@@Tab-pn5qw Not really, what I did was construct a perpendicular from the intersection to the base. Then looking only at the blue triangle, you can see the new smaller triangle to the right is congruent to the larger blue triangle in the ratio of 1:2, thus the height is 1.5u. The base (i.e. 4u) is divided in half since it's an isosceles triangle. thus one side is 2, so the area is 1.5u and you have 2 of those so 3u total for the shaded region. I assume others did it the same way.
@@Tab-pn5qw You can show that it is in the middle. Just like in the video, since the side angles are equal, then it is an isosceles triangle, so a line perpendicular to the base that passes through the other vertex divides it in half and is the height.
I thought the bit about staying hydrated was going to be a segue into a sponsored segment for LMNT or something and I just immediately started bracing myself
The diagonal of the rectangle approach is great. Another path: At 0:50 you establish that the triangle is isosceles with base angles theta, defined such that its tangent is 3/4 (since it is opposite the shortest side of the 3-4-5). We know that a line segment from the apex to the base perpendicularly bisects it (i.e., into two segments of length 2) and thus represents the triangle's height (true for all isosceles triangles). Hence the height is 2 * 3/4 or 1.5, and the area of the triangle is 1.5 * 4/2 = 3.
Advent Day 15 : R : radius of the semi-circle (and side length of the equilateral triangle too) r : radius of the circle inscribed in the equilateral triangle (Pi*r^2=8) x : radius of the other circle H : height of the equilateral triangle ; H=R*sin(60°)=sqrt(3)/2*R "other circle" is inscribed between a chord and an arch (center angle for the chord is 180°-60°=120°=2*60°) R=R*cos(60°)+2*x=R/2+2*x R-R/2=R/2=2*x R=4*x R^2=16*x^2 H=r/sin(60°/2)+r=r/(1/2)+r=3*r sqrt(3)/2*R=3*r R=sqrt(12)*r R^2=12*r^2 16*x^2=12*r^2 16*Pi*x^2=12*Pi*r^2=12*8=96 Pi*x^2=96/16=6 answer : area of the "other circle" = 6
Advent Day 22 : Overlapping Pink quadrilateral area=15 can be split into 2 right-angled triangles. One of these 2 right-angled triangle has area = 1/4*square area (=1/2*(side square)*(1/2*side square)) The other right-angled triangle has : one side = width of rectangle ; other side=1/2*length of rectangle Then, "other right-angled triangle" area = 1/4*rectangle area 15=1/4*(square area+rectangle area) 60=square area+rectangle area answer : total yellow area=square area + rectangle area - Pink area total yellow area=60-15=45
I dropped a vertical from the apex of the green area. This gave 2 congruent triangles which are similar to the original 3-4-5 (AAA) at a ratio 4:2 so the triangle sides became 1.5,2,2.5. Green area is then twice the area of the 2 congruent green triangles = twice area of 1 triangle. Area of one is 1/2.b.h so area of 2 triangles is twice area = b.h =2x1.5 =3. Green area is therefore 3 sq units.
Advent Day 18 : Point O : center of the circle Point A : Top Left vertex of the orange square Point B : Bottom Right vertex of the orange square = Bottom Left vertex of the pink square Point C : Top Left vertex of the pink square Point D : Top Right vertex of the pink square A, B, D are on the circle (C is inside the circle) a : side length of the orange square b : side length of the pink square r=4 : radius of the circle angle(ABD)=angle(ABC)+angle(CBD)=45°+45°=90° angle at the center = angle (AOD)=2*angle(ABD)=2*90°=180° Then : [AD] is a diameter of the circle ; AD=2*4=8 AD^2=(a+b)^2+(b-a)^2 AD^2=2*(a^2+b^2) 8^2=64=2*(a^2+b^2) 32=a^2+b^2 answer : Combined area of the 2 squares = a^2+b^2 = 32
Advent Day 24 : We can see that the 3 right-angled triangles are "similar" : they have the same angles. (a, 90°-a, 90°) Then, if the smaller right-angled triangle, where is inscribed the circle of diameter=3, has its smallest length equal to "b", then the other adjacent side is equal to : "4/3*b" and, then , the hypothénuse H equals to : H=sqrt(b^2+(4/3*b)^2)=5/3*b Diameter of inscribed circle=3 3=b+4/3*b-H 3=b+4/3*b-5/3*b 3=2/3*b b=9/2 H is the width of the rectangle. H=5/3*b=5/3*9/2=15/2 L is the length of the rectangle. Because of the similarities, L=4/3*H=4/3*15/2=10 area of rectangle=H*L=15/2*10 area of rectangle=75
Advent Day 30 : Triangle can be split into 2 right-angled triangles. H is the hypothenuse of the left triangle and h is the hypothenuse of the right triangle. Left triangle : (3+H-4)^2+4^2=H^2 then H=17/2 Right triangle : (2+h-4)^2+4^2=h^2 then h=5 area of the triangle=(3+H-4+2+h-4)*4/2 area of the triangle=(15/2+3)*4/2 area of the triangle=21
Advent Day 19 : Quarter Circle : r=radius ; area=Pi/4*r^2=Pi then r=2 Semi-circle : R=radius ; area=Pi/2*R^2=Pi then R=sqrt(2) a= side length of the square Diagonal of the square : sqrt(2)*a=r+sqrt(2)*R=2+sqrt(2)*sqrt(2)=4 a=4/sqrt(2) a^2=8 answer : area of the square=a^2=8
The bottom of the shaded area is 4. If you draw a vertical line from the apex of the shaded area, the left hand side triangle is similar to the blue triangle, hence the height is 1.5. The area is therefore 1/2 * 4 * 1.5, which is 3u^2.
Advent Day 23 : both the regular area=6 (a=side length of hexagons) 6=3/2*a*sqrt(3)*a 6=3*sqrt(3)/2*a^2 a^2=4/sqrt(3) L : length of rectangle ; w : width of rectangle L=7/2*a w=3*sqrt(3)/2*a Total shaded area = rectangle area - 2*6 =L*w-12 =21*sqrt(3)/4*a^2-12 =21*sqrt(3)/4*4/sqrt(3)-12 =21-12=9 Total shaded area =9
Advent Day 17 : Yellow equilateral triangle : side length=a ; area=sqrt(3)/4*a^2 Blue equilateral triangle : side length=b ; area=sqrt(3)/4*b^2 Green triangle area = 1/2*a*b*sin(180°-60°-60°)=sqrt(3)/4*a*b Pink equilateral triangle : side length=c ; area=sqrt(3)/4*c^2 Cosine law (al-kashi) : c^2=a^2+b^2-2*a*b*cos(60°) c^2=a^2+b^2-a*b Pink area + Green area = sqrt3)/4*(a^2+b^2-a*b)+sqrt(3)/4*a*b Pink area + Green area = sqrt(3)/4*(a^2+b^2) Pink area + Green area = Yellow area + Blue area = 100
Advent Day 25 : a=side length of each of the 6 purple squares Right Black square : horizontally : a+b+b+a=12 vertically : a-b+sqrt(2)*a=12 Solution of this system : a=9*(2-sqrt(2)) (and b=6-a but we don't care about b) Left Black square : Missing length=a*(2+sqrt(2)) Missing length=18
Advent Day 16 : r : radius of the circle inscribed in the equilateral triangle (Pi*r^2=16) R : radius of the circle inscribed in the rhombus H : Height of the equilateral triangle H=2*R=r/sin(60°/2)+r=r/(1/2)+r=3*r 4*R^2=9*r^2 Pi*R^2=9/4*Pi*r^2 Pi*R^2=9/4*16=36 answer : 36
The green triangle is isosceles because both corners are made from the counter-3 corner of a 3/4/5 triangle. The unshaded portion of the blue triangle is isosceles because the top angle is 180*-90*-counter3angle, and the bottom angle is 90*-counter3angle. Both of these isosceles triangles share a side, so the intersection point is in the middle of the blue hypotenuse. that means that its projection onto the blue "3" edge is at its midpoint, separating it into 2 sections of 3/2 each. This is also the height of the shaded triangle. So the shaded triangle has base 4 and height 3/2, and its area is 3.
Ventriloquism at its finest! I was surprised when you didn't start off like the other videos giving a formula to find what was asked, in this case the area of the triangle = 1/2bh, and then solve for the variables. These videos have shown how important it is to have that base knowledge of stuff in your toolbox and use critical thinking to apply them in isolation or in combination to solve a problem. Math and me don't agree, but I've come to really enjoy these videos.
Advent Day 28 : Angle at the centre of a circle is twice the angle at the circumference. Then, we can draw a quadrilateral with the sum of the angles is obviously equal to 360°. 90°+2*a+2*b+90°=360° a+b=90°
Advent Day 21 : (Yellow+Blue) area = 8^2*sqrt(3)/4 Yellow equilateral triangle : a=side length Yellow area=1/2*(Yellow+Blue) area=1/2*8^2*sqrt(3)/4 a^2*sqrt(3)/4=1/2*8^2*sqrt(3)/4 a^2=1/2*8^2 a=4*sqrt(2) Height of the Yellow equilateral triangle=R=a*sqrt(3)/2= radius of the semi-circle R=4*sqrt(2)*sqrt(3)/2=sqrt(24) R^2=24 area of the semi-circle=Pi/2*R^2 =Pi/2*24=12*Pi
Advent Day 12 : small yellow circle : radius=r and area=Pi*r^2=6 semi-circle radius = R Height of equilateral triangle = H H=R/sin(60°/2)=R+r+r/sin(60°/2) R/sin(30°)=R+r+r/sin(30°) 2*R=R+r+2*r R=3*r 1/2*Pi*R^2=1/2*Pi*(3*r)^2=9/2*Pi*r^2=9/2*6=27 answer : area of semi-circle = 27
The triangle whose side length is 2 has two angles measuring 40°, and hence this triangle is isosceles. The length of the other side of this triangle is equal to 2, which represents the length of the side of the large polygon. Therefore, the perimeter of the large polygon is equal to 2*13=26.
❤ if base angles of green triangle are t , t then angles of other small triangle in original triangle are 90 - t , 90 - t , 2 t hence it is also isosceles and left side of green triangle is a median of original triangle hence area of green triangle = (1/2)(1/2)(3)(4) = 3 sq units Second method hypotenuse of original triangle is a diameter of circumcircle and since green triangle is isosceles left side of green triangle is a radius of this circle and median of original triangle area of green triangle = 3 sq units (as in method 1)
Advent Day 13 : D : diameter of the semi-circle 1/2 * (D*sin(k))*cos(k) * (D*sin(k))*sin(k)=32 1/2 * (D*cos(k))*cos(k) * (D*cos(k))*sin(k)=8 Let's divide equation 1 by equation 2 below : (sin(k)/cos(k))^2=32/8=4 tan(k)=2 then cos(k)=1/sqrt(5) and sin(k)=2/sqrt(5) 1/2*(D*cos(k))*(D*sin(k))=8+32=40 D^2=2*40/(cos(k)*sin(k)) D^2=80/(1/sqrt(5)*2/sqrt(5))=80/(2/5)=80*5/2=200 area of semi-circle=1/2*Pi*D^2/4=Pi/8*200 area of semi-circle=25*Pi
Advent Day 14 : What fraction of the semi-circle is not covered by those 3 identical circles ? That is to say : 1-(area of the 3 white squares)/(semi-circle area) Let's decide that radius of semi-circle = 1 (then, semi-circle area=Pi/2) radius of each of the 3 circles = x We notice that center angle = 180° = 6*30° =6*arcsin(x/(1-x)) Indeed, we can draw 6 identical right-angled triangles in this semi-circle with : sin(30°)=1/2=x/(1-x) then x=1/3 area of the 3 white squares=3*Pi*x^2=3*Pi*(1/3)^2=Pi/3 1-(area of the 3 white squares)/(semi-circle area)=1-(Pi/3)/(Pi/2)=1-2/3=1/3 answer : 1/3
Advent Day 31 : w : width of the rectangle L=length of the rectangle ; L=6*radius=6*1=6 w^2+L^2=(w-1+L-1)^2 w^2+6^2=(w+4)^2 w=5/2 area of the rectangle=L*w=6*5/2=15
You could also after finding the mid point of the hypotenuse draw a parallel line from it and it would be half of 4 and it would also be height of the non shaded part of blue triangle with base 3. Then u get its area as half of 3x2 = 3. Then subtract it from area of the blur triangle half of 3x4 = 6. So u get 3 sq units
I would have used coordinate geometry, the blue hypotenuse follows the equation y = -3/4 x + 3, and the green hypotenuse follows the equation y = 3/4 x. Setting them equal to each other to find the intersection point gives 3 = 6/4 x, x = 2. At x = 2, y = 3/4 (2) = 3/2. You know the base of the triangle = 4, and we just worked out the perpendicular height is 3/2 (highest point of triangle is at y = 3/2) so the area = 1/2 x 4 x 3/2 = 3 units squared.
Hey Andy if you want to look at it another way, look at the isosceles triangle with the base of 4. Draw the height of the triangle with base 4, now you have two smaller triangles with base 2. They are the same ratio sides as the 3-4-5 triangles since you have theta and the 90. Set up the ratio and you get the height of the triangles as 1.5. 2 triangles * [ 1/2 * base(2) * height(1.5) ] = 3
I probably would have gone the route of turning the two crossing lines into slopes with y = -3/4 x + 3 and y = 3/4 x. And then taking the integral to get the area below.
i also found those 2 lines, but instead i found their intersection point since its y-coordinate is the height of the shaded triangle and we know the base is 4
@@Andres183_ Right. That is a bit easier than my way. You get -3/4 x + 4 = 3/4x, which turns to 6/4 x = 3 and that results in x = 2, which is exactly the half way on the baseline, which shows, that we have an isosceles triangle. Then you only need to input the x = 2 into one of the slope equations like y = 3/4 x = 3/4 * 2 = 3/2 and multiply with the base 3/2 * 1/2 * 4 = 3.
I'm posting this here at around 9:00am GMT, before the new day's vid comes out (just so no one complains that I wrote AFTER it came out, and sorry for sounding arrogant or whatever, not intentional) The new day's problem is simple angle seeking that gets us to the common side length of the hexagon and nonagon being exactly 2units. Angle seeking in the small triangle that has a side of the hexagon and another of the nonagon (which are equal by looking at the bottom side of both), so it's an isosceles triangle with vertex angle being the difference between the vertex angle of a nonagon and that of a hexagon (hexagon has total angles = 180×(6 - 2) = 720,deg for a regular hexagon each vertex angle = 720/6 = 120deg, and nonagon total angles = 180×(9 - 2) = 1260deg, each vertex angle = 1260/9 = 140deg) which is 140 - 120 = 20deg. Then the base angles would be 80deg each. Then we go for the triangle with a side of 2 and the hexagon side (the drawn extension without the part inside). The top angle = 180 - 80 = 100deg, the bottom left angle is 180 - (120 + 20) = 40deg (the 120 is the hexagon vertex angle). That makes the bottom right angle 180 - (100 + 40) = 40deg,making it an isosceles triangle. That makes the left side = the right side, hexagon side length = 2 = nonagon side length. Perimeter of the nonagon = 9 × side length = 9 × 2 = 18. And let's put a box around it.
Lol I used trig to find the angles inside the shaded triangle and then used that with sine rule to find out the other length and used that to find the area.
26 I have never felt so dumb when getting one of these right angle of a regular hexagon corner is 120, angle of a regular nonagon is 140, making the angle between them, 20 both shapes have the same length sides so they create an isosceles triangle, making the other two angles both 80 working on the triangle outside the nonagon: the exterior angle of the nonagon is 40 and the top angle of the triangle is 180-80=100 the remaining side must then be 40 then I used length of one side=(2sin40)/sin100 and the calculator said the length was 2, a nonagon has 9 sides so the perimeter is 18 and I am glad I typed this out because what I should have done was see that the bottom two angles were the same so it's an isosceles triangle again and the length of one side is 2 and then I need to realize that the shaded region is not the nonagon and is, in fact, bound by 13 lines, making the answer 26
This is the first one I was able to do completely on my own without using any hints from his video, and I ended up solving it with a completely different approach. Cool.
Could've gone another direction. If we label the blue 3-4-5 triangle vertices as A, B, and C (starting from the right angle vertex and going anticlockwise), then D, A, and E for the other 3-4-5 triangle (same naming as the first, note that 1 vertex is common with the first triangleand hence already named), then label the upper and last unnamed vertex of the shaded triangle as M. Instead of drawing a rectangle around and such, we can just find that
Advent Day 20 : Total shaded area = Pi/4*R^2-Pi*r^2=Pi/4*(R^2-4*r^2) We can see that : R^2=12^2+(2*r)^2 R^2-4*r^2=12^2 Total shaded area = Pi/4*12^2=36*Pi th-cam.com/video/a_rODCw_DSw/w-d-xo.html
I like this solution BUT you could have calculated the height of the shaded triangle since you can draw a line segment from the vertex angle to bisect the base and produce a triangle similar to the blue 3-4-5.
Fun fact: Since there are no accurate indicators on which sides of each triangle measure 3 or 4, and you just assingned the lengths base on 'what seemed to be the shortes/longest side', one could argue that there are actually 3 different answers depending on which length one applies to what sides of each right triangle (Area = 3; 2.16; or 3.84)
You know the base is 4. Since it is Isosceles triangle, the height splits base in half. The height will be h=2*tan([theta]) or h=2*tan(arcsin(3/5)) = 3
Advent Day 29 : 6=1/2*x*y*sin(a)=1/2*x*z*sin(b)=1/2*y*z*sin(c) area of red triangle=1/2*x*(2*y)*sin(180°-a)+1/2*(2*x)*z*sin(180°-b)+1/2*y*(2*z)*sin(180°-c)+6 area of red triangle=1/2*x*(2*y)*sin(a)+1/2*(2*x)*z*sin(b)+1/2*y*(2*z)*sin(c)+6 area of red triangle=2*6+2*6+2*6+6 area of red triangle=42
Anyone plz tell me how i am wrong After we have made out that the smaller triangle is an Isoceles triangle ABC with Smaller side(BC),base=3 Base angles= the smaller angle of a 345 right triangle, lets call it=x From the 345 triangle we can establish tan(x)=3÷4 In the isoceles triangle, Drop an altitude AD from the top vertex A to the base BC. Being an isoceles triangle altitude=median. Implying the joining point (D) bisects BC. i.e., BD=DC=3/2 Now the triangle ABD is a right triangle. With angle B= angle x, i.e., tanx=TanB We know tanx=3/4 Tan B=AD/BD 3/4=AD/(3/2) AD=9/8 Area of triangle ABC=1/2 basexheight =1/2 x BC x AD =1/2 x 3 x 9/8 =27/16. I am really confused how this doesnt correspond to Andy's answer
The way you drew the left triangle, it is not clear which side is 3 and which side is 4 If the bottom side is 4, the area is 4(3/2)/2 = 3 If the bottom side is 3, the area is (3)(3/5)(4)(3/5)/2 = (108/25)/2 = 54/25
Advent Day 10 : Area of the rectangle = (6/2)/cos(k) * 6*cos(k) = 6*6/2 = 18 th-cam.com/video/wARGaYjZJ0c/w-d-xo.html th-cam.com/video/OQAqKFBNxX4/w-d-xo.html
By the end of this, he'll have completely lost his mind.
how exciting.
@@tapio_m6861 LOL
ah, yes, new character
adding the comedic element really makes the video 10x more entertaining because your humor is deadpan and i love it
How exciting! (and the animation was very fun)
The aprox with the 1/4 of the area of the rectangle was pretty clever. I use congruent triangles to calculate de height of the green área.
That’s what I did also.
But then aren't you assuming that the intersection of both triangles is exactly in the middle of the 4cm length of the blue triangle?
@@Tab-pn5qw Not really, what I did was construct a perpendicular from the intersection to the base. Then looking only at the blue triangle, you can see the new smaller triangle to the right is congruent to the larger blue triangle in the ratio of 1:2, thus the height is 1.5u. The base (i.e. 4u) is divided in half since it's an isosceles triangle. thus one side is 2, so the area is 1.5u and you have 2 of those so 3u total for the shaded region. I assume others did it the same way.
@thisismalhar ah I see it now, thanks for the explanation!
@@Tab-pn5qw You can show that it is in the middle. Just like in the video, since the side angles are equal, then it is an isosceles triangle, so a line perpendicular to the base that passes through the other vertex divides it in half and is the height.
I thought the bit about staying hydrated was going to be a segue into a sponsored segment for LMNT or something and I just immediately started bracing myself
The diagonal of the rectangle approach is great. Another path: At 0:50 you establish that the triangle is isosceles with base angles theta, defined such that its tangent is 3/4 (since it is opposite the shortest side of the 3-4-5). We know that a line segment from the apex to the base perpendicularly bisects it (i.e., into two segments of length 2) and thus represents the triangle's height (true for all isosceles triangles). Hence the height is 2 * 3/4 or 1.5, and the area of the triangle is 1.5 * 4/2 = 3.
Very good consistency🎉🎉.Keep going and “LETS PUT A BOX AROUND IT”😂😂
how exciting
Advent Day 15 :
R : radius of the semi-circle (and side length of the equilateral triangle too)
r : radius of the circle inscribed in the equilateral triangle (Pi*r^2=8)
x : radius of the other circle
H : height of the equilateral triangle ; H=R*sin(60°)=sqrt(3)/2*R
"other circle" is inscribed between a chord and an arch
(center angle for the chord is 180°-60°=120°=2*60°)
R=R*cos(60°)+2*x=R/2+2*x
R-R/2=R/2=2*x
R=4*x
R^2=16*x^2
H=r/sin(60°/2)+r=r/(1/2)+r=3*r
sqrt(3)/2*R=3*r
R=sqrt(12)*r
R^2=12*r^2
16*x^2=12*r^2
16*Pi*x^2=12*Pi*r^2=12*8=96
Pi*x^2=96/16=6
answer : area of the "other circle" = 6
Advent Day 22 :
Overlapping Pink quadrilateral area=15 can be split into 2 right-angled triangles.
One of these 2 right-angled triangle has area = 1/4*square area (=1/2*(side square)*(1/2*side square))
The other right-angled triangle has :
one side = width of rectangle ; other side=1/2*length of rectangle
Then, "other right-angled triangle" area = 1/4*rectangle area
15=1/4*(square area+rectangle area)
60=square area+rectangle area
answer : total yellow area=square area + rectangle area - Pink area
total yellow area=60-15=45
I dropped a vertical from the apex of the green area. This gave 2 congruent triangles which are similar to the original 3-4-5 (AAA) at a ratio 4:2 so the triangle sides became 1.5,2,2.5. Green area is then twice the area of the 2 congruent green triangles = twice area of 1 triangle. Area of one is 1/2.b.h so area of 2 triangles is twice area = b.h =2x1.5 =3. Green area is therefore 3 sq units.
Gotta hydrate when solving math problems.
Advent Day 18 :
Point O : center of the circle
Point A : Top Left vertex of the orange square
Point B : Bottom Right vertex of the orange square = Bottom Left vertex of the pink square
Point C : Top Left vertex of the pink square
Point D : Top Right vertex of the pink square
A, B, D are on the circle (C is inside the circle)
a : side length of the orange square
b : side length of the pink square
r=4 : radius of the circle
angle(ABD)=angle(ABC)+angle(CBD)=45°+45°=90°
angle at the center = angle (AOD)=2*angle(ABD)=2*90°=180°
Then : [AD] is a diameter of the circle ; AD=2*4=8
AD^2=(a+b)^2+(b-a)^2
AD^2=2*(a^2+b^2)
8^2=64=2*(a^2+b^2)
32=a^2+b^2
answer : Combined area of the 2 squares = a^2+b^2 = 32
Advent Day 24 :
We can see that the 3 right-angled triangles are "similar" : they have the same angles. (a, 90°-a, 90°)
Then, if the smaller right-angled triangle, where is inscribed the circle of diameter=3, has its smallest length equal to "b", then the other adjacent side is equal to : "4/3*b" and, then , the hypothénuse H equals to :
H=sqrt(b^2+(4/3*b)^2)=5/3*b
Diameter of inscribed circle=3
3=b+4/3*b-H
3=b+4/3*b-5/3*b
3=2/3*b
b=9/2
H is the width of the rectangle.
H=5/3*b=5/3*9/2=15/2
L is the length of the rectangle. Because of the similarities,
L=4/3*H=4/3*15/2=10
area of rectangle=H*L=15/2*10
area of rectangle=75
Advent Day 30 :
Triangle can be split into 2 right-angled triangles.
H is the hypothenuse of the left triangle and h is the hypothenuse of the right triangle.
Left triangle : (3+H-4)^2+4^2=H^2 then H=17/2
Right triangle : (2+h-4)^2+4^2=h^2 then h=5
area of the triangle=(3+H-4+2+h-4)*4/2
area of the triangle=(15/2+3)*4/2
area of the triangle=21
Advent Day 19 :
Quarter Circle : r=radius ; area=Pi/4*r^2=Pi then r=2
Semi-circle : R=radius ; area=Pi/2*R^2=Pi then R=sqrt(2)
a= side length of the square
Diagonal of the square :
sqrt(2)*a=r+sqrt(2)*R=2+sqrt(2)*sqrt(2)=4
a=4/sqrt(2)
a^2=8
answer : area of the square=a^2=8
The bottom of the shaded area is 4. If you draw a vertical line from the apex of the shaded area, the left hand side triangle is similar to the blue triangle, hence the height is 1.5. The area is therefore 1/2 * 4 * 1.5, which is 3u^2.
Advent Day 23 :
both the regular area=6 (a=side length of hexagons)
6=3/2*a*sqrt(3)*a
6=3*sqrt(3)/2*a^2
a^2=4/sqrt(3)
L : length of rectangle ; w : width of rectangle
L=7/2*a
w=3*sqrt(3)/2*a
Total shaded area = rectangle area - 2*6
=L*w-12
=21*sqrt(3)/4*a^2-12
=21*sqrt(3)/4*4/sqrt(3)-12
=21-12=9
Total shaded area =9
Advent Day 17 :
Yellow equilateral triangle : side length=a ; area=sqrt(3)/4*a^2
Blue equilateral triangle : side length=b ; area=sqrt(3)/4*b^2
Green triangle area = 1/2*a*b*sin(180°-60°-60°)=sqrt(3)/4*a*b
Pink equilateral triangle : side length=c ; area=sqrt(3)/4*c^2
Cosine law (al-kashi) :
c^2=a^2+b^2-2*a*b*cos(60°)
c^2=a^2+b^2-a*b
Pink area + Green area = sqrt3)/4*(a^2+b^2-a*b)+sqrt(3)/4*a*b
Pink area + Green area = sqrt(3)/4*(a^2+b^2)
Pink area + Green area = Yellow area + Blue area = 100
Advent Day 25 :
a=side length of each of the 6 purple squares
Right Black square :
horizontally : a+b+b+a=12
vertically : a-b+sqrt(2)*a=12
Solution of this system : a=9*(2-sqrt(2)) (and b=6-a but we don't care about b)
Left Black square :
Missing length=a*(2+sqrt(2))
Missing length=18
Advent Day 16 :
r : radius of the circle inscribed in the equilateral triangle (Pi*r^2=16)
R : radius of the circle inscribed in the rhombus
H : Height of the equilateral triangle
H=2*R=r/sin(60°/2)+r=r/(1/2)+r=3*r
4*R^2=9*r^2
Pi*R^2=9/4*Pi*r^2
Pi*R^2=9/4*16=36
answer : 36
The green triangle is isosceles because both corners are made from the counter-3 corner of a 3/4/5 triangle.
The unshaded portion of the blue triangle is isosceles because the top angle is 180*-90*-counter3angle, and the bottom angle is 90*-counter3angle.
Both of these isosceles triangles share a side, so the intersection point is in the middle of the blue hypotenuse. that means that its projection onto the blue "3" edge is at its midpoint, separating it into 2 sections of 3/2 each. This is also the height of the shaded triangle.
So the shaded triangle has base 4 and height 3/2, and its area is 3.
Ventriloquism at its finest!
I was surprised when you didn't start off like the other videos giving a formula to find what was asked, in this case the area of the triangle = 1/2bh, and then solve for the variables. These videos have shown how important it is to have that base knowledge of stuff in your toolbox and use critical thinking to apply them in isolation or in combination to solve a problem. Math and me don't agree, but I've come to really enjoy these videos.
In math class they will made you prove the smallest side is 3 and not just because "it's smallest".
Unfortunately the way these puzzles are designed you have to make assumptions. But that's more like real life than the classroom, after all.
That shaded area for tomorrow looks like a afro, so don't forget the thetas and a semicircle there, too.
I want to see that one talk, too.
Advent Day 9 :
Perimeter of the shaded region = 26
I have this solution too.
I got the same answer!
My approach:
After proving isosceles, draw the perpendicular height of the triangle, makes base= 4 divide into 2 equal parts.
Then with tan(©) =3/4=h/2, we can get height, h=1.5
Area =0.5 * 4* 1.5 =3
Advent Day 26 :
Semi-circle : diameter=8 then radius=4
Shaded Area=4*4/2=8
Got to stay hydrated and (mathated)! Many thanks for an excellent puzzle vid.
Advent Day 28 :
Angle at the centre of a circle is twice the angle at the circumference.
Then, we can draw a quadrilateral with the sum of the angles is obviously equal to 360°.
90°+2*a+2*b+90°=360°
a+b=90°
Advent Day 21 :
(Yellow+Blue) area = 8^2*sqrt(3)/4
Yellow equilateral triangle : a=side length
Yellow area=1/2*(Yellow+Blue) area=1/2*8^2*sqrt(3)/4
a^2*sqrt(3)/4=1/2*8^2*sqrt(3)/4
a^2=1/2*8^2
a=4*sqrt(2)
Height of the Yellow equilateral triangle=R=a*sqrt(3)/2= radius of the semi-circle
R=4*sqrt(2)*sqrt(3)/2=sqrt(24)
R^2=24
area of the semi-circle=Pi/2*R^2
=Pi/2*24=12*Pi
Advent Day 12 :
small yellow circle : radius=r and area=Pi*r^2=6
semi-circle radius = R
Height of equilateral triangle = H
H=R/sin(60°/2)=R+r+r/sin(60°/2)
R/sin(30°)=R+r+r/sin(30°)
2*R=R+r+2*r
R=3*r
1/2*Pi*R^2=1/2*Pi*(3*r)^2=9/2*Pi*r^2=9/2*6=27
answer : area of semi-circle = 27
0:52 looks like they triangle got mad at him for calling its sides equal
with this and the 0^0 video, I gotta love when you engage in a little roleplay here and there. Keep up the good work, Andy!
The triangle whose side length is 2 has two angles measuring 40°, and hence this triangle is isosceles. The length of the other side of this triangle is equal to 2, which represents the length of the side of the large polygon. Therefore, the perimeter of the large polygon is equal to 2*13=26.
❤ if base angles of green triangle are t , t then angles of other small triangle in original triangle are 90 - t , 90 - t , 2 t hence it is also isosceles and
left side of green triangle is a median of original triangle
hence area of green triangle = (1/2)(1/2)(3)(4) = 3 sq units
Second method
hypotenuse of original triangle is a diameter of circumcircle and since green triangle is isosceles left side of green triangle is a radius of this circle and median of original triangle
area of green triangle = 3 sq units (as in method 1)
Advent Day 13 :
D : diameter of the semi-circle
1/2 * (D*sin(k))*cos(k) * (D*sin(k))*sin(k)=32
1/2 * (D*cos(k))*cos(k) * (D*cos(k))*sin(k)=8
Let's divide equation 1 by equation 2 below :
(sin(k)/cos(k))^2=32/8=4
tan(k)=2 then cos(k)=1/sqrt(5) and sin(k)=2/sqrt(5)
1/2*(D*cos(k))*(D*sin(k))=8+32=40
D^2=2*40/(cos(k)*sin(k))
D^2=80/(1/sqrt(5)*2/sqrt(5))=80/(2/5)=80*5/2=200
area of semi-circle=1/2*Pi*D^2/4=Pi/8*200
area of semi-circle=25*Pi
Advent Day 14 :
What fraction of the semi-circle is not covered by those 3 identical circles ?
That is to say : 1-(area of the 3 white squares)/(semi-circle area)
Let's decide that radius of semi-circle = 1 (then, semi-circle area=Pi/2)
radius of each of the 3 circles = x
We notice that center angle = 180° = 6*30° =6*arcsin(x/(1-x))
Indeed, we can draw 6 identical right-angled triangles in this semi-circle with :
sin(30°)=1/2=x/(1-x) then x=1/3
area of the 3 white squares=3*Pi*x^2=3*Pi*(1/3)^2=Pi/3
1-(area of the 3 white squares)/(semi-circle area)=1-(Pi/3)/(Pi/2)=1-2/3=1/3
answer : 1/3
Advent Day 31 :
w : width of the rectangle
L=length of the rectangle ; L=6*radius=6*1=6
w^2+L^2=(w-1+L-1)^2
w^2+6^2=(w+4)^2
w=5/2
area of the rectangle=L*w=6*5/2=15
You could also after finding the mid point of the hypotenuse draw a parallel line from it and it would be half of 4 and it would also be height of the non shaded part of blue triangle with base 3. Then u get its area as half of 3x2 = 3. Then subtract it from area of the blur triangle half of 3x4 = 6. So u get 3 sq units
Thats super neat, never thought of that. I love seeing cool geometric proofs appear out of the fog
I would have used coordinate geometry, the blue hypotenuse follows the equation y = -3/4 x + 3, and the green hypotenuse follows the equation y = 3/4 x. Setting them equal to each other to find the intersection point gives 3 = 6/4 x, x = 2. At x = 2, y = 3/4 (2) = 3/2. You know the base of the triangle = 4, and we just worked out the perpendicular height is 3/2 (highest point of triangle is at y = 3/2) so the area = 1/2 x 4 x 3/2 = 3 units squared.
Tomorrow's puzzle is either a trick question or I misread it. Watch where the arrow points.
Love your videos.
Thanks for making me laugh. That was so unexpected 😂
Hey Andy if you want to look at it another way, look at the isosceles triangle with the base of 4.
Draw the height of the triangle with base 4, now you have two smaller triangles with base 2. They are the same ratio sides as the 3-4-5 triangles since you have theta and the 90.
Set up the ratio and you get the height of the triangles as 1.5.
2 triangles * [ 1/2 * base(2) * height(1.5) ] = 3
1:06 it's a face
Yes, and it will talk to you when you are doing math and smoking crack at the same time.
I probably would have gone the route of turning the two crossing lines into slopes with y = -3/4 x + 3 and y = 3/4 x. And then taking the integral to get the area below.
i also found those 2 lines, but instead i found their intersection point since its y-coordinate is the height of the shaded triangle and we know the base is 4
@@Andres183_ Right. That is a bit easier than my way. You get -3/4 x + 4 = 3/4x, which turns to 6/4 x = 3 and that results in x = 2, which is exactly the half way on the baseline, which shows, that we have an isosceles triangle. Then you only need to input the x = 2 into one of the slope equations like y = 3/4 x = 3/4 * 2 = 3/2 and multiply with the base 3/2 * 1/2 * 4 = 3.
gosh i finally do it right, bro you rly made me use my brain again ,and thats feel good ,appreciate!
bro you really making riddle by simple math
I'm posting this here at around 9:00am GMT, before the new day's vid comes out (just so no one complains that I wrote AFTER it came out, and sorry for sounding arrogant or whatever, not intentional)
The new day's problem is simple angle seeking that gets us to the common side length of the hexagon and nonagon being exactly 2units.
Angle seeking in the small triangle that has a side of the hexagon and another of the nonagon (which are equal by looking at the bottom side of both), so it's an isosceles triangle with vertex angle being the difference between the vertex angle of a nonagon and that of a hexagon (hexagon has total angles = 180×(6 - 2) = 720,deg for a regular hexagon each vertex angle = 720/6 = 120deg, and nonagon total angles = 180×(9 - 2) = 1260deg, each vertex angle = 1260/9 = 140deg) which is 140 - 120 = 20deg.
Then the base angles would be 80deg each.
Then we go for the triangle with a side of 2 and the hexagon side (the drawn extension without the part inside).
The top angle = 180 - 80 = 100deg, the bottom left angle is 180 - (120 + 20) = 40deg (the 120 is the hexagon vertex angle).
That makes the bottom right angle 180 - (100 + 40) = 40deg,making it an isosceles triangle.
That makes the left side = the right side, hexagon side length = 2 = nonagon side length.
Perimeter of the nonagon = 9 × side length = 9 × 2 = 18.
And let's put a box around it.
I was struggling so I decided to go all out and use Calculus. Slope-intercept formula and some integration got me to 3 units².
Lol I used trig to find the angles inside the shaded triangle and then used that with sine rule to find out the other length and used that to find the area.
26
I have never felt so dumb when getting one of these right
angle of a regular hexagon corner is 120, angle of a regular nonagon is 140, making the angle between them, 20
both shapes have the same length sides so they create an isosceles triangle, making the other two angles both 80
working on the triangle outside the nonagon:
the exterior angle of the nonagon is 40 and the top angle of the triangle is 180-80=100
the remaining side must then be 40
then I used length of one side=(2sin40)/sin100 and the calculator said the length was 2, a nonagon has 9 sides so the perimeter is 18
and I am glad I typed this out because what I should have done was see that the bottom two angles were the same so it's an isosceles triangle again and the length of one side is 2
and then I need to realize that the shaded region is not the nonagon and is, in fact, bound by 13 lines, making the answer 26
0:54 it looks like an angry isosceles triangle, how exciting
edit: i commented just before the comedic relief xD
Semicircle guy NEEDS to make another appearance :D
This is the first one I was able to do completely on my own without using any hints from his video, and I ended up solving it with a completely different approach. Cool.
Could've gone another direction.
If we label the blue 3-4-5 triangle vertices as A, B, and C (starting from the right angle vertex and going anticlockwise), then D, A, and E for the other 3-4-5 triangle (same naming as the first, note that 1 vertex is common with the first triangleand hence already named), then label the upper and last unnamed vertex of the shaded triangle as M.
Instead of drawing a rectangle around and such, we can just find that
Advent Day 20 :
Total shaded area = Pi/4*R^2-Pi*r^2=Pi/4*(R^2-4*r^2)
We can see that :
R^2=12^2+(2*r)^2
R^2-4*r^2=12^2
Total shaded area = Pi/4*12^2=36*Pi
th-cam.com/video/a_rODCw_DSw/w-d-xo.html
Much easier than Heron's formula. Why do I always do things the hard way?
I like this solution BUT you could have calculated the height of the shaded triangle since you can draw a line segment from the vertex angle to bisect the base and produce a triangle similar to the blue 3-4-5.
I used the Heron’s formula as I got the length of the side of the triangle is equal and would be 2.5 and base is 4
Bro gave up a career in ventriloquism to bring us math instead. Thank you Andy
This is the first one where I really didn't know what to do... Nice theorem
Fun fact: Since there are no accurate indicators on which sides of each triangle measure 3 or 4, and you just assingned the lengths base on 'what seemed to be the shortes/longest side', one could argue that there are actually 3 different answers depending on which length one applies to what sides of each right triangle (Area = 3; 2.16; or 3.84)
we did it using tangent rule in right triangles to determine the height of the shaded triangle (3/2)
this one kinda took me by surprise
You know the base is 4. Since it is Isosceles triangle, the height splits base in half. The height will be h=2*tan([theta]) or h=2*tan(arcsin(3/5)) = 3
wonder if the new character introduced is canon to the channel or not
Thanks a lot Andy 🎉
How exciting!
I might have done this with 2 intersecting graphs, but your way is interesting
we need more math characters, please
Gold! I love it
Advent Day 29 :
6=1/2*x*y*sin(a)=1/2*x*z*sin(b)=1/2*y*z*sin(c)
area of red triangle=1/2*x*(2*y)*sin(180°-a)+1/2*(2*x)*z*sin(180°-b)+1/2*y*(2*z)*sin(180°-c)+6
area of red triangle=1/2*x*(2*y)*sin(a)+1/2*(2*x)*z*sin(b)+1/2*y*(2*z)*sin(c)+6
area of red triangle=2*6+2*6+2*6+6
area of red triangle=42
Anyone plz tell me how i am wrong
After we have made out that the smaller triangle is an
Isoceles triangle ABC with
Smaller side(BC),base=3
Base angles= the smaller angle of a 345
right triangle, lets call it=x
From the 345 triangle we can establish tan(x)=3÷4
In the isoceles triangle, Drop an altitude AD from the top vertex A to the base BC.
Being an isoceles triangle altitude=median. Implying the joining point (D) bisects BC.
i.e., BD=DC=3/2
Now the triangle ABD is a right triangle. With angle B= angle x,
i.e., tanx=TanB
We know tanx=3/4
Tan B=AD/BD
3/4=AD/(3/2)
AD=9/8
Area of triangle ABC=1/2 basexheight
=1/2 x BC x AD
=1/2 x 3 x 9/8
=27/16.
I am really confused how this doesnt correspond to Andy's answer
BC is 4 not 3 so BD and BC are 2 not 3/2. It seems you just switched the 3 and 4 sides on one of the triangles
Wow!!! ❤
waiting for 13 dec
its will be my birthday
Why does it look visually that 3u^2 is way too large?
for day 9 is EZ 13x2= 26
It’s 12:44 am, I have school tomorrow, I don’t care, I will keep watching your math videos
My solution is that the overlapping area is half the 3-4-5 triangle's area
The way you drew the left triangle, it is not clear which side is 3 and which side is 4
If the bottom side is 4, the area is 4(3/2)/2 = 3
If the bottom side is 3, the area is (3)(3/5)(4)(3/5)/2 = (108/25)/2 = 54/25
Day 8 of asking where you solve your problems. Like in google docs?
Stay hydrated.
Why tribangle angy 😔
The next day will be 14,42?
Advent Day 10 :
Area of the rectangle = (6/2)/cos(k) * 6*cos(k) = 6*6/2 = 18
th-cam.com/video/wARGaYjZJ0c/w-d-xo.html
th-cam.com/video/OQAqKFBNxX4/w-d-xo.html
Advent Day 11 :
angle = 135°
Same method and same result than Andy's previous video :
th-cam.com/video/RKYie9VHNDs/w-d-xo.html
Damn, you went from 0 to 100 in like 3 seconds. What's that rule about the squares again?
Wicked solve
Say "hello" to the little friend...
Are you spesial for geometri ?
New character!
(possible schizophrenia friend?)
🎉😊❤
He is losing his mind already!
Advent Day 27 :
Shaded Area = 5^2= 25
th-cam.com/video/GqrZ56j-rK8/w-d-xo.html
Who is Katrina Ag?
Solved the day 9 one it's 26 units😁
😂😂😂😂
LOL!!!
hi