As a pretend engineer, I knew that the tan 60 was sqrt(3) - so that made it easy. But I am glad you showed this method as well - thanks for all your posts. $2 on the way!
Yay I solved this one pretty quickly. I skipped the pythagorean calculations to get root3 because I instantly saw the central right triangle was 2x on the hypotenuse and x on the shortest side, meaning it's a 30-60-90 triangle (hypotenuse is always double the shortest length), so the length of 3 is equivalent to root3 x, therefore x is 3/root3 which simplifies to root3.
One at end is simpler than it appears, and a fun ride too. The base of the new triangle is 4, and the congruent angles make it isosceles. Half the base is 2, and this is the long leg of a scaled-down 3-4-5 so the height is 1.5. The area is (4 * 1.5)/2 = 3.
Here is an alternative to the two 3-4-5 triangles problem using coordinate geometry: Place the origin at the left lower corner and the blue triangle has 3 vertices (0, 0), (4, 0) and (0, 3). Step 1: Find the equation of the blue hypotenuse. The blue hypotenuse is the line connecting (4, 0) and (0, 3), so its slope = -3/4, and its equation is: y = -3/4 • x + 3 Step 2: Find the equation of the longer side of the green triangle. Given the angle between the green line and the blue base = atan(3/4), the slope of the green line = tan(atan(3/4)) = 3/4, and its equation is: y = 3/4 • x Step 3: Find the intersection of the two lines by solving simultaneous equations. The intersection = (1.125, 1.5). So, we know the shaded triangle has vertices (0, 0), (4, 0), and (1.125, 1.5). Step 4: Directly calculate the area. The area = 1/2 • || 0 0 || || 4 0 || ||1.125 1.5|| = 1/2 • 6 = 3 Although this method may appear a bit slower, it is still super fast, simple, and powerful. It can be generalized for any two overlapping triangles with any side lengths, even though they may not be right triangles! 😎😎
I got A=3 using a different method. the shaded triangle is isosceles, and would be the same if the green 345 triangle were oriented with the "4" side colinear with the blue triangles "4" side. From this point it's easy to intuit the area of the shaded triangle is 1/4 the area of a 3x4 rectangle. I'm glad we both got the same answer!
I originally got four, but after reading your comments I redid the same math that I done before, Henry is that at some point since I did all the math in my head, I got the answer wrong. I did use a valid method to get the cancer originally, but I just messed up when calculating the height of shaded triangle.
The fun part about tomorrow's puzzle: not only is the shaded area an isosceles triangle, if you bisect it (draw a vertical line cutting it in half), it makes 2 congruent right triangles, with sides 1.5, 2 and 2.5 -- half-sized 3-4-5 triangles! Anyways, the height is 1.5, the base is 4, so the area is 3 square units.
I cut the image in half and looked at the right half as a 30-60-90 triangle in a quarter circle, where the height is rt3*base. If the height is 3 (radius of semicircle, the distance), the base is 3*(1/rt3), which simplifies to rt3. I then brought back the rest and imagined it as a semicircle on a graph, where the equation is sqrt(9-x^2). if x is the distance of rt3 we just found, the height where the square touches is sqrt(9-[rt3]^2). That is just sqrt(6), and the area of the square is just that but squared. It's almostexactly what you did, but you proved each step by pythagorean theorem
3 the highlighted triangle is an isosceles one with base 4 the two identical angles are sin^-1(3/5) split the isosceles triangle in half vertically to get a right angles triangle and tan(sin^-1(3/5))=height/2 calculator says tan(sin^-1(3/5)) is 0.75 so 0.75=height/2 therefore height=1.5 area=(basexheight)/2 area=(4*1.5)/2 area=3 am not doing too badly, am 5/7 correct so far
If you flip the r8ght hand traingle so they share the 4 as the base and the area in question stays the same, you have a quarter of a 3 by 4 rectangle shaded, so its 3.
I realised I could avoid the trig by the observation that two right triangles mirrored left to right will intersect on the diagonals at exactly half way, as you're essentially drawing a rectangle and connecting opposite diagonals. Because the angle is the same if you flip one of the triangles upside down, then the length of intersecting point is the same for the left of the two triangles, and therefore 5/2. You can then use either Pythagoras or the area of the isosceles triangle formula to work out the area. Edit to add that I got the same answer you did.
Solution to the 8th day of the Aggvent Calendar: The Green triangle is ABC. B is the left corner of the shaded triangle and D is the right corner of the shaded triangle and lastly O is the top point of the shaded triangle or the intersection between the two sides of the green and blue triangle. The Blue triangle is EBD. Now start solving, Now,OB is the height of triangle EBD. Therefore,O is the midpoint of ED. In EBD, EB = 3 BD = 4 ED = 5 Therefore,0.5×ED=2.5 OE = OD = 2.5 units. Now,the base of the shaded triangle is BD or 4 units. Take a midpoint F on EB. Therefore,OF will be the height of triangle EBO. OF = 0.5×BD = 2 units. Since,F is the midpoint of side EB = 3 units. Therefore,EF = BF = 1.5 units. In triangle OFB, Let OB = x. Using Pythagorean theorem, (OF)²+(BF)² = x² OF = 2 units. BF = 1.5 units. 2²+(1.5)² = x² (Write in fractions) 4+(9/4) = x² x² = 25/4 x = 5/2 x = 2.5 units. Therefore,OB = 2.5 units. In triangle OBD, Side OB = 2.5 units. Side OD = 2.5 units. Side BD = 4 units. Plug in the values in Heron's Formula. Therefore, The area of shaded part is 3 units² Ans:Area is 3 units²
I'm so happy that I got one of these correct, although, I only used a fraction of the math you did so I can't be sure its not just a coincidence that I got 3 units²
A lot of mistakes, dude: 1) OB is NOT the height of triangle EBD 2) Even if it were, the height of a triangle does not divide the base into 2 equal parts, this only occurs if the triangle is isosceles and the base is the different side. 3) You assumed that OF will be the height of triangle EBO without knowing for sure. You can't assume that, even if this time you're right (luckily) 4) I can't believe you used Heron's formula at the end The answer is indeed 3, and it is much simpler than it seems: just make a triangle similarity! - Note that triangle OBD is isosceles - The height will divide the base into 2 parts of 2 units. - The new triangles formed are similar to triangles 3, 4 and 5. - Therefore, h/2 = 3/4 (h is the height of the shaded triangle) Therefore h = 3/2 and A = 3.
I noticed the area of the square was the same as the diameter of the semicircle, and was curious if that was a coincidence or not. I generalized the diameter as 2r, and worked through the same steps using r instead of 3. The area of the square: s² = r² - (r²/3). It is a coincidence. When r is 3, s² just happens to be r.
Solution: The height of an equilateral triangle is √3/2 times its base, so: 3 = √3/2 * a |*2/√3 a = 6/√3 = 2 * √3 * √3/√3 = 2√3 But we only need a/2 = √3 for the next part Thales Theorem tells us, that the triangle formed from the diameter of the half circle and the top left corner of the sqare is a right angle triangle. With the geometric mean theorem, we now have: h² = p * q where p = 3 + √3 and q = 3 - √3, therefore: h² = (3 + √3)(3 - √3) h² = 3² - (√3)² h² = 9 - 3 h² = 6 Which is what we were looking for, as h is one side of the square.
this is cool, but the problem's text should be more detailed, there is not enough information to deduce the area of the square, you needed to make a bunch of assumptions based on the image, which are not always correct.
OK; you are good at math but ... do you know an advent calendar does not have 31 days ?🤣😁 I hope you correct that error and give us a big problem for christmas day ! this would be ... exciting ! 😀😀
This is way better than the usual advent calendar with candy
c(Andy)
I hope this Pythagoras guy is collecting royalties
As a pretend engineer, I knew that the tan 60 was sqrt(3) - so that made it easy. But I am glad you showed this method as well - thanks for all your posts. $2 on the way!
Yay I solved this one pretty quickly. I skipped the pythagorean calculations to get root3 because I instantly saw the central right triangle was 2x on the hypotenuse and x on the shortest side, meaning it's a 30-60-90 triangle (hypotenuse is always double the shortest length), so the length of 3 is equivalent to root3 x, therefore x is 3/root3 which simplifies to root3.
Thanks!
One at end is simpler than it appears, and a fun ride too.
The base of the new triangle is 4, and the congruent angles make it isosceles. Half the base is 2, and this is the long leg of a scaled-down 3-4-5 so the height is 1.5. The area is (4 * 1.5)/2 = 3.
I was half way through brute forcing it when I realized that it’s just a quarter of the 3-4 rectangle
Hey spoilers
@@gdash.dasher-y7e Don't look then! :D
There are actually 3 different answers depending on what sides of each triangle you consider to be either 3 or 4
Ah, finally a geometry problem I can solve by myself before watching the video 😆
Here is an alternative to the two 3-4-5 triangles problem using coordinate geometry:
Place the origin at the left lower corner and the blue triangle has 3 vertices (0, 0), (4, 0) and (0, 3).
Step 1: Find the equation of the blue hypotenuse.
The blue hypotenuse is the line connecting (4, 0) and (0, 3), so its slope = -3/4, and its equation is:
y = -3/4 • x + 3
Step 2: Find the equation of the longer side of the green triangle.
Given the angle between the green line and the blue base = atan(3/4), the slope of the green line = tan(atan(3/4)) = 3/4, and its equation is:
y = 3/4 • x
Step 3: Find the intersection of the two lines by solving simultaneous equations.
The intersection = (1.125, 1.5).
So, we know the shaded triangle has vertices (0, 0), (4, 0), and (1.125, 1.5).
Step 4: Directly calculate the area. The area
= 1/2 • || 0 0 ||
|| 4 0 ||
||1.125 1.5||
= 1/2 • 6
= 3
Although this method may appear a bit slower, it is still super fast, simple, and powerful. It can be generalized for any two overlapping triangles with any side lengths, even though they may not be right triangles! 😎😎
interesting solution
I got A=3 using a different method. the shaded triangle is isosceles, and would be the same if the green 345 triangle were oriented with the "4" side colinear with the blue triangles "4" side. From this point it's easy to intuit the area of the shaded triangle is 1/4 the area of a 3x4 rectangle.
I'm glad we both got the same answer!
There are actually 3 different answers depending on what sides of each triangle you consider to be either 3 or 4
@@P4RH3L10N point but I was too lazy so I just referred the diagram
Area of shaded triangle = 3 unit square
I originally got four, but after reading your comments I redid the same math that I done before, Henry is that at some point since I did all the math in my head, I got the answer wrong. I did use a valid method to get the cancer originally, but I just messed up when calculating the height of shaded triangle.
There are actually 3 different answers depending on what sides of each triangle you consider to be either 3 or 4
man, i hope these videos stay online for my kids as they grow up.
The fun part about tomorrow's puzzle:
not only is the shaded area an isosceles
triangle, if you bisect it (draw a vertical
line cutting it in half), it makes 2 congruent
right triangles, with sides 1.5, 2 and 2.5 --
half-sized 3-4-5 triangles!
Anyways, the height is 1.5, the base is 4,
so the area is 3 square units.
the base is 4 and the height is 1.5. the area of a triangle is base times height divided by 2. so the shaded area should be 3
There are actually 3 different answers depending on what sides of each triangle you consider to be either 3 or 4
If you live streamed yourself solving these I would 100% tune in
I cut the image in half and looked at the right half as a 30-60-90 triangle in a quarter circle, where the height is rt3*base. If the height is 3 (radius of semicircle, the distance), the base is 3*(1/rt3), which simplifies to rt3. I then brought back the rest and imagined it as a semicircle on a graph, where the equation is sqrt(9-x^2). if x is the distance of rt3 we just found, the height where the square touches is sqrt(9-[rt3]^2). That is just sqrt(6), and the area of the square is just that but squared. It's almostexactly what you did, but you proved each step by pythagorean theorem
3
the highlighted triangle is an isosceles one with base 4
the two identical angles are sin^-1(3/5)
split the isosceles triangle in half vertically to get a right angles triangle and tan(sin^-1(3/5))=height/2
calculator says tan(sin^-1(3/5)) is 0.75
so 0.75=height/2 therefore height=1.5
area=(basexheight)/2
area=(4*1.5)/2
area=3
am not doing too badly, am 5/7 correct so far
If you flip the r8ght hand traingle so they share the 4 as the base and the area in question stays the same, you have a quarter of a 3 by 4 rectangle shaded, so its 3.
I realised I could avoid the trig by the observation that two right triangles mirrored left to right will intersect on the diagonals at exactly half way, as you're essentially drawing a rectangle and connecting opposite diagonals. Because the angle is the same if you flip one of the triangles upside down, then the length of intersecting point is the same for the left of the two triangles, and therefore 5/2. You can then use either Pythagoras or the area of the isosceles triangle formula to work out the area. Edit to add that I got the same answer you did.
Oh damn, didn't even notice I was *this* early
got it right before seeing the solution im proud of myself
love the videos, keep it up!
After pausing the video at 3:00 (the 3 4 5 triangles) and visualizing it in my head I think the answer is 3.
Great content 👏
Each side of the triangle = 2√3
Each side of the square = √[(3^2 - (√3)^2] = √6
Area of the square = 6
Solution to the 8th day of the Aggvent Calendar:
The Green triangle is ABC.
B is the left corner of the shaded triangle and D is the right corner of the shaded triangle and lastly O is the top point of the shaded triangle or the intersection between the two sides of the green and blue triangle.
The Blue triangle is EBD.
Now start solving,
Now,OB is the height of triangle EBD.
Therefore,O is the midpoint of ED.
In EBD,
EB = 3
BD = 4
ED = 5
Therefore,0.5×ED=2.5
OE = OD = 2.5 units.
Now,the base of the shaded triangle is BD or 4 units.
Take a midpoint F on EB.
Therefore,OF will be the height of triangle EBO.
OF = 0.5×BD = 2 units.
Since,F is the midpoint of side EB = 3 units.
Therefore,EF = BF = 1.5 units.
In triangle OFB,
Let OB = x.
Using Pythagorean theorem,
(OF)²+(BF)² = x²
OF = 2 units.
BF = 1.5 units.
2²+(1.5)² = x²
(Write in fractions)
4+(9/4) = x²
x² = 25/4
x = 5/2
x = 2.5 units.
Therefore,OB = 2.5 units.
In triangle OBD,
Side OB = 2.5 units.
Side OD = 2.5 units.
Side BD = 4 units.
Plug in the values in Heron's Formula.
Therefore,
The area of shaded part is 3 units²
Ans:Area is 3 units²
I'm so happy that I got one of these correct, although, I only used a fraction of the math you did so I can't be sure its not just a coincidence that I got 3 units²
A lot of mistakes, dude:
1) OB is NOT the height of triangle EBD
2) Even if it were, the height of a triangle does not divide the base into 2 equal parts, this only occurs if the triangle is isosceles and the base is the different side.
3) You assumed that OF will be the height of triangle EBO without knowing for sure. You can't assume that, even if this time you're right (luckily)
4) I can't believe you used Heron's formula at the end
The answer is indeed 3, and it is much simpler than it seems: just make a triangle similarity!
- Note that triangle OBD is isosceles
- The height will divide the base into 2 parts of 2 units.
- The new triangles formed are similar to triangles 3, 4 and 5.
- Therefore, h/2 = 3/4 (h is the height of the shaded triangle) Therefore h = 3/2 and A = 3.
@@wubleu_w same here use trigonometry in rough sheet
6 unit^2
edit: niceee ✔
1:20 why x is half the distance?
I love this guy
I think the next one is six has the area looks like half of the other triangle but IDK how to proof if the the line is infact cutting it equally.
You are a genius
I noticed the area of the square was the same as the diameter of the semicircle, and was curious if that was a coincidence or not.
I generalized the diameter as 2r, and worked through the same steps using r instead of 3.
The area of the square:
s² = r² - (r²/3).
It is a coincidence.
When r is 3, s² just happens to be r.
I have not skipped a video so far
Will a playlist be made for this along with the website?
Good idea. I made a playlist for it!
are the triangles congruent or similar for day8?
edit:
SPOILER
nvm the answer is 3 unit^2 by simple trig
Solution:
The height of an equilateral triangle is √3/2 times its base, so:
3 = √3/2 * a |*2/√3
a = 6/√3 = 2 * √3 * √3/√3 = 2√3
But we only need a/2 = √3 for the next part
Thales Theorem tells us, that the triangle formed from the diameter of the half circle and the top left corner of the sqare is a right angle triangle.
With the geometric mean theorem, we now have:
h² = p * q
where p = 3 + √3 and q = 3 - √3, therefore:
h² = (3 + √3)(3 - √3)
h² = 3² - (√3)²
h² = 9 - 3
h² = 6
Which is what we were looking for, as h is one side of the square.
I *LOVE* all these problems. But it's 2:00 a.m., are you falling behind?
Oh wow. Tomorrow's is way too easy
Day 7 of asking where you solve your problems. Do you solve them in google slides/docs?
That was easy peasy
day 8
right triangle aera is 3x4/2=6
→A=6/2=3😊
Shouldn’t it be day 8?
why did you square root x? it was going to be squared in the pythagorean theorem anyway
dude simping with Cat Agg
Nice
U could've just used the H = (L x sqrt3)/2
this is cool, but the problem's text should be more detailed, there is not enough information to deduce the area of the square, you needed to make a bunch of assumptions based on the image, which are not always correct.
Everything flowed from the regular triangle being centered, which was a given per the problem’s accompanying text.
-(tan(30°)-3)+3
=6
r = 6/2 = 3 = h
a = 2h/√3 = 3 · 2 / √3 = 2√3
b² = (2r - a)/2 · [(2r - a)/2 + a]
= (r - a/2) · (r + a/2)
= (3 - √3) · (3 + √3)
= 9 - 3
= 6.
Hi
OK; you are good at math but ... do you know an advent calendar does not have 31 days ?🤣😁 I hope you correct that error and give us a big problem for christmas day ! this would be ... exciting ! 😀😀
First❤
Next problem: A = 3
Correct
Fun fact: no one has finished watching the video in it's entirety as of right now
Edit: and now they could have
I watched it in 2x ez
Then you didn't watch half of the video @@Ludaris1
Watched in 600x speed ez
Didn't watch it. Solved it mentally just by looking at the thumbnail ez
first