Shifting the green square to the left of the rectangle produces two hypotenuses of 5, and shows that the square takes up half the area of the rectangle. As the square takes up half the horizontal distance of the rectangle, the a^2 + b^2 = c^2 may be rewritten as a^2 + (2a)^2 = c^2. Therefore, 5a^2 = 25 As the square has sides of 2a, its area is 4a^2. Square's area is 25*(4/5) = 20 un^2. I have now watched the video. It seemed unnecessarily complex, but I assume that was because it's meant for teaching about similar triangles rather than getting to the answer as quickly as possible. Shifting the square to the left simplifies things immensely, especially as it's exactly half of the rectangle. Thank you.
Maybe you can help explain where I went wrong because I used a similar approach and got 24 square units I said that the hypotenuse of the entire rectangle was 10 and since there is a Pythagorean triple (3,4,5) with a factor of 10 (5) then I could just multiply the entire triple to get the sides of the rectangle which gave me a rectangle of 6 by 8 or 48sq units. Then I said since the combined length of parts of the hypotenuse that are outside the shaded square is equal to the part that is inside and therefore the square must be half the rectangle. Therefore if the rectangle is 48sq units the square would have to be 24.
@@yehoshuabedziner2314 You went wrong by assuming that the two short sides are 6 and 8. They are, in fact, sqrt(20) and 2*sqrt(20) (the length is twice the height). With the large rectangle, a^2 + b^2 can be re-written as a^2 + (2a)^2 = 5a^2. 5a^2 = 100 (due to the hypotenuse of 10 being squared). As 5a^2 = 100, a = sqrt(20). Whole rectangle is sqrt(20) * 2*sqrt(20) = 40. Halve it for 20. Although using Pythagoras Theorem to help solve, it doesn't necessarily mean that it's a Pythagorean triple. Pythagorean triple is when all sides are integers, but the theorem is just as applicable when they aren't. I hope that helps.
This is exactly how I did it. The key is that the diagonal of the rectangle creates two triangles with proportions 1, 2, sqrt5. Knowing the hypotenuse makes everything else simple. I thought this looked too easy, but it really was easy. Just shift the green square to the right, and the rectangle becomes 1x tall and 2x wide with a diagonal (hypotenuse) of sqrt5x, where x = 2*sqrt5.
It has been pointed out by several commenters that the figure represents an impossible situation. Too much information is given. If the green figure is a square then there is no STRAIGHT diagonal line of 2, 5, 3 sub sections. You can prove almost anything in that situation. Like: equilateral triangles have side length equal to altitude length. Imo, this problem is not salvagable. If fixed up, it is no longer charming. This channel is great. It doesn't need to display a bogus physical situation.
Simple solution. Since all the inscribed triangles are similar. Divide the x-axis into 2x, 5x, and 3x. The y-axis is 5x. Now solve for Pythagoras. 5x^2 + 10x^2 = 10^2. x = sqrt(4/5) or approx .8944. To find the area of the green square, A = (5x)^2 or 25x^2 which should be 25 * 4/5 = 20. Alot lot easier than all the work the Premath's technique.
@@PreMath I took a shortcut in assuming that the triangles would be ratiometric (2/5/3) vs "proving" it with similar triangles, but I got the same result. Even short-circuited that by noting that 2+3=5, sliding the square to the left, and noting that the sides would be x and 2x. A little fast'n'loose, but it worked, and I still could've "proven" it if I had to. So figure out the area of the big rectangle, halve it, and you got the area of the square.
H and K the orthogonal projections of E and F on (BC). BH/AE= HK/EF = KC/FC = cos(angleBCA) = k, so BH = 2.k, HK = 5.k which is the side length of the squate or the height AB of the rectangle ABCD, KC = 3.k, and BC = 10.k. In ABC: 100 = AC^2 = BC^2 + AB^2 = (10.k)^2 + (5.k)^2, So 100 = 125.k^2 The area of the square is (5.k)^2 = 25.k^2= (1/5).(125.k^2) = (1/5).100 = 20.
∆ AEG ~∆CFH EG/FH=2/3 So EG=2a ; FH=3a Let x is side of square ∆ AEG ~ ∆ ACD 2a/x=2/10=1/5 So x=10a FH=3a; FI=10a-3a=7a Connect to E to G (G on HI and EG right HI) FJ=7a-2a=5a GI=x=10a In ∆ EJ F (5a)^2+(10a)^2=5^2 So a=√5/5 So x=10√5/5=2√5 So Green square area=(2√5)^2=20 square units.❤❤❤
it's a repeated calculation considering the proportions and the 90 degree angle: 10 print "premath-can you find area of the green shaded square" 20 l1=2:l2=5:l3=3:n=l1+l2+l3:sw=l1^2/10/n:l5=sw:dim x(2,3),y(2,3):goto 60 30 dl=(l2+l3)*l5/l3:l4=dl-l5:l6=(l4+l5)*sqr(l3^2-l5^2)/l5 40 l7=l1*l5/l3:dgu1=(l7+l4+l5)^2/n^2:dgu2=l4^2/n^2:dgu3=(l1+l2+l3)^2/n^2 50 dg=dgu1+dgu2-dgu3:return 60 gosub 30 70 dg1=dg:l51=l5:l5=l5+sw:gosub 30:if l5>30*l1 then stop 80 l52=l5:if dg1*dg>0 then 70 90 l5=(l51+l52)/2:gosub 30:if dg1*dg>0 then l51=l5 else l52=l5 100 if abs(dg)>1E-10 then 90 110 print l4,"die gruene flaeche=",l4^2:masx=1200/(l7+l4+l5):masy=850/l4 120 x(0,0)=0:y(0,0)=0:x(0,1)=l7:y(0,1)=0:x(0,2)=x(0,1):y(0,2)=l4:x(0,3)=0:y(0,3)=y(0,2) 130 x(1,0)=l7:y(1,0)=0:x(1,1)=l7+l4:y(1,1)=0:x(1,2)=x(1,1):y(1,2)=l4:x(1,3)=x(1,0):y(1,3)=y(1,2) 140 x(2,0)=x(1,1):y(2,0)=0:x(2,1)=x(2,0)+l5:y(2,1)=0:x(2,2)=x(2,1):y(2,2)=l4:x(2,3)=x(2,0) 150 y(2,3)=y(2,2): if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
20 Let the length of the rectangle = m then, since the two small triangles with hypotenuses 2 and 3 are similar to the one with hypotenuse 10, then 10/m = 2/? and 10/m = 3/? Hence, then? = 2/10 or 1/5 m and 3/10m or 0.3 m 0.3 + 0.2 = 0.5 Hence, the length of the square is one-half the length of the rectangle Since the width of the rectangle = the length of the square, then the width of the rectangle = 0.5 m note if the length of the square =0.5m , then its area = 0.25 m^2 Using Pythagorean (m)^2 + (0.5 m)^2 = 10^2 m^2 + 0.25m^2 =100 1.25m^2 = 100 1.25/5 = 100/5 0.25 m^2 = 20 Answer
I disagree with 20 but 24 as the area of the green area. The rectangle ABCD with diagonal 10m implies that the sides lengths are 8m and 6m. The 6m forms a side of the green area and 4m, a part of the 8m, forms the other side. Therefore area of the green portion is 6x4=24.
I think the assumption that the green area is a SQUARE led to the wrong calculation. The green area is a RECTANGLE based on the DIAGONAL of the whole shape, which is a RECTANGLE and not a SQUARE.
You are correct. Based on the numbers in this scenario, 24 is the correct answer. The hypotenuse is 10, therefore the sides are 8 and 6. There's no arguing with that. The entire rectangle has an area of 48. No matter where that green square is, it's always going to be 50% because it is half the total hypotenuse. 24 is the correct answer.
a^2 + (1/4)a^2 = 25 as 5 is a hypotenuse. opposite of triangle equal to 1/2 a due to congruency of sides on diagonal side. Another side of triangle = a. Love math. ❤ Will subscribe.
The answer is 20 unita square. I muat admit that this video is simialr to yesterday's Math Booster video which talked used AS similarity. I reallt would like to see a playlist of problems that make use of AA similarity. And if understand correctly, the AA similarity works if similar sides can be set equal. Because not all similar triangles are congruent!!!
Let the corners of the green square be GHJK, counterclockwise from top left (corresponding to ABCD). Let the side length of GHJK (and height of ABCD) be s, and let the width of ABCD be w. Draw FM, where M is the point on GH where FM is parallel to BC and AD. As FM is parallel to HJ, JF is parallel to MH, and ∠MHJ = ∠HJF = 90°, FMHJ is a rectangle, and FM = HJ = s. As ∠GAE = ∠JCF, as alternate interior angles, and ∠EGA = ∠FJC = 90°, ∆EGA and ∆FJC are similar triangles. As ∠MFE = ∠JCF by corresponding angles and ∠EMF = ∠FJC = 90°, ∆EMF is also similar to the above triangles. GA/AE = JC/CF = FM/EF GA/2 = s/5 GA = 2s/5 JC/3 = s/5 JC = 3s/5 BC = BH + HJ + JC BC = GA + FM + JC w = 2s/5 + s + 3s/5 w = 2s Triangle ∆ABC: AB² + BC² = CA² s² + (2s)² = (5+3+2)² s² + 4s² = 100 5s² = 100 s² = 100/5 = 20 sq units
My way of solution ▶ H ∈ [AD] G ∈ [AD] the small thriangle is similar with the medium and the large triangle, let's write these similarities: ΔAEH ~ ΔAFG ( I ) ⇒ x/[FG]= y/[AG]= 2/7 ΔAEH ~ ΔACD (II) ⇒ x/[CD]= y/[DA]= 2/10 [HA]= y [EH]= x ⇒ from equation (I) we get: x/[FG]= 2/7 ⇒ 7x= 2[FG] [FG]= 7x/2 y/[AG]= 2/7 ⇒ 7y= 2[AG] [AG]= 7y/2 [HG]= 7y/2 -y [HG]= 5y/2 from equation (II) we get: x/[CD]= 2/10 ⇒ [CD]= 5x while the green area is a square ❗ [HG]= [CD] 5y/2= 5x y= 2x from the Pythagorean theorem for the small triangle ΔAEH: [HA]²+[EH]²= [AE]² [HA]= y [EH]= x [AE]= 2 ⇒ y²+x²= 4 y= 2x ⇒ (2x)²+x²= 4 5x²= 4 x²= 4/5 x= 2/√5 x= 2√5/5 ⇒ y= 2x y= 4√5/5 Agreen= [HG]*[CD], we can also write [HG]² or [CD]² [HG]= 5y/2 [CD]= 5x ⇒ [HG]= 5*4√5/5/2 [HG]= 2√5 [CD]= 5*2√5/5 [CD]= 2√5 Agreen= (2√5)² Agreen= 20 square units
1/ Just draw HK perpendicular to both AD BC ( the points H and K are on AD and BC such that AH=BK=a) We have AHKB is a square of which side= a HK intersects AC at point G By Thales theorem AG/BK=EF/a=5/a --> AG= 5-> G is the midpoint of AC--> BC=2a Focus on the triangle ADC SqAD+ sqCD= sq AC sqa+ 4sqa=100 -> sqa=20 Area= 20 sq units😅😅😅
Let's find the area: . .. ... .... ..... May P and Q be the upper left and the lower right corner of the green square, respectively, and may R be located on the right side of the green square such that EFR is a right triangle. The right triangles AEP, CFQ and EFR are obviously similar, so we can conclude: EP:FQ:FR = AE:CF:EF = 2:3:5 Since EP+FQ+FR=AB, we obtain: EP + FQ + FR = AB (2/5)*FR + (3/5)*FR + FR = AB 2*FR = AB ⇒ FR = AB/2 The triangle EFR is a right triangle, so we can apply the Pythagorean theorem. With s=AB being the side length of the green square we obtain: ER² + FR² = EF² s² + (s/2)² = EF² s² + s²/4 = EF² 5*s²/4 = EF² ⇒ s² = 4*EF²/5 = 4*5²/5 = 20 Therefore the area of the green square turns out to be: A(green square) = s² = 20 Best regards from Germany
By Thales, we can immediately see that the length of the rectangle is twice his height. Thus : h² + (2h)² = 10² => h²=20, which is the area of the square.
The length of the rectangle is not twice its height according to the given information, diagonal is 10m. Apply Pythagoras and you'll find the length as 8m and the height as 6m. So, 20 is not the area of the green portion.
@@apooia2006 There are many ways to get a diagonal of 10m, not only 8 and 6 as side lenghts ; here, 2.sqrt(5) and 4.sqrt(5) give also a diagonal of sqrt(20++80) = 10
if triangle EPF is a right triangle with hypotenuse 5 then surely PF must b e4 and PE 3 if this is the case then a must be 4 and the area of the square is 16 what am i doing wrong)
I believe you are assuming 3:4:5 right triangles. If you do, then the shape in the rectangle will not be a square. Try drawing it out and you will see. Start with a piece of paper and mark off 6 units from the same side, and connect that line straight across. From the corner, draw a diagonal to the line you just drew that is 10 units long, and mark off 2, 5, and 3 unit lengths on the diagonal. Now create a rectangle from that point of intersection, and you will see that the shape in the middle will not be a square. It will be a 6x4 rectangle.
The answer should be 36. The diagonal, AC = 10. But AC is the hypotenuse of the right triangle ABC. Right triangle sides ratio=3:4:5. Therefore, AB should be 6. BC should be 8. If AB=6, then the side of the green square=6. Therefore, the area of the green square =(6)(6)=36. The answer of 20 is a possible answer. But it is not the final answer. Its possible that AB=sq.rt. 20 & BC=sq.rt.80.
@@user-yc3yy4sf1x It will be a square of 6 units on each side. I have proved it. Please prove your own statement that the green area CANNOT be a square of side 6.
@@anpadh5850 I cannot draw it in the comment, but you can draw the picture yourself. Start with a piece of paper and mark off 6 units from the same side, and connect that line straight across. From the corner, draw a diagonal to the line you just drew that is 10 units long, and mark off 2, 5, and 3 unit lengths on the diagonal. Now create a rectangle from that point of intersection, and you will see that the shape in the middle will not be a square. It will be a 6x4 rectangle. You may be able to draw a 6x6 square inside a rectangle, but it will not have a diagonal with the 2, 5, 3 breakdown like the original drawing. Just think, the segment in the middle has to be 5 units long. How can there be a segment 5 units long going from one end to the other of a 6x6 square?
STEP-STEP RESOLUTION PROPOSAL : 01) As anyone can see, the Rectangle [ABCD] is equal to two Green Squares. 5 = 2 + 3 02) So : AB = a and BC = 2a 03) a^2 + 4a^2 = 100 ; 5a^2 = 100 ; a^2 = 100 / 5 ; a^2 = 20 ; a = sqrt(20) 04) Area = sqrt(20) * sqrt(20) ; A = 20 OUR ANSWER: Green Square Area equal to 20 Square Units.
Math is like a puzzle to me, too trick but i love it😂
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It is a 3 4 5 triangle, so the grean base is 4, the height is 6, so the area is 24.
Shifting the green square to the left of the rectangle produces two hypotenuses of 5, and shows that the square takes up half the area of the rectangle.
As the square takes up half the horizontal distance of the rectangle, the a^2 + b^2 = c^2 may be rewritten as a^2 + (2a)^2 = c^2.
Therefore, 5a^2 = 25
As the square has sides of 2a, its area is 4a^2.
Square's area is 25*(4/5) = 20 un^2.
I have now watched the video. It seemed unnecessarily complex, but I assume that was because it's meant for teaching about similar triangles rather than getting to the answer as quickly as possible.
Shifting the square to the left simplifies things immensely, especially as it's exactly half of the rectangle.
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Maybe you can help explain where I went wrong because I used a similar approach and got 24 square units
I said that the hypotenuse of the entire rectangle was 10 and since there is a Pythagorean triple (3,4,5) with a factor of 10 (5) then I could just multiply the entire triple to get the sides of the rectangle which gave me a rectangle of 6 by 8 or 48sq units. Then I said since the combined length of parts of the hypotenuse that are outside the shaded square is equal to the part that is inside and therefore the square must be half the rectangle. Therefore if the rectangle is 48sq units the square would have to be 24.
@@yehoshuabedziner2314 You went wrong by assuming that the two short sides are 6 and 8. They are, in fact, sqrt(20) and 2*sqrt(20) (the length is twice the height). With the large rectangle, a^2 + b^2 can be re-written as a^2 + (2a)^2 = 5a^2. 5a^2 = 100 (due to the hypotenuse of 10 being squared).
As 5a^2 = 100, a = sqrt(20). Whole rectangle is sqrt(20) * 2*sqrt(20) = 40. Halve it for 20.
Although using Pythagoras Theorem to help solve, it doesn't necessarily mean that it's a Pythagorean triple. Pythagorean triple is when all sides are integers, but the theorem is just as applicable when they aren't.
I hope that helps.
This is exactly how I did it. The key is that the diagonal of the rectangle creates two triangles with proportions 1, 2, sqrt5. Knowing the hypotenuse makes everything else simple. I thought this looked too easy, but it really was easy. Just shift the green square to the right, and the rectangle becomes 1x tall and 2x wide with a diagonal (hypotenuse) of sqrt5x, where x = 2*sqrt5.
obvious and more simple
Mentally move the square to the left until its side aligns with AB.
According to the Pythagorean theorem:
a^2+(a/2)^2=5^2;
5a^2=100;
S=a^2=100/5=20
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It has been pointed out by several commenters that the figure represents an impossible situation.
Too much information is given.
If the green figure is a square then there is no STRAIGHT diagonal line of 2, 5, 3 sub sections.
You can prove almost anything in that situation.
Like: equilateral triangles have side length equal to altitude length.
Imo, this problem is not salvagable. If fixed up, it is no longer charming.
This channel is great. It doesn't need to display a bogus physical situation.
Nice observation 🎉
Thank you! Cheers!
Simple solution. Since all the inscribed triangles are similar. Divide the x-axis into 2x, 5x, and 3x. The y-axis is 5x. Now solve for Pythagoras. 5x^2 + 10x^2 = 10^2. x = sqrt(4/5) or approx .8944. To find the area of the green square, A = (5x)^2 or 25x^2 which should be 25 * 4/5 = 20. Alot lot easier than all the work the Premath's technique.
DC= x, sin(
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@@PreMath I took a shortcut in assuming that the triangles would be ratiometric (2/5/3) vs "proving" it with similar triangles, but I got the same result.
Even short-circuited that by noting that 2+3=5, sliding the square to the left, and noting that the sides would be x and 2x.
A little fast'n'loose, but it worked, and I still could've "proven" it if I had to. So figure out the area of the big rectangle, halve it, and you got the area of the square.
16. 3 4 5 pitagorean
5=2+3--->Base verde =a =BC/2---> Área verde =ABCD/2---> a²+(2a)²=(2+5+3)²=100---> a²=20 ud² =Área verde.
Gracias y saludos.
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H and K the orthogonal projections of E and F on (BC).
BH/AE= HK/EF = KC/FC = cos(angleBCA) = k, so BH = 2.k, HK = 5.k which is the side length of the squate or the height AB of the rectangle ABCD, KC = 3.k, and BC = 10.k.
In ABC: 100 = AC^2 = BC^2 + AB^2 = (10.k)^2 + (5.k)^2, So 100 = 125.k^2
The area of the square is (5.k)^2 = 25.k^2= (1/5).(125.k^2) = (1/5).100 = 20.
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∆ AEG ~∆CFH
EG/FH=2/3
So EG=2a ; FH=3a
Let x is side of square
∆ AEG ~ ∆ ACD
2a/x=2/10=1/5
So x=10a
FH=3a; FI=10a-3a=7a
Connect to E to G (G on HI and EG right HI)
FJ=7a-2a=5a
GI=x=10a
In ∆ EJ F
(5a)^2+(10a)^2=5^2
So a=√5/5
So x=10√5/5=2√5
So Green square area=(2√5)^2=20 square units.❤❤❤
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x+√(25-l^2)+3x/2=l...arccos(x/2)=arcsin(l/5)...l,lato del quadrato..x=EK(K,proiezione su AD)..le due equazioni portano a l^2=20,x=2/√5
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it's a repeated calculation considering the proportions and the 90 degree angle:
10 print "premath-can you find area of the green shaded square"
20 l1=2:l2=5:l3=3:n=l1+l2+l3:sw=l1^2/10/n:l5=sw:dim x(2,3),y(2,3):goto 60
30 dl=(l2+l3)*l5/l3:l4=dl-l5:l6=(l4+l5)*sqr(l3^2-l5^2)/l5
40 l7=l1*l5/l3:dgu1=(l7+l4+l5)^2/n^2:dgu2=l4^2/n^2:dgu3=(l1+l2+l3)^2/n^2
50 dg=dgu1+dgu2-dgu3:return
60 gosub 30
70 dg1=dg:l51=l5:l5=l5+sw:gosub 30:if l5>30*l1 then stop
80 l52=l5:if dg1*dg>0 then 70
90 l5=(l51+l52)/2:gosub 30:if dg1*dg>0 then l51=l5 else l52=l5
100 if abs(dg)>1E-10 then 90
110 print l4,"die gruene flaeche=",l4^2:masx=1200/(l7+l4+l5):masy=850/l4
120 x(0,0)=0:y(0,0)=0:x(0,1)=l7:y(0,1)=0:x(0,2)=x(0,1):y(0,2)=l4:x(0,3)=0:y(0,3)=y(0,2)
130 x(1,0)=l7:y(1,0)=0:x(1,1)=l7+l4:y(1,1)=0:x(1,2)=x(1,1):y(1,2)=l4:x(1,3)=x(1,0):y(1,3)=y(1,2)
140 x(2,0)=x(1,1):y(2,0)=0:x(2,1)=x(2,0)+l5:y(2,1)=0:x(2,2)=x(2,1):y(2,2)=l4:x(2,3)=x(2,0)
150 y(2,3)=y(2,2): if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
20
Let the length of the rectangle = m
then, since the two small triangles with hypotenuses 2 and 3 are similar to the one with hypotenuse 10,
then 10/m = 2/? and 10/m = 3/?
Hence, then? = 2/10 or 1/5 m
and 3/10m or 0.3 m
0.3 + 0.2 = 0.5
Hence, the length of the square is one-half the length of the rectangle
Since the width of the rectangle = the length of the square, then the width of the rectangle = 0.5 m
note if the length of the square =0.5m , then its area = 0.25 m^2
Using Pythagorean
(m)^2 + (0.5 m)^2 = 10^2
m^2 + 0.25m^2 =100
1.25m^2 = 100
1.25/5 = 100/5
0.25 m^2 = 20 Answer
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The green square area is half the area of the rectangle.
area of the rectangle = (10/✓5) (20/✓5) = 40
The green square area is 20
I disagree with 20 but 24 as the area of the green area.
The rectangle ABCD with diagonal 10m implies that the sides lengths are 8m and 6m. The 6m forms a side of the green area and 4m, a part of the 8m, forms the other side. Therefore area of the green portion is 6x4=24.
I think the assumption that the green area is a SQUARE led to the wrong calculation. The green area is a RECTANGLE based on the DIAGONAL of the whole shape, which is a RECTANGLE and not a SQUARE.
You are correct. Based on the numbers in this scenario, 24 is the correct answer. The hypotenuse is 10, therefore the sides are 8 and 6. There's no arguing with that. The entire rectangle has an area of 48. No matter where that green square is, it's always going to be 50% because it is half the total hypotenuse. 24 is the correct answer.
I agree. The figure is impossible/OVER-DEFINED.
The points of the "diagonal" are NOT co-linear (or the lengths 2, 5, 3 are not possible .)
Simple but great. I love it and feel like going back to school. I am 56 now!!
Super! Stay blessed🙏
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a^2 + (1/4)a^2 = 25 as 5 is a hypotenuse. opposite of triangle equal to 1/2 a due to congruency of sides on diagonal side. Another side of triangle = a. Love math. ❤ Will subscribe.
The answer is 20 unita square. I muat admit that this video is simialr to yesterday's Math Booster video which talked used AS similarity. I reallt would like to see a playlist of problems that make use of AA similarity. And if understand correctly, the AA similarity works if similar sides can be set equal. Because not all similar triangles are congruent!!!
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Let the corners of the green square be GHJK, counterclockwise from top left (corresponding to ABCD). Let the side length of GHJK (and height of ABCD) be s, and let the width of ABCD be w.
Draw FM, where M is the point on GH where FM is parallel to BC and AD. As FM is parallel to HJ, JF is parallel to MH, and ∠MHJ = ∠HJF = 90°, FMHJ is a rectangle, and FM = HJ = s.
As ∠GAE = ∠JCF, as alternate interior angles, and ∠EGA = ∠FJC = 90°, ∆EGA and ∆FJC are similar triangles. As ∠MFE = ∠JCF by corresponding angles and ∠EMF = ∠FJC = 90°, ∆EMF is also similar to the above triangles.
GA/AE = JC/CF = FM/EF
GA/2 = s/5
GA = 2s/5
JC/3 = s/5
JC = 3s/5
BC = BH + HJ + JC
BC = GA + FM + JC
w = 2s/5 + s + 3s/5
w = 2s
Triangle ∆ABC:
AB² + BC² = CA²
s² + (2s)² = (5+3+2)²
s² + 4s² = 100
5s² = 100
s² = 100/5 = 20 sq units
side of Green square : x 5/(2+5+3)=5/10=1/2
x²+(x/2)²=5² 5x²/4=25 5x²=100
Green Square area = x * x = x² = 20
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My way of solution ▶
H ∈ [AD]
G ∈ [AD]
the small thriangle is similar with the medium and the large triangle, let's write these similarities:
ΔAEH ~ ΔAFG ( I )
⇒
x/[FG]= y/[AG]= 2/7
ΔAEH ~ ΔACD (II)
⇒
x/[CD]= y/[DA]= 2/10
[HA]= y
[EH]= x
⇒
from equation (I) we get:
x/[FG]= 2/7
⇒
7x= 2[FG]
[FG]= 7x/2
y/[AG]= 2/7
⇒
7y= 2[AG]
[AG]= 7y/2
[HG]= 7y/2 -y
[HG]= 5y/2
from equation (II) we get:
x/[CD]= 2/10
⇒
[CD]= 5x
while the green area is a square ❗
[HG]= [CD]
5y/2= 5x
y= 2x
from the Pythagorean theorem for the small triangle ΔAEH:
[HA]²+[EH]²= [AE]²
[HA]= y
[EH]= x
[AE]= 2
⇒
y²+x²= 4
y= 2x
⇒
(2x)²+x²= 4
5x²= 4
x²= 4/5
x= 2/√5
x= 2√5/5
⇒
y= 2x
y= 4√5/5
Agreen= [HG]*[CD], we can also write [HG]² or [CD]²
[HG]= 5y/2
[CD]= 5x
⇒
[HG]= 5*4√5/5/2
[HG]= 2√5
[CD]= 5*2√5/5
[CD]= 2√5
Agreen= (2√5)²
Agreen= 20 square units
1/ Just draw HK perpendicular to both AD BC ( the points H and K are on AD and BC such that AH=BK=a)
We have AHKB is a square of which side= a
HK intersects AC at point G
By Thales theorem
AG/BK=EF/a=5/a
--> AG= 5-> G is the midpoint of AC--> BC=2a
Focus on the triangle ADC
SqAD+ sqCD= sq AC
sqa+ 4sqa=100
-> sqa=20
Area= 20 sq units😅😅😅
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@@PreMathI often solve the problem on my cell phone so I make lots of typo errors.
So sorry 😢😢😢!
@@phungpham1725
No worries!
Take care😀
Interesting and creative puzzle, 5s^2=10^2, the area is s^2=100/5=20.😊
Yes, you are right
Thanks for the feedback ❤️
Let's find the area:
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...
....
.....
May P and Q be the upper left and the lower right corner of the green square, respectively, and may R be located on the right side of the green square such that EFR is a right triangle. The right triangles AEP, CFQ and EFR are obviously similar, so we can conclude:
EP:FQ:FR = AE:CF:EF = 2:3:5
Since EP+FQ+FR=AB, we obtain:
EP + FQ + FR = AB
(2/5)*FR + (3/5)*FR + FR = AB
2*FR = AB
⇒ FR = AB/2
The triangle EFR is a right triangle, so we can apply the Pythagorean theorem. With s=AB being the side length of the green square we obtain:
ER² + FR² = EF²
s² + (s/2)² = EF²
s² + s²/4 = EF²
5*s²/4 = EF²
⇒ s² = 4*EF²/5 = 4*5²/5 = 20
Therefore the area of the green square turns out to be:
A(green square) = s² = 20
Best regards from Germany
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By Thales, we can immediately see that the length of the rectangle is twice his height. Thus : h² + (2h)² = 10² => h²=20, which is the area of the square.
The length of the rectangle is not twice its height according to the given information, diagonal is 10m. Apply Pythagoras and you'll find the length as 8m and the height as 6m. So, 20 is not the area of the green portion.
@@apooia2006 You are assuming that the large triangle is 3-4-5. It is not, Thales is correct.
@@egillandersson1780 Good point a/5 = b/10 -> b = 2a
@@apooia2006 There are many ways to get a diagonal of 10m, not only 8 and 6 as side lenghts ; here, 2.sqrt(5) and 4.sqrt(5) give also a diagonal of sqrt(20++80) = 10
Can also solve via application of trig identities.
S = 20
20
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if triangle EPF is a right triangle with hypotenuse 5 then surely PF must b e4 and PE 3 if this is the case then a must be 4 and the area of the square is 16 what am i doing wrong)
I believe you are assuming 3:4:5 right triangles. If you do, then the shape in the rectangle will not be a square. Try drawing it out and you will see. Start with a piece of paper and mark off 6 units from the same side, and connect that line straight across. From the corner, draw a diagonal to the line you just drew that is 10 units long, and mark off 2, 5, and 3 unit lengths on the diagonal. Now create a rectangle from that point of intersection, and you will see that the shape in the middle will not be a square. It will be a 6x4 rectangle.
20 sq. units
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I'd Say A = 20
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The answer should be 36. The diagonal, AC = 10. But AC is the hypotenuse of the right triangle ABC. Right triangle sides ratio=3:4:5. Therefore, AB should be 6. BC should be 8. If AB=6, then the side of the green square=6. Therefore, the area of the green square =(6)(6)=36.
The answer of 20 is a possible answer. But it is not the final answer. Its possible that AB=sq.rt. 20 & BC=sq.rt.80.
One cannot assume 3:4:5 right triangles. If you do, the green shape inside the rectangle will not be a square.
@@user-yc3yy4sf1x It will be a square of 6 units on each side. I have proved it. Please prove your own statement that the green area CANNOT be a square of side 6.
@@anpadh5850 I cannot draw it in the comment, but you can draw the picture yourself. Start with a piece of paper and mark off 6 units from the same side, and connect that line straight across. From the corner, draw a diagonal to the line you just drew that is 10 units long, and mark off 2, 5, and 3 unit lengths on the diagonal. Now create a rectangle from that point of intersection, and you will see that the shape in the middle will not be a square. It will be a 6x4 rectangle. You may be able to draw a 6x6 square inside a rectangle, but it will not have a diagonal with the 2, 5, 3 breakdown like the original drawing. Just think, the segment in the middle has to be 5 units long. How can there be a segment 5 units long going from one end to the other of a 6x6 square?
@@user-yc3yy4sf1x The diagonal goes through the larger rectangle, not through the green area. Just think of the definition of a diagonal!
@@user-yc3yy4sf1x what kinds of right triangles are there where the sides aren't a Pythagorean triplet?
Somthing is wrong here. With a hypo of 5, legs can olny be 3, 4, thus 4^2=16, not 20.
STEP-STEP RESOLUTION PROPOSAL :
01) As anyone can see, the Rectangle [ABCD] is equal to two Green Squares. 5 = 2 + 3
02) So : AB = a and BC = 2a
03) a^2 + 4a^2 = 100 ; 5a^2 = 100 ; a^2 = 100 / 5 ; a^2 = 20 ; a = sqrt(20)
04) Area = sqrt(20) * sqrt(20) ; A = 20
OUR ANSWER:
Green Square Area equal to 20 Square Units.