One small comment I would make at 3:38 regarding your solving for (3*sqrt5)^2 We already knew that (3*sqrt5) is the value of r, so we can rewrite that as r^2, which we already knew is 45 from when we solved for r back at 1:07 in the video. So there was really no need to go through that just to, essentially, retrace our earlier steps.
All you need to know is it's two squares inside a perfect arc. The only way to get the two squares to intersect the arc ate those three vertices is if the side length of the small square is 1/2 the large square. The large square has to be 6x6 so the small one must be 3x3.
Without doing any calculation, if you divide the main square in four smaller squares, each with side=3 and then you rotate the image by 90 degrees you see by symmetry that if the yellow square has side=3 you have exactly the same picture where the small blue square on the bottom and the yellow square on top are exactly half the size of the big blue square. So that is the only option, yellow square with side=3 like the smaller blue squares.
You can solve this problem by drawing the radius to the corner of the yellow square, and slide that radius segment to the right and find that it will touch the corner and mid-point of the blue square. Voila, the side of the small square is 3 or half of the larger square, thus its area is 1/4 of the larger square.
You'd have to assume the drawing is to scale, which they don't have to be in these sort of problems. But if you add the fact its circumscribed in the semi circle then you can get there without the Pythagorean theorem.
Ok, the large length has side length 6. So it's top-left corner is at (x, y) = (-3, 6). Therefore the circle radius is r = sqrt(3^2 + 6^2) = sqrt(45) Let s be the side length of the small square. We need to satisfy (s+3)^2 + s^2 = 45 to put the top left corner of the small square on the circle. So, 2*s^2 + 6*s - 36 = 0 The roots are 3 and -6; the -6 can't be a side length, so s = 3. That means the area of the small square is 9.
Using the coordinates, it's easy to see. Suppose the origine at the center of the circle and (x0 , y ) , (x1 , y) two points of square touching the circle. Then x0*x0 + y*y = x1*x1 + y*y (circle equation). So x0 = x1 or x0 = -x1.
Side of small =3 hence area 3^2 =9 Did somewhat a similar problem in the past. The length of the blue square = 6 since its area = 36 Finding the radius of the semi-circle: Draw a straight line from the center of the semi-circle (also the center of the blue square) to the edge of the blue square, which touches the circumference of the semi-circle. A triangle will be formed with sides 6 and 3. Hence the circle's radius is sqrt (6^2 + 3^2) or sqrt 45. The radius of the semi-circle is sqrt 45. Draw a straight line from the semi-circle's center to the top edge of the yellow square, forming another triangle. Let the side of the yellow square = P, then the sides of the triangles are sqrt 45, P, and P + 3 ( The 3 = half the length of the blue square) Since sqrt = the hypotenuse of the triangle, then (sqrt 45)^2 = P^2 + (P+3)^2 45 = P^2 + P^2 + 6P + 9 0 = 2P^2 + 6P - 36 0 = P^2 + 3P - 18 { divide both sides by 2 } 0 = (P+ 6)(P-3) P =3 And P= -6, since the length cannot be -6, the length of P= 3 Side of small =3
Solution: The larger square has side length a=√36=6.The radius r of the semicircle can then be calculated using the Pythagorean theorem as the hypotenuse from the center of the semicircle to the upper left corner of the large square: r=√(6²+3²)=√45. Another right triangle: Hypotenuse: from the center of the semicircle to the upper left corner of the small square with side length b: r=√45 vertical leg: b horizontal leg: b+3 Theorem of Pythagoras: b²+(b+3)² = 45 ⟹ b²+b²+6b+9 = 45 |-45 ⟹ 2b²+6b-36 = 0 |/2 ⟹ b²+3b-18 = 0 |p-q formula ⟹ b1/2 = -1,5±√(1,5²+18) = -1,5±4,5 b1 = -1,5+4,5 = 3 und b2 = -1,5-4,5 = -6[invalid in this problem because it is negative] the area of the small square = b² = 3² = 9
В вашей геометрии по всей видимости не хватат правил равенства треугольников, как треугольники равны если у них равна одна сторона и противолежащий угол. Сразу становится очевидно решение.
it's more fun if you generalize. Put y = 3. Then (x+y)**2 + x**2 = 5y**2. So x = y . Note x = -2y is the other solution that would represent the blue square itself or his mirror copy in respect of the horizontal line.
If the middle point of the semicircle is also the middle point of the blue square’s side, the question should’ve mentioned so. Other than that, great video!
That much is understood. It is by definition impossible to completely inscribe one symmetrical figure within another symmetrical figure without maintaining symmetry.
The área of the small square ever is 1/4 the area of the central square area= 1/4 Area area = 36/4 = 9 cm² You don't need to complicate it, this is a theorem !!! Too easy !!!
x^2 + (x+3)^2 = 3^2 + 6^2 by observation, 3 is at least one of the root(s) of x So the area is 9. To see if there are other solutions, one may solve the quadratic equation
The perpendicular bisector of any chord always travels through the centre of the circle. Since it cuts the chord exactly in half, it also cuts the base of blue square exactly in half at the centre of the circle. As a side note, you can use any two chords and their perpendicular bisectors to find the centre of a circle, but in construction it is sensible to use three in case of error.
36=6*6 side of blue square :6 radius of semicircle : √[3²+6²]=√45=3√5 side of smaller square = x x²+(x+3)²=(3√5)² x²+x²+6x+9=45 2x²+6x-36=0 x²+3x-18=0 (x+6)(x-3)=0 x=-6 , 3 x>0 , thus x=3 3*3=9 area of the smaller square : 9
Using the geometric layout, finding the solution is easier: Through a 90º rotation around the center of the circumference, the small square comes to occupy the upper right quadrant of the large square, that is, we obtain a surface of ¼*36=9. The reasoning is as follows: If we apply the aforementioned rotation to the lower right vertex, its new position will be the center of the large square → Now we apply the rotation to the diagonal that starts from that lower right vertex and to the radius whose end coincides with that of the previous diagonal. → The new position of the intersection point of both rotated lines necessarily coincides with the upper right vertex of the large square. → The two points obtained after the previous steps define the ends of a segment of length equal to the diagonal of the small square and also of length equal to ½ of the diagonal of the large square → Area of the small square = (1/2)² (large square area)= ¼*36=9. Thanks and greetings to all
Exactly! Rotate the drawn diameter and the blue square together for 90 degrees CCW and the diameter will now exactly bisect the original placement of the blue square. So the yellow square's area must be exactly 1/4th of the blue square. So much simpler because they are squares. The algrebra and trig would be needed if they were rectangles, though.
Otra posibilidad por la simetría era asumir que estaba a la mitad del cuadrado su altura y se hallaba el lado del cuadrado amarillo, claro que en otra oportunidad de no ser así no se asume directamente
When you assume the big square is centre of the semi Circle the side of the small square will be half of big square. Hence it's side will be 3 units. Therefore the area of small square will be 9 sq. units.
Why calculate r? Simply, in blue triangle, r^2 =6^2 + 3^2 .....1 In the other triangle, r^2 =(x+3)^2 + x^2 ......2 Equating ...1 n .. .2, 45 = 2x^2 + 6x +9 x^2 + 3x - 18 = 0 Solve to get x = 3, -6 of which only 3 is valid.
R^2 = a^2 + ( a/2)^2 = 5 a^2/4 R^2 = ( b +a/2 ) ^2 + b^2 5 a^2 / 4 = 2 b^2 + a^2 /4 + b a 2 b^2 + b a - a^2 = 0 ( b + a) ( 2 b - a) = 0 , i.e. b = a /2
Somethin I don't get the original picture has the area of the blue square = to 36. 36 what it does not say. Is it sq inches sq ft yards? Also what is a sq unit? If I am not mistaken the area of a square =4xthe side Therefore the length of 1 side is 9 .4x9=36 4x6=24 not 36. or am I missing something?
If the area of the big square is 36, then the smaller square in the picture is 1/4 that, so it's 9. Because it's only half the height of the other square.. and each successive square arranged in the white space between these 2 squares would be 1/4 the previous one... You don't need any amazing math knowledge... The semicircle doesn't even matter.
This is a ridiculous example of "show your steps" because going from a somewhat complicated 36 area find the other square area to breaking down a simple thing like (x+3)^2 or rooting 45 and squaring the result just to show it's still 45 is just a waste
My solution (of course way faster): Let's imagine first that we don't have a square but just a rectangle. We can see that if the point on the circle moves, the lengths of the two sides of the rectangle move on opposite directions. Therefore there is only one position where we obtain a square. That being said, let's draw a 3x3 square in the right top angle of the blue square and let's a apply a rotation of 90° to the left, centered on the center of the circle. We know that we have a square. The image of the bottom side is the bottom half of the left side of the blue square. We also know that the point on the circle is still on the circle. Which means that the image of the 3x3 square is the yellow square! (since we know that only one of the rectlangles that can be drawn the same way is a square) Then the area is 9.
Wow, all of that to find out that the side of the small square is equal to 3? You can look at it and know immediately that it is half of 6. Took about 2 seconds to figure out.
تمرين جميل. رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين
I don't think there was any need to break down the square root of 45. You end up having to square the square root of 45, and you end up with 45 again.
You get 3root5 naturally if you factor 3^2 out of 3^2*2^2+3^2 like I did lol.
Agreed.. just to the left of all that extra work was written "r^2 = 45"
One small comment I would make at 3:38 regarding your solving for (3*sqrt5)^2
We already knew that (3*sqrt5) is the value of r, so we can rewrite that as r^2, which we already knew is 45 from when we solved for r back at 1:07 in the video.
So there was really no need to go through that just to, essentially, retrace our earlier steps.
Yes, there was no need to change sqrt 45 to 3 sqrt 9
Since (sqrt 45)^2 = 45, and thus easier to see.
All you need to know is it's two squares inside a perfect arc. The only way to get the two squares to intersect the arc ate those three vertices is if the side length of the small square is 1/2 the large square. The large square has to be 6x6 so the small one must be 3x3.
if we continue with another square in this pattern would the sides be 0.75 x 0.75? thus being a quarter of 3?
@@captainhamburgers8571 No
Without doing any calculation, if you divide the main square in four smaller squares, each with side=3 and then you rotate the image by 90 degrees you see by symmetry that if the yellow square has side=3 you have exactly the same picture where the small blue square on the bottom and the yellow square on top are exactly half the size of the big blue square. So that is the only option, yellow square with side=3 like the smaller blue squares.
You can solve this problem by drawing the radius to the corner of the yellow square, and slide that radius segment to the right and find that it will touch the corner and mid-point of the blue square.
Voila, the side of the small square is 3 or half of the larger square, thus its area is 1/4 of the larger square.
You'd have to assume the drawing is to scale, which they don't have to be in these sort of problems. But if you add the fact its circumscribed in the semi circle then you can get there without the Pythagorean theorem.
Też tak zrobiłem, obrót o 90st rozwiązuje problem bez liczenia
Ok, the large length has side length 6. So it's top-left corner is at (x, y) = (-3, 6). Therefore the circle radius is
r = sqrt(3^2 + 6^2) = sqrt(45)
Let s be the side length of the small square. We need to satisfy
(s+3)^2 + s^2 = 45
to put the top left corner of the small square on the circle. So,
2*s^2 + 6*s - 36 = 0
The roots are 3 and -6; the -6 can't be a side length, so s = 3. That means the area of the small square is 9.
Very well explained and illustrated. I just don't see the point or purpose of splitting the sqr(45) into 3(sqr(5)). Any explanation for that?
I see eye to eye with you. I think the explanation is a lenght of the video.
By just taking a quick look at it, I was able to tell that the area was 9.
Great, thank you for watching.
well you kinda have to prove it
And the scales are not correctly designed?
How? The diagram is not necessarily to scale.
Is it generally true that the smaller square must be of sides half the larger square?
It's a best teaching !
How did you know the centre of the circle is at the midpoint of the square ?
Draw the hypotenuse from the center of semicircle to the right upper corner of the blue square, then it’ll be clear the bottom parts are equal.
Using the coordinates, it's easy to see. Suppose the origine at the center of the circle and (x0 , y ) , (x1 , y) two points of square touching the circle. Then x0*x0 + y*y = x1*x1 + y*y (circle equation). So x0 = x1 or x0 = -x1.
It just is. In order for a square to be inscribed in a semicircle it has to be in the center.
Side of small =3 hence area 3^2 =9
Did somewhat a similar problem in the past.
The length of the blue square = 6 since its area = 36
Finding the radius of the semi-circle:
Draw a straight line from the center of the semi-circle (also the center of the blue square) to the edge of the blue square, which touches the circumference
of the semi-circle. A triangle will be formed with sides 6 and 3. Hence the circle's radius is sqrt (6^2 + 3^2) or sqrt 45.
The radius of the semi-circle is sqrt 45.
Draw a straight line from the semi-circle's center to the top edge of the yellow square, forming
another triangle. Let the side of the yellow square = P, then the sides of the
triangles are sqrt 45, P, and P + 3 ( The 3 = half the length of the blue square)
Since sqrt = the hypotenuse of the triangle, then
(sqrt 45)^2 = P^2 + (P+3)^2
45 = P^2 + P^2 + 6P + 9
0 = 2P^2 + 6P - 36
0 = P^2 + 3P - 18 { divide both sides by 2 }
0 = (P+ 6)(P-3)
P =3 And P= -6, since the length cannot be -6, the length of P= 3
Side of small =3
Wonderful.❤❤
Great mind analysis challenge. At age 85, I can still anticipate steps to solve.
Ego enhancement.
Can we not just infer that the yellow is just a quarter of the blue, since both squares are within the context of a single semicircle?
This was indeed amazing congratulations
Solution:
The larger square has side length a=√36=6.The radius r of the semicircle can then be calculated using the Pythagorean theorem as the hypotenuse from the center of the semicircle to the upper left corner of the large square:
r=√(6²+3²)=√45.
Another right triangle:
Hypotenuse:
from the center of the semicircle to the upper left corner of the small square with side length b: r=√45
vertical leg: b
horizontal leg: b+3
Theorem of Pythagoras:
b²+(b+3)² = 45 ⟹
b²+b²+6b+9 = 45 |-45 ⟹
2b²+6b-36 = 0 |/2 ⟹
b²+3b-18 = 0 |p-q formula ⟹
b1/2 = -1,5±√(1,5²+18) = -1,5±4,5
b1 = -1,5+4,5 = 3 und b2 = -1,5-4,5 = -6[invalid in this problem because it is negative]
the area of the small square = b² = 3² = 9
Side of the blue square: a = √36 = 6
Radius of the semicircle: r = √(6² + (6/2)²) = √45 = 3√5
Side of the yellow square: b² = r² - (b + 3)²
b² = 45 - b² - 6b - 9
2b² + 6b - 36 = 0
b² + 3b - 18 = 0
b₁,₂ = - 1,5 ± √ (2,25 + 18)
b₁,₂ = - 1,5 ± 4,5
b₁ = - 6 ∨ b₂ = 3
b can't be negative ⇒ b = 3
Area of the yellow square: A = b² = 3² = 9.
В вашей геометрии по всей видимости не хватат правил равенства треугольников, как треугольники равны если у них равна одна сторона и противолежащий угол. Сразу становится очевидно решение.
note x=-6 corresponds to the blue square.
negative because its width is in the opposite direction to the way x is drawn.
it's more fun if you generalize. Put y = 3. Then (x+y)**2 + x**2 = 5y**2. So x = y . Note x = -2y is the other solution that would represent the blue square itself or his mirror copy in respect of the horizontal line.
If the middle point of the semicircle is also the middle point of the blue square’s side, the question should’ve mentioned so. Other than that, great video!
Good point, thank you.
Not really. You can easily proof this must be the case by drawing two radii and a perpendicular line to the top side of the square.
That much is understood. It is by definition impossible to completely inscribe one symmetrical figure within another symmetrical figure without maintaining symmetry.
do you know what is the geometrical solution for x=-6 ???
It would be the mirror copy of the blue square in respect of the horizontal line.
The área of the small square ever is 1/4 the area of the central square
area= 1/4 Area
area = 36/4 = 9 cm²
You don't need to complicate it, this is a theorem !!!
Too easy !!!
Chord theorem
One chord is through upper yelow side, the other through right yellow side:
(6 - x)*(6 + x) = x*(6 +x),
i.e. x = 3
It's a rule and a common sense that the area of the small square must be 1/4 of the big square under such situation.
x^2 + (x+3)^2 = 3^2 + 6^2
by observation, 3 is at least one of the root(s) of x
So the area is 9. To see if there are other solutions, one may solve the quadratic equation
Why is the center of the circle the middle of the square side?
The perpendicular bisector of any chord always travels through the centre of the circle. Since it cuts the chord exactly in half, it also cuts the base of blue square exactly in half at the centre of the circle.
As a side note, you can use any two chords and their perpendicular bisectors to find the centre of a circle, but in construction it is sensible to use three in case of error.
How could it be anything else? Think it through.
36=6*6 side of blue square :6
radius of semicircle : √[3²+6²]=√45=3√5
side of smaller square = x x²+(x+3)²=(3√5)²
x²+x²+6x+9=45 2x²+6x-36=0 x²+3x-18=0
(x+6)(x-3)=0 x=-6 , 3 x>0 , thus x=3 3*3=9
area of the smaller square : 9
Blue square side = 6. Yellow side half that = 3. Area = 9.
(a+3)^2+a^2=45 2a^2+6a-36=0 a=(-6+-18)/4=3;-6 a=3 S=9
Dude... you can put 4 yellow squares inside the big green. Of course that square has 9 as area.
Using the geometric layout, finding the solution is easier:
Through a 90º rotation around the center of the circumference, the small square comes to occupy the upper right quadrant of the large square, that is, we obtain a surface of ¼*36=9.
The reasoning is as follows: If we apply the aforementioned rotation to the lower right vertex, its new position will be the center of the large square → Now we apply the rotation to the diagonal that starts from that lower right vertex and to the radius whose end coincides with that of the previous diagonal. → The new position of the intersection point of both rotated lines necessarily coincides with the upper right vertex of the large square. → The two points obtained after the previous steps define the ends of a segment of length equal to the diagonal of the small square and also of length equal to ½ of the diagonal of the large square → Area of the small square = (1/2)² (large square area)= ¼*36=9.
Thanks and greetings to all
Exactly! Rotate the drawn diameter and the blue square together for 90 degrees CCW and the diameter will now exactly bisect the original placement of the blue square. So the yellow square's area must be exactly 1/4th of the blue square. So much simpler because they are squares. The algrebra and trig would be needed if they were rectangles, though.
Nice solution
Or, one can see the box is 1/2 the size of 36 so the area will be 1/4 of 36, or 9
What about:
a = 1/2 Pi×r² - A
= 1/2 (Pi×45) - 36
Otra posibilidad por la simetría era asumir que estaba a la mitad del cuadrado su altura y se hallaba el lado del cuadrado amarillo, claro que en otra oportunidad de no ser así no se asume directamente
When you assume the big square is centre of the semi Circle the side of the small square will be half of big square. Hence it's side will be 3 units.
Therefore the area of small square will be 9 sq. units.
That doesn't require an assumption. A square inscribed in a semicircle will by definition be in the center.
Why calculate r?
Simply, in blue triangle,
r^2 =6^2 + 3^2 .....1
In the other triangle,
r^2 =(x+3)^2 + x^2 ......2
Equating ...1 n .. .2,
45 = 2x^2 + 6x +9
x^2 + 3x - 18 = 0
Solve to get x = 3, -6
of which only 3 is valid.
R^2 = a^2 + ( a/2)^2 = 5 a^2/4
R^2 = ( b +a/2 ) ^2 + b^2
5 a^2 / 4 = 2 b^2 + a^2 /4 + b a
2 b^2 + b a - a^2 = 0
( b + a) ( 2 b - a) = 0 , i.e. b = a /2
Somethin I don't get the original picture has the area of the blue square = to 36. 36 what it does not say. Is it sq inches sq ft yards? Also what is a sq unit? If I am not mistaken the area of a square =4xthe side Therefore the length of 1 side is 9 .4x9=36 4x6=24 not 36. or am I missing something?
?area=36/4=9 cm ^2
seems kind of pointless to breakdown the √45 to 3√5 when you turned around and remade it back into 45. would have been faster to (√45)² = 45.
If the area of the big square is 36, then the smaller square in the picture is 1/4 that, so it's 9. Because it's only half the height of the other square.. and each successive square arranged in the white space between these 2 squares would be 1/4 the previous one...
You don't need any amazing math knowledge... The semicircle doesn't even matter.
Muito bom!!!!
Power of a point!
x*(6+x) =(6-x)(6+x) x=3 A =9
🙏🙏🙏🙏🙏🙏🙏
36scrooth 6+6=12×12=144+36=179scrooth=13.3790881603÷2=6.6895440801×6.6895440801=44.75×3.14159268=140.5862724294÷2=70.2931362147half round aria
这种题目不用算，大方块的面积是36，那么小方块的面积就等于9。
❤
Did anybody else just did 36/4 = 9?
Not sound idea because you are assuming that the little square's side is crossing the big square's side right in the middle which is not given.
A=9,la relazione per trovare il lato l è:(l+3)^2+l^2=45, cioè l=3
Thank you for the feedback, I really appreciate it.
My initial estimate is 9. Let's see...
Update: OK
36 / 4 = 9 very easy
Area=9sq. unit
This is a ridiculous example of "show your steps" because going from a somewhat complicated 36 area find the other square area to breaking down a simple thing like (x+3)^2 or rooting 45 and squaring the result just to show it's still 45 is just a waste
9.
My solution (of course way faster):
Let's imagine first that we don't have a square but just a rectangle. We can see that if the point on the circle moves, the lengths of the two sides of the rectangle move on opposite directions. Therefore there is only one position where we obtain a square.
That being said, let's draw a 3x3 square in the right top angle of the blue square and let's a apply a rotation of 90° to the left, centered on the center of the circle. We know that we have a square. The image of the bottom side is the bottom half of the left side of the blue square. We also know that the point on the circle is still on the circle.
Which means that the image of the 3x3 square is the yellow square! (since we know that only one of the rectlangles that can be drawn the same way is a square)
Then the area is 9.
It means that those two triangles are congruent. Interesting!
Since corresponding sides of congruent triangles are equal, Therefore x + 6 = 3.
DAYUMN
9
answer 9, 6*6=36 3*3=9 I like nonsense. math answer. sin?=cos(90-?) ?=30 r*cos30=6 rsin30=3 =3 x*x=9
please get some one eles to speake zo yu kan eksplain it eh
Wow, all of that to find out that the side of the small square is equal to 3? You can look at it and know immediately that it is half of 6. Took about 2 seconds to figure out.
But that's not mathematics, is it?
cord theorem
(6-x)(6+x)=x(6+x)
therefore(6-x)=x
x=3
Voila
tedious
I really can't stand it anymore, this person's calculation skills are too bad! Simple calculations are so complicated
unbelievable tripe ..you can tell it's 9 just by looking at the diagram.
But that's not mathematics, is it?
The answer was obvious without doing all that proof.
No need to explain those 45 if you got it before already 🤷♂️
Simply measure one side of the small square and X by itself . Never mind all this Bolloxology
'measure' 🤣🤣
🙏🙏🙏🙏🙏🙏🙏
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