Simpler solution: OA is || to DB. Triangles with the same base and vertex on a line parallel to the base have equal area. Hence, triangles ODB and ADB have equal areas. The area of ADB is 8*8/4 = 16 sq units.
@@hanswust6972Good observation! And since BDO will always be the same area, although BDO tends to be collinear as the radius of the circle increases, this can never be.
Calculating the Side Length of Square [ABCD] 1) Diagonal = 8 2) Any Square Diagonal = Side * sqrt(2) 3) 8 = Side * sqrt(2) 4) Side = 8 / sqrt(2) 5) Side = 8 * sqrt(2) / 2 6) Side = 4*sqrt(2) 7) Area of Square = [4*sqrt(2)]^2 = 16 * 2 = 32 If we draw a Straight Line passing the points O and A one can easily see that this Line is parallel to Diagonal Line BD = 8. The Square [OPAA'] (being A' the Point of Tangency between the Square and the Circle) has Side equal to Radius of the Circle = R Any Square adjacent to another Square, its Diagonal are Perpendicular to each other, in this case Diagonal OA is perpendicular to Diagonal AC. Conclusion: Diagonal OA is parallel to Diagonal BD, as we said before! And we this knowledge we can state that the distance between the Perpendicular Line of Line BD passing point O and is equal to the Semi Diagonal AC = 4 So, we can be sure that the Green Triangle Area is Half the Area of the Given Square [ABCD] A = (4 * 8) / 2 = 32 / 2 = 16 Answer: The Green Triangle Area is equal to 16 Square Units, wich is the same Area of Half Square [ABCD].
Solución general: El círculo es tangente a las alineaciones PA y DA → Su centro equidista de ambas alineaciones → Su centro siempre pertenecerá a la recta que partiendo del vértice "A" es paralela a la diagonal BD → Si consideramos que todos los triángulos posibles tienen como base la diagonal BD, su vértice "O"puede desplazarse hasta el vértice "A" sin variar la altura del triángulo y obtenemos siempre el triángulo de área equivalente ADB, cuya área es la mitad del cuadrado ABCD → Área verde =OBD =8²/4=16.
Extemely easy: Area of triangle; A = ½ b . h A = ½ 8 . 4 Á = 16 cm² ( Solved √ ) This figure is not defined, can't be drawn. Is missing radius of circle. This means We can choose this radius value I chose R=0 to simplify the calculations. Point O becomes point A.
1/ Because the distance from O to AD and AB= r so OA is the bisector of the right angle PAD. Therefore OA //DB so, the distance (or the height of the green triangle) from O to the base BD=8/2 =4 Area of the green triangle = 1/2 x4x8= 16 sq units
Since size of circle is not specified, we can calculate this quickly if there is no need to show work. For circle with radius = 0, point O = A, so triangle is half the area of square: Area(△OBD) = 1/2 * (d^2/2) = 1/4 * 8^2 = 16 -------- Since I haven't seen this used here, I'll use coordinate geometry to find solution for all cases. Let x = side length of square, r = radius of circle. x^2 + x^2 = 8^2 → 2x^2 = 64 → x^2 = 32 → *x = 4√2* Using coordinate geometry with A at origin, we get: A = (0, 0), B = (4√2, 0), C = (4√2, 4√2), D = (0, 4√2), O = (−r, r) Line BD: x + y = 4√2 Triangle OBD has base *b* = BD *= 8* and height h = Perpendicular distance from O (−r, r) to line BD (x + y − 4√2 = 0) *h =* |−r + r − 4√2| / √(1^2 + 1^2) *= 4* *Area(△OBD) = 1/2 * b * h = 1/2 * 8 * 4 = 16*
Area of the trapezoid=1/2(r+r+4√2)4√2=2√2(2r+4√2) Area of the ODE triangle=1/2(r)(4√2-r) Area of the OPB triangle=1/2(r)(r+4√2) Area of the green shaded region=Area of the trapezoid -(Area of the ODE triangle + Area the OPB triangle =4√2r+16-r^2/2-2√2r-2√2r+ r^2/2=16 square units .❤❤❤ Thanks.
Construct OA. Note that OA and BD are parallel, meeting AB at a 45° angle. Drop a perpendicular from O to BC. Note that the distance along this perpendicular from O to AD is the same as the distance from the intersection with the diagonal to side BC. Both equal the radius of the circle. Therefore, we can find that the distance from the center of the circle to the diagonal is the length of a side of the square. We have divided the green triangle into two triangles. Use the segment from O to the diagonal as a common base. The heights h₁ and h₂ sum to the side of the square. The area is (1/2)bh₁ for the two triangle, (1/2)bh₂, total area = (1/2)b(h₁ + h₂) = (1/2)bh. Both b and h are equal to the side of the square, which is 4√2. Total area = (1/2)(4√2)(4√2) = 16 square units.
The quickest solution : the inital conditions say nothing about the radius of the center. So, let"s set it to zero. The green shaded triangle becames half the square and the solution is trivial.
This is a tricky one! Since the center point of the circle on change of radius moves parallel to the BD line, the height of the triangle over the BD line never changes, thus its area stays the same. When i acknowledge that fact, i can immediately spot the height of that triangle by making the circle radius zero, shifting its edge O into A, giving height = 4, thus area = 4 * 8 /2 = 16. Just by this little fact, the task at hand goes from "somewhat complicated" to "outright trivial". Well: I still need a few more exercises like this to learn to immediately look at those principle laws.
Another solution: 1. Let r= radius of the circle; a= AD = AB = 8/sqrt(2)=4*sqrt(2); 2. Let us draw OM parallel to AB, point K is the crossing of DB and OM, M belongs to BC; 3. let b = OK 4. So our green triangle is split in two ODK and OBK, both have the same base b; 5. Area of ADK = b * (4*sqrt(2)-4)/2; Area of AKB = b*r/2; 6. Area of ADB = b * (4*sqrt(2)-r)/2 + b*r/2 = b/2*(4*sqrt(2)-r+r)= b*2*sqrt(2); 7. triangle AKM is an isosles right triangle with cathetus r => KM = r => b=a= 4*sqrt(2); 8. Let us return to the area of the green triangle: Agreen=b*2*sqrt(2)= 4*sqrt(2)*2*sqrt(2)=16 sq units.
I loved this problem. What i did was figure out the area of triangle DOP where P is the intersection of AD and OB. Then I figured out the area of ABP. Both are equal! A = 16*r/(r+4*2^1/2). Since this are equal. The area will always be A = 1/2*(4*2^1/2)^2 = 16. Interestingly, you can assume r = 0 so again A = 16. Premath, i thought your approach was too complicated so i thought this approach comparing the triangles was easier (for my simple mind). This is an amazing problem of stability and equilibrium. It took me about 6 attempts to get my arithmetic right. Thanks Premath for your brain puzzles. My high school math teacher Mr. Delbert Meitz would be so flattered I am working these problems. Keep supplying them.
Let _r_ be the radius of the circle and _s_ be the side of the square. When O = A then r = 0 and the area of the green triangle is ½ s². When O = E then r = s and the area of the green triangle is ½ s². But can we draw a conclusion from two particular cases? 🤔
Let OHE a line parallel to PB, H intersection with AD and E intersection with BD sOBD = sODE + sOBE = OE*HD/2 + OE*HA/2 = OE*(HD + HA)/2 = OE*AD/2 EB is parallel to OA because EBP = OAP = 45° So OE = AB sOBD = AB*AD/2 = sADB = (8/sqrt(2))**2/2 = 16 sOBD is always equal to sADB The radius of the circle does not matter
If radius of circle has not been defined then it will likely cancel out in the answer. So make the radius of the circle to any convenient value that makes the diagram much simpler, and see what the answer is. Obvious test values are to set it to either the same as the side length of the square or to zero. Easiest is to set the radius to zero and see what the diagram looks like then. The green shaded triangle is thus half of the square. If you set the radius to the same as side length then area (half base times height) has the same value.
8*8/2=32= OBD Desconocemos el radio pero sabemos que su valor está en el intervalo r=0 y r=AB. Para ambos valores, el resultado es la mitad del área del cuadrado y para cualquier otro valor intermedio. Gracias por el vídeo. Un saludo cordial.
I'm sure I would have got there eventually, but I spent half an hour writing total bollocks before watching the video :) Actually, I was on the right lines but went down a couple of dead ends before giving up. Thank you.
Umm...this is an interesting problem, but the radius of the circle doesn't matter (which is why it is not given). The area is half the area of the square.
Wow! Soooooo cool. 🙂
Thanks ❤️
Simpler solution: OA is || to DB. Triangles with the same base and vertex on a line parallel to the base have equal area. Hence, triangles ODB and ADB have equal areas. The area of ADB is 8*8/4 = 16 sq units.
And so you prove that r can be any size even longer than a.
@@hanswust6972Good observation! And since BDO will always be the same area, although BDO tends to be collinear as the radius of the circle increases, this can never be.
Very good.
Very good. It is simple method.
Yes, it is parallel.
But prove it.
Calculating the Side Length of Square [ABCD]
1) Diagonal = 8
2) Any Square Diagonal = Side * sqrt(2)
3) 8 = Side * sqrt(2)
4) Side = 8 / sqrt(2)
5) Side = 8 * sqrt(2) / 2
6) Side = 4*sqrt(2)
7) Area of Square = [4*sqrt(2)]^2 = 16 * 2 = 32
If we draw a Straight Line passing the points O and A one can easily see that this Line is parallel to Diagonal Line BD = 8.
The Square [OPAA'] (being A' the Point of Tangency between the Square and the Circle) has Side equal to Radius of the Circle = R
Any Square adjacent to another Square, its Diagonal are Perpendicular to each other, in this case Diagonal OA is perpendicular to Diagonal AC. Conclusion: Diagonal OA is parallel to Diagonal BD, as we said before!
And we this knowledge we can state that the distance between the Perpendicular Line of Line BD passing point O and is equal to the Semi Diagonal AC = 4
So, we can be sure that the Green Triangle Area is Half the Area of the Given Square [ABCD]
A = (4 * 8) / 2 = 32 / 2 = 16
Answer:
The Green Triangle Area is equal to 16 Square Units, wich is the same Area of Half Square [ABCD].
Solución general: El círculo es tangente a las alineaciones PA y DA → Su centro equidista de ambas alineaciones → Su centro siempre pertenecerá a la recta que partiendo del vértice "A" es paralela a la diagonal BD → Si consideramos que todos los triángulos posibles tienen como base la diagonal BD, su vértice "O"puede desplazarse hasta el vértice "A" sin variar la altura del triángulo y obtenemos siempre el triángulo de área equivalente ADB, cuya área es la mitad del cuadrado ABCD → Área verde =OBD =8²/4=16.
Extemely easy:
Area of triangle;
A = ½ b . h
A = ½ 8 . 4
Á = 16 cm² ( Solved √ )
This figure is not defined, can't be drawn.
Is missing radius of circle.
This means We can choose this radius value
I chose R=0 to simplify the calculations. Point O becomes point A.
Other convenient value for R is:
R = Side of square
R = 8/√2 = 4√2 cm
A = ½ b.h
A = ½ (4√2)²
A = 16 cm² (Solved √ )
Other convenient value for R is:
R = ½ Side of square
R = 4/√2 = 2√2 cm
A = ½ b.h
A = ½ 8 .(2√2√2) = ½ 8 . 4
A = 16 cm² (Solved √ )
@@marioalb9726 With these three special cases, we can strongly suspect that the radius of the circle doesn't matter. All that remains is to prove it 😉
1/ Because the distance from O to AD and AB= r so OA is the bisector of the right angle PAD. Therefore OA //DB
so, the distance (or the height of the green triangle) from O to the base BD=8/2 =4
Area of the green triangle = 1/2 x4x8= 16 sq units
I learn new things from here everyday.
Since size of circle is not specified, we can calculate this quickly if there is no need to show work.
For circle with radius = 0, point O = A, so triangle is half the area of square:
Area(△OBD) = 1/2 * (d^2/2) = 1/4 * 8^2 = 16
--------
Since I haven't seen this used here, I'll use coordinate geometry to find solution for all cases.
Let x = side length of square, r = radius of circle.
x^2 + x^2 = 8^2 → 2x^2 = 64 → x^2 = 32 → *x = 4√2*
Using coordinate geometry with A at origin, we get:
A = (0, 0), B = (4√2, 0), C = (4√2, 4√2), D = (0, 4√2), O = (−r, r)
Line BD: x + y = 4√2
Triangle OBD has base *b* = BD *= 8* and height
h = Perpendicular distance from O (−r, r) to line BD (x + y − 4√2 = 0)
*h =* |−r + r − 4√2| / √(1^2 + 1^2) *= 4*
*Area(△OBD) = 1/2 * b * h = 1/2 * 8 * 4 = 16*
Area of the trapezoid=1/2(r+r+4√2)4√2=2√2(2r+4√2)
Area of the ODE triangle=1/2(r)(4√2-r)
Area of the OPB triangle=1/2(r)(r+4√2)
Area of the green shaded region=Area of the trapezoid -(Area of the ODE triangle + Area the OPB triangle
=4√2r+16-r^2/2-2√2r-2√2r+ r^2/2=16 square units .❤❤❤ Thanks.
Construct OA. Note that OA and BD are parallel, meeting AB at a 45° angle. Drop a perpendicular from O to BC. Note that the distance along this perpendicular from O to AD is the same as the distance from the intersection with the diagonal to side BC. Both equal the radius of the circle. Therefore, we can find that the distance from the center of the circle to the diagonal is the length of a side of the square. We have divided the green triangle into two triangles. Use the segment from O to the diagonal as a common base. The heights h₁ and h₂ sum to the side of the square. The area is (1/2)bh₁ for the two triangle, (1/2)bh₂, total area = (1/2)b(h₁ + h₂) = (1/2)bh. Both b and h are equal to the side of the square, which is 4√2. Total area = (1/2)(4√2)(4√2) = 16 square units.
Love it!
Amazing how the radius length is not a factor here.
Beautiful sharing❤❤❤❤
Marvelous!!! 👍👍👍👍❤️🔥
Glad to hear that!
Thanks ❤️
Best solution...
The quickest solution : the inital conditions say nothing about the radius of the center. So, let"s set it to zero. The green shaded triangle becames half the square and the solution is trivial.
Enjoyed the video, a bit of surprise at the end the way it simplifies
This is a tricky one! Since the center point of the circle on change of radius moves parallel to the BD line, the height of the triangle over the BD line never changes, thus its area stays the same. When i acknowledge that fact, i can immediately spot the height of that triangle by making the circle radius zero, shifting its edge O into A, giving height = 4, thus area = 4 * 8 /2 = 16.
Just by this little fact, the task at hand goes from "somewhat complicated" to "outright trivial".
Well: I still need a few more exercises like this to learn to immediately look at those principle laws.
I enlarged the radius to the length of the square:
b(triangle) = 8/√2
h(triangle) = 8/√2
A(triangle) = 1/2 * 64/2 = 1/2 * 32 = 16 square units
Another solution:
1. Let r= radius of the circle;
a= AD = AB = 8/sqrt(2)=4*sqrt(2);
2. Let us draw OM parallel to AB, point K is the crossing of DB and OM, M belongs to BC;
3. let b = OK
4. So our green triangle is split in two ODK and OBK, both have the same base b;
5. Area of ADK = b * (4*sqrt(2)-4)/2;
Area of AKB = b*r/2;
6. Area of ADB = b * (4*sqrt(2)-r)/2 + b*r/2 = b/2*(4*sqrt(2)-r+r)= b*2*sqrt(2);
7. triangle AKM is an isosles right triangle with cathetus r => KM = r => b=a= 4*sqrt(2);
8. Let us return to the area of the green triangle: Agreen=b*2*sqrt(2)= 4*sqrt(2)*2*sqrt(2)=16 sq units.
Nice!
I loved this problem. What i did was figure out the area of triangle DOP where P is the intersection of AD and OB. Then I figured out the area of ABP. Both are equal! A = 16*r/(r+4*2^1/2). Since this are equal. The area will always be A = 1/2*(4*2^1/2)^2 = 16. Interestingly, you can assume r = 0 so again A = 16. Premath, i thought your approach was too complicated so i thought this approach comparing the triangles was easier (for my simple mind). This is an amazing problem of stability and equilibrium. It took me about 6 attempts to get my arithmetic right. Thanks Premath for your brain puzzles. My high school math teacher Mr. Delbert Meitz would be so flattered I am working these problems. Keep supplying them.
Let _r_ be the radius of the circle and _s_ be the side of the square.
When O = A then r = 0 and the area of the green triangle is ½ s².
When O = E then r = s and the area of the green triangle is ½ s².
But can we draw a conclusion from two particular cases? 🤔
Let OHE a line parallel to PB, H intersection with AD and E intersection with BD
sOBD = sODE + sOBE = OE*HD/2 + OE*HA/2 = OE*(HD + HA)/2 = OE*AD/2
EB is parallel to OA because EBP = OAP = 45°
So OE = AB
sOBD = AB*AD/2 = sADB = (8/sqrt(2))**2/2 = 16
sOBD is always equal to sADB
The radius of the circle does not matter
Just set R=0. The area of the triangle is equal to half the square...
Thanks ❤️
Could you elaborate
If radius of circle has not been defined then it will likely cancel out in the answer. So make the radius of the circle to any convenient value that makes the diagram much simpler, and see what the answer is. Obvious test values are to set it to either the same as the side length of the square or to zero. Easiest is to set the radius to zero and see what the diagram looks like then. The green shaded triangle is thus half of the square. If you set the radius to the same as side length then area (half base times height) has the same value.
Assume or prove that the problem has only one solution. Of these two options, which one is preferable? 🤔
@@ybodoN hmm... I don't think you have to prove that, because the area obviously has a defined value for a given triangle.
8*8/2=32= OBD
Desconocemos el radio pero sabemos que su valor está en el intervalo r=0 y r=AB.
Para ambos valores, el resultado es la mitad del área del cuadrado y para cualquier otro valor intermedio.
Gracias por el vídeo. Un saludo cordial.
Excellent!
You are very welcome!
Thanks ❤️
OA angle 45 degree, so triangle height is 4.
8 x 4 / 2 = 16
Thanks ❤️
Impressive teacher. I thought was missing some date... kkkkkkkk Very good!!!
(8)^2 =64. (8)^2 =64 {64+64}=128 {128-180°}=√52° √4^√13 2^2√13^1 √2^2 √1^√1 1^2 (x+1x-2)
Surprising!😵💫
I'm sure I would have got there eventually, but I spent half an hour writing total bollocks before watching the video :)
Actually, I was on the right lines but went down a couple of dead ends before giving up.
Thank you.
area of triangle depends on r values
If I have a question where can I show you it because I can’t find the solution 😭
the triangle abd is equal to the green one
Umm...this is an interesting problem, but the radius of the circle doesn't matter (which is why it is not given). The area is half the area of the square.
0
Thanks ❤️
Another way to solve this problem: th-cam.com/users/shortsYfmYsLoYE8k
ABCD est un carré ???? ce n'est pas dit ! Bonne journée !
C'est dit dans les cinq premières secondes de la vidéo 😉
@@ybodoNAh ! oui ! je ne regarde que le croquis, je réfléchis et ne regarde la video qu'après !!! 🤗
Il le dit au début de la vidéo!