i like your videos very much😁!!
You're using 120 radians for the calculation instead of 120 degrees.
Solve using area of parallelogram
@@sohamjain8599explain how please
You’re mistaken; division is the same for each of these. If you have 10 sets of 5 apples and split them in half that is the same as splitting the 10 sets in half and saying they have 5 apples in each set. You don’t change the number of apples.
First time ever i can do qes like this without hint 🤑
I solved this using area of parallelogram
same
Yeah isn't it easier cause you don't need the angle?
Saaame much easier imo and relies on less theorems
I totally don't know what you mean by that
hi im confused about whyd you use that formula instead of pitagoras for the first triangle
Its a special right triangle, you dont have to solve that way
that formula is just a pattern used when given a specific scenario. It is possible to use pythagoras, but in this case it was possible to use prior knowledge of this specific scenario of a 30 60 90 triangle.
hi Andy sorry to point it out but sin 120 is not √3/2 . it is actually sin 60
I have this little confusion in my head, let's assume that that red triangle is half of a parallelogram in which the area is 18sqm so the area of that said triangle should be 9sqm because it's the half of it can someone please enlighten me 😅😅
(9sqrt(3))/2