Working out the sides and the hypotenuse of triangle ABC was easy. 4:46 was the point at which I stopped and realized that extending the upper side of the green triangle across the yellow one, then up to meet point C, generated a second right-angled triangle with the same hypotenuse as the first and with side-lengths (a+b) and (a-b). The rest is basic algebra.
I find it astonishing that we can work out the combined area of both squares without knowing the side length of either of them! The power of mathematics! It's also curious that the combined area of the squares turns out to be exactly twice the area of the isosceles triangle.
Let the side of the green square be 0. Then the diagonal of the yellow square is equal to the hypotenuse of the triangle. Then the area of the square is twice the area of the triangle. So 2*88 is 176.
I thought the problem was to find the area of the green square and the area of the yellow square. I got to the point a^2 + b^2 = 176, but got stuck there. Little did I know that 176 is the answer to what the actual problem is, the combined area of the green and yellow squares.
The problem statement implies that the angle of the hypotenuse of the isosceles right triangle does not change the combined area, as long as it produces a valid figure. So, the solution to a special case should be the same as that for the general case. From the area, we find that the AB and BC have length √(176) and, from Pythagoras, AC = √(352). Let's make AC have positive slope but be almost horizontal. The sides of the green and yellow triangle are almost the same. As we reduce the slope, the sides become closer to being equal. So, let's take the special case where they are equal. The side of each square is (√(352))/2. Each square has area 352/4. Combined area is 2(352/4) = 176. Another special case is AC being almost 45°. The side of the green square is almost 0 and the side of the yellow square is almost AB. As we increase the slope to 45°, the green square has side 0 and yellow square side equal to the right triangle, or √(176). The area of the green square is 0 and yellow square (√(176))² = 176. If this problem were on a multiple choice test, we could safely mark 176 as the correct choice. If we had to show our work and knew how to solve the general case, we could use the special cases to check our work. If we didn't know how to solve the general case and were marked wrong, we could argue that the problem statement implies that the solution to the special case applies to the general case!
It might be helpful to add to the problem statement that the upper left corner of the yellow square must lie within ΔABC. Both of my special cases will be ruled out, but they can be treated as limiting cases. There is no discontinuity between the limiting cases and valid cases, so the limiting cases must provide the correct combined area if it is the same for all locations of that upper left corner within ΔABC. I see no way to stop students from using the special cases to find the correct answer on a multiple choice test. If the student must show his or her work, the problem could add "prove that the combined area is the same regardless of the position of the upper left corner of the yellow square within ΔABC".
The side lengths of the green & yellow squares are not fixed, so can be made congruent. The problem is then very simple since length AC becomes twice the lengths of 2 equal squares below.
Solution can be much easier. Use the same labelling of the squares as a and b. Then connect a diagonal from the green square from top left corner to bottom right corner. This diagonal will have length bsqrt(2). Connect another diagonal from the yellow square from from top right corner to bottom left corner. This diagonal will have a value asqrt(2). Notice that these two diagonals form a right angled triangle because the sum of the angles at the diagonals is 45 + 45 = 90. So the hypotenuse of this triangle is the hypotenuse of the other triangle. Then 2a^2 + 2b^2 = 2x^2, a^2 + b^2 = x^2 and x^2 =176, so the area is 176.
I can see a few different approaches to this problem presented and a lot of nice math performed. And doing all that math for the sake of practicing math skills is perfectly fine. Finding side leghts and areas of all different triangles eventually leads to the right answer. I'd like to point out that none of it is necessary to solve this problem. The answer is as trivial as double the area of the given triangle. And it's obvious just by looking at the drawing. The tiangle ABC with AB = BC and ABC being a right angle is simply a quarter of a square. If you take diagonals of the green and yellow squares oroginating in A and C the will join forming a right angle. Each diagonal cuts its respective square in half making that half a quarter of a square built on that diagonal. The sum of the areas of such squares is equal to the area of a square built on AB. The areas of the green and yellow squares are half the areas of the squres built on their diagonals, therefore their sum is falf the area of thr square built on AB. And as said in the begining the area of the triangle ABC is a quarter of the area of the square built on AB, so half of that area i double the area of the triangle ABC. No calculations necessary!
But you are doing calculations; you are just doing them in prose rather than symbols. And... "...forming a right angle. Each diagonal cuts its respective square in half making that half a quarter of a square built on that diagonal. The sum of the areas of such squares is equal to the area of a square built on AB." So... the Pythagorean Theorem.
I know my comment is rather long and descriptive. But I would call those descritions observations rather than calculations. A calculation would be the 2 x 88 = 176. And to that extend my statement "no calculations necessary" proves not to be true :) And I also agree that Pythagorean Theorem is inescapable. But again, it's an observation...
You can imagine the squares inscribed inside a semicircle to get the answer. Whenever squares are adjacent like this, they can be placed along a diameter of a semicircle with the points A and C assigned to the arc. From there you can resize either square (larger or smaller) and the other square will adjust accordingly so that the combined area remains constant. And A and C will remain along the arc. With this, we can imagine the squares to be the same size. In this example, the side lengths of the squares will be half the hypotenuse. The hypotenuse is 4root22cm. Half is 2root22cm. Square result to solve for one square's area and get 88cm^2. Double to get 176cm^2 for combined total. However, like many others, I interpreted the question to mean to find each square individually and I had no idea how to proceed. Normally, you shade the unknown the same color and ask us to find that.
let AB=BC=a Area of triangle ABC= 1/2(a)(a)=88 a^2=176 a=√176=4√11cm BC^2=2(4√11)^2 BC=4√22cm Let x is side of the yellow square and y is green square Connect A to D (D on the side of the yellow square (x+y)^2+(x-y)^2=(4√22)^2 x^2-2xy+y^2+x^2+2xy+y^2=352 2(x^2+y^3)=352 x^2+y^2=352/2=176cm^2 Area of yellow and green square =176 cm^2.❤❤❤
Cuadrado amarillo CDEF ; cuadrado verde AGEH. El triángulo ABC es la mitad de un cuadrado→ Trazando su simétrico respecto al eje AC obtenemos el cuadrado ABCS→ El vértice S está sobre ED puesto que los triángulos rectángulos CDS, SHA, BGA y CFB son congruentes→ Verde+Amarillo =ABCS =2*ABC =2*88=176 cm². Buen acertijo. Gracias y saludos.
Ok, Now it's my turn. 1) Green Square Area = G = a^2 ; Side of Green Square = a 2) Yellow Square Area = Y = b^2 ; Side o Yellow Square = b 3) Diagonal of Green Square = a*sqrt(2) 4) Diagonal of Yellow Square = b*sqrt(2) 5) Diagonal of Green Square ia Perpendicular to Diagonal of Yellow Square 8) [a*sqrt(2)]^2 + [b*sqrt(2)]^2 = [sqrt(352)]^2 ; 2a^2 + 2b^2 = 352 ; 2*(a^2 + b^2) = 352 ; a^2 + b^2 = 352/2 ; a^2 + b^2 = 176 Square Centimeters But I did a little bit more, calculating each Area separately. 9) sqrt(352) - sqrt(176) = Diagonal of Green Square ~ 5,495 cm 10) Diagonal of Green Square = a * sqrt(2) ; 5,495 = a * sqrt(2) ; a = 5,495 / sqrt(2) ; a = 3,89 cm 11) Green Square Area ~ 15 Square Centimeters 12) Yellow Square Area = 176 - 15 ~ 161 Square Centimeters 13) Answer : The Area of Yellow & Green Squares is equal to 176 Square Centimeters. NOTE: Observe that the Area of Yellow & Green Squares is equal to the Area of Square [ABCD], where D os not depicted here, or is equal to twice the Triangle [ABC] = 88 * 2 = 176.
Let's find the area: . .. ... .... ..... Since ABC is an isosceles right triangle, we can easily calculate the length of AC (by applying the Pythagorean theorem) and the area of the triangle: A(ABC) = (1/2)*AB*BC = (1/2)*AB² AC² = AB² + BC² = AB² + AB² = 2*AB² = 4*(1/2)*AB² = 4*A(ABC) = 4*88cm² = 352cm² Now we can build another right triangle with the long cathetus y+g, the short cathetus y−g and the hypotenuse AC (y and g are the side lengths of the yellow and the green square, respectively). By applying the Pythagorean theorem again we obtain: (y + g)² + (y − g)² = AC² y² + 2*y*g + g² + y² − 2*y*g + g² = AC² 2*y² + 2*g² = AC² y² + g² = AC²/2 ⇒ A(yellow square) + A(green square) = AC²/2 = 352cm²/2 = 176cm² Best regards from Germany
Some proofs are impossible without drawing auxiliary lines and like this problem it's hard to conclude anything from the drawing unless of course one knows the properties of geometric shapes. So drawing transversal lines to find congruent angles...alternate interior angles...ect...or in this case Line AD makes the problem obvious to solve. The tricky part is finding the Not So Obvious Auxiliary Construct to solve any problem. 🙂
My solution is ▶ for this isosceles triangle ΔACB: AB=BC= a AC= b ⇒ a*a/2= 88 cm² a²= 176 according to the Pythagorean theorem: a²+a²= b² 176+176= b² b²= 352 b= √352 If we apply the Pythagorean theorem using the lengths of the sides of the triangle, we can determine that the hypotenuse would be equal to the square of the sum of the squares of the other two sides of the triangle: the side of the green square: c A= c² the side of the yellow square: d A= d² ⇒ (c+d)²+(d-c)²= b² c²+2cd+d²+d²-2cd+d²= 352 2(c²+d²)= 352 c²+d²= 176 cm² is the solution ✅
Pas la peine de se fatiguer. Rien n'empêche d'imaginer la base du triangle rectangle comme la diagonale du carré jaune (l'aire du carré vert est alors égal à zéro !). On voit que l'aire du triangle est égale à la moitié du carré jaune. Il suffit donc de multiplier....88 X 2 = 176 !!!! 😅
AB=BC=w w＊w/2=88 w^2=176 w=4√11 AC=4√11＊√2=4√22 side of the Green square : x side of the Yellow square : y (x+y)^2+(y-x)^2=(4√22)^2 2x^2+2y^2=352 x^2+y^2=176 area of the Yellow & Green squares: 176cm^2
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Cool question!! 👍👍
Glad to hear that!
Thanks ❤️
Excellent informative video. New friend here.
Working out the sides and the hypotenuse of triangle ABC was easy. 4:46 was the point at which I stopped and realized that extending the upper side of the green triangle across the yellow one, then up to meet point C, generated a second right-angled triangle with the same hypotenuse as the first and with side-lengths (a+b) and (a-b). The rest is basic algebra.
Thanks for the feedback❤️
I find it astonishing that we can work out the combined area of both squares without knowing the side length of either of them! The power of mathematics! It's also curious that the combined area of the squares turns out to be exactly twice the area of the isosceles triangle.
Thanks for the feedback❤️
Let the side of the green square be 0. Then the diagonal of the yellow square is equal to the hypotenuse of the triangle. Then the area of the square is twice the area of the triangle. So 2*88 is 176.
2a^2=c^2, a^2/2=88, a^2=176/, c^2=352=(x+y)^2+(x-y)^2, x^2+2xy+y^2+x^2-2xY+y^2=352, 2(x^2+y^2)=352, x^2+y^2=176.
Thanks for sharing ❤️
I thought the problem was to find the area of the green square and the area of the yellow square. I got to the point a^2 + b^2 = 176, but got stuck there. Little did I know that 176 is the answer to what the actual problem is, the combined area of the green and yellow squares.
Thanks for the feedback ❤️
A(blue triangle) = 88 cm²
A(imagined blue square) = 176 cm²
AC² = 2 * 176 cm² = 352 cm²
a = side (yellow square)
b = side (green square)
(a + b)² + (a - b)² = AC²
a² + 2ab + b² + a² - 2ab + b² = 352 cm²
2a² + 2b² = 352 cm²
2 (a² + b²) = 352 cm²
a² + b² = 176 cm²
The problem statement implies that the angle of the hypotenuse of the isosceles right triangle does not change the combined area, as long as it produces a valid figure. So, the solution to a special case should be the same as that for the general case. From the area, we find that the AB and BC have length √(176) and, from Pythagoras, AC = √(352). Let's make AC have positive slope but be almost horizontal. The sides of the green and yellow triangle are almost the same. As we reduce the slope, the sides become closer to being equal. So, let's take the special case where they are equal. The side of each square is (√(352))/2. Each square has area 352/4. Combined area is 2(352/4) = 176. Another special case is AC being almost 45°. The side of the green square is almost 0 and the side of the yellow square is almost AB. As we increase the slope to 45°, the green square has side 0 and yellow square side equal to the right triangle, or √(176). The area of the green square is 0 and yellow square (√(176))² = 176.
If this problem were on a multiple choice test, we could safely mark 176 as the correct choice. If we had to show our work and knew how to solve the general case, we could use the special cases to check our work. If we didn't know how to solve the general case and were marked wrong, we could argue that the problem statement implies that the solution to the special case applies to the general case!
Thanks for the feedback ❤️
It might be helpful to add to the problem statement that the upper left corner of the yellow square must lie within ΔABC. Both of my special cases will be ruled out, but they can be treated as limiting cases. There is no discontinuity between the limiting cases and valid cases, so the limiting cases must provide the correct combined area if it is the same for all locations of that upper left corner within ΔABC.
I see no way to stop students from using the special cases to find the correct answer on a multiple choice test. If the student must show his or her work, the problem could add "prove that the combined area is the same regardless of the position of the upper left corner of the yellow square within ΔABC".
wow, amazingly smart. I tried doing one of your other questions and got it right, but the way you think is amazing! I love your videos, keep it up!
The side lengths of the green & yellow squares are not fixed, so can be made congruent.
The problem is then very simple since length AC becomes twice the lengths of 2 equal squares below.
Thanks for the feedback ❤️
Thanks a lot.
Solution can be much easier. Use the same labelling of the squares as a and b. Then connect a diagonal from the green square from top left corner to bottom right corner. This diagonal will have length bsqrt(2). Connect another diagonal from the yellow square from from top right corner to bottom left corner. This diagonal will have a value asqrt(2). Notice that these two diagonals form a right angled triangle because the sum of the angles at the diagonals is 45 + 45 = 90. So the hypotenuse of this triangle is the hypotenuse of the other triangle. Then 2a^2 + 2b^2 = 2x^2, a^2 + b^2 = x^2 and x^2 =176, so the area is 176.
I propose to make this problem more interesting/difficult to change the area of ABC from 88 to n, with n
Thanks for the feedback ❤️
I can see a few different approaches to this problem presented and a lot of nice math performed. And doing all that math for the sake of practicing math skills is perfectly fine. Finding side leghts and areas of all different triangles eventually leads to the right answer.
I'd like to point out that none of it is necessary to solve this problem. The answer is as trivial as double the area of the given triangle. And it's obvious just by looking at the drawing.
The tiangle ABC with AB = BC and ABC being a right angle is simply a quarter of a square. If you take diagonals of the green and yellow squares oroginating in A and C the will join forming a right angle. Each diagonal cuts its respective square in half making that half a quarter of a square built on that diagonal. The sum of the areas of such squares is equal to the area of a square built on AB. The areas of the green and yellow squares are half the areas of the squres built on their diagonals, therefore their sum is falf the area of thr square built on AB. And as said in the begining the area of the triangle ABC is a quarter of the area of the square built on AB, so half of that area i double the area of the triangle ABC. No calculations necessary!
But you are doing calculations; you are just doing them in prose rather than symbols. And... "...forming a right angle. Each diagonal cuts its respective square in half making that half a quarter of a square built on that diagonal. The sum of the areas of such squares is equal to the area of a square built on AB." So... the Pythagorean Theorem.
I know my comment is rather long and descriptive. But I would call those descritions observations rather than calculations. A calculation would be the 2 x 88 = 176. And to that extend my statement "no calculations necessary" proves not to be true :)
And I also agree that Pythagorean Theorem is inescapable. But again, it's an observation...
Thanks for the feedback ❤️
You can imagine the squares inscribed inside a semicircle to get the answer.
Whenever squares are adjacent like this, they can be placed along a diameter of a semicircle with the points A and C assigned to the arc.
From there you can resize either square (larger or smaller) and the other square will adjust accordingly so that the combined area remains constant. And A and C will remain along the arc.
With this, we can imagine the squares to be the same size. In this example, the side lengths of the squares will be half the hypotenuse.
The hypotenuse is 4root22cm. Half is 2root22cm. Square result to solve for one square's area and get 88cm^2. Double to get 176cm^2 for combined total.
However, like many others, I interpreted the question to mean to find each square individually and I had no idea how to proceed. Normally, you shade the unknown the same color and ask us to find that.
Thanks for the feedback ❤️
Thanks Sir
That’s nice and useful
With my respects
❤❤❤
let AB=BC=a
Area of triangle ABC=
1/2(a)(a)=88
a^2=176
a=√176=4√11cm
BC^2=2(4√11)^2
BC=4√22cm
Let x is side of the yellow square and y is green square
Connect A to D (D on the side of the yellow square
(x+y)^2+(x-y)^2=(4√22)^2
x^2-2xy+y^2+x^2+2xy+y^2=352
2(x^2+y^3)=352
x^2+y^2=352/2=176cm^2
Area of yellow and green square =176 cm^2.❤❤❤
Thanks for sharing ❤️
Cuadrado amarillo CDEF ; cuadrado verde AGEH. El triángulo ABC es la mitad de un cuadrado→ Trazando su simétrico respecto al eje AC obtenemos el cuadrado ABCS→ El vértice S está sobre ED puesto que los triángulos rectángulos CDS, SHA, BGA y CFB son congruentes→ Verde+Amarillo =ABCS =2*ABC =2*88=176 cm².
Buen acertijo. Gracias y saludos.
Excellent!
Thanks for sharing ❤️
Stealthy math problem... The answer sneaks up on you using the very effective Pythagorean theorem. Thank you!
You are very welcome!
Thanks for the feedback❤️
Ok, Now it's my turn.
1) Green Square Area = G = a^2 ; Side of Green Square = a
2) Yellow Square Area = Y = b^2 ; Side o Yellow Square = b
3) Diagonal of Green Square = a*sqrt(2)
4) Diagonal of Yellow Square = b*sqrt(2)
5) Diagonal of Green Square ia Perpendicular to Diagonal of Yellow Square
8) [a*sqrt(2)]^2 + [b*sqrt(2)]^2 = [sqrt(352)]^2 ; 2a^2 + 2b^2 = 352 ; 2*(a^2 + b^2) = 352 ; a^2 + b^2 = 352/2 ; a^2 + b^2 = 176 Square Centimeters
But I did a little bit more, calculating each Area separately.
9) sqrt(352) - sqrt(176) = Diagonal of Green Square ~ 5,495 cm
10) Diagonal of Green Square = a * sqrt(2) ; 5,495 = a * sqrt(2) ; a = 5,495 / sqrt(2) ; a = 3,89 cm
11) Green Square Area ~ 15 Square Centimeters
12) Yellow Square Area = 176 - 15 ~ 161 Square Centimeters
13) Answer : The Area of Yellow & Green Squares is equal to 176 Square Centimeters.
NOTE: Observe that the Area of Yellow & Green Squares is equal to the Area of Square [ABCD], where D os not depicted here, or is equal to twice the Triangle [ABC] = 88 * 2 = 176.
Got that but mis read the question and then went on to try and find the individual areas, which I believe is impossible with the information given.
Yes, I was not also sure of the exact question.
Not possible because the side lengths are not fixed, but the sum of their areas will be constant.
Small square 25.77
Big square 150.23
Angle BAC=45
Angle between ac and the square then is 22.5
Then use trigo the rest is easy.
My apologies if there is any confusion...
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
Since ABC is an isosceles right triangle, we can easily calculate the length of AC (by applying the Pythagorean theorem) and the area of the triangle:
A(ABC) = (1/2)*AB*BC = (1/2)*AB²
AC² = AB² + BC² = AB² + AB² = 2*AB² = 4*(1/2)*AB² = 4*A(ABC) = 4*88cm² = 352cm²
Now we can build another right triangle with the long cathetus y+g, the short cathetus y−g and the hypotenuse AC (y and g are the side lengths of the yellow and the green square, respectively). By applying the Pythagorean theorem again we obtain:
(y + g)² + (y − g)² = AC²
y² + 2*y*g + g² + y² − 2*y*g + g² = AC²
2*y² + 2*g² = AC²
y² + g² = AC²/2
⇒ A(yellow square) + A(green square) = AC²/2 = 352cm²/2 = 176cm²
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Some proofs are impossible without drawing auxiliary lines and like this problem it's hard to conclude anything from the drawing unless of course one knows the properties of geometric shapes. So drawing transversal lines to find congruent angles...alternate interior angles...ect...or in this case Line AD makes the problem obvious to solve. The tricky part is finding the Not So Obvious Auxiliary Construct to solve any problem. 🙂
Excellent!
Thanks for the feedback ❤️
S=176
My solution is ▶
for this isosceles triangle ΔACB:
AB=BC= a
AC= b
⇒
a*a/2= 88 cm²
a²= 176
according to the Pythagorean theorem:
a²+a²= b²
176+176= b²
b²= 352
b= √352
If we apply the Pythagorean theorem using the lengths of the sides of the triangle, we can determine that the hypotenuse would be equal to the square of the sum of the squares of the other two sides of the triangle:
the side of the green square: c
A= c²
the side of the yellow square: d
A= d²
⇒
(c+d)²+(d-c)²= b²
c²+2cd+d²+d²-2cd+d²= 352
2(c²+d²)= 352
c²+d²= 176 cm² is the solution ✅
Am I the only one who took the trouble to calculate the area of each of the two squares ?
Green square = 25.77
Yellow square = 150.23
Total = 176
Pas la peine de se fatiguer. Rien n'empêche d'imaginer la base du triangle rectangle comme la diagonale du carré jaune (l'aire du carré vert est alors égal à zéro !). On voit que l'aire du triangle est égale à la moitié du carré jaune. Il suffit donc de multiplier....88 X 2 = 176 !!!! 😅
First comment
Excellent!
Thanks ❤️
AB=BC=w
w＊w/2=88 w^2=176 w=4√11
AC=4√11＊√2=4√22
side of the Green square : x
side of the Yellow square : y
(x+y)^2+(y-x)^2=(4√22)^2
2x^2+2y^2=352 x^2+y^2=176
area of the Yellow & Green squares: 176cm^2
Excellent!
Thanks for sharing ❤️