Can you find area of the Green shaded Square? | (Rectangle) | #math #maths | #geometry

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Can you find area of the Green shaded Square? | (Rectangle) | #math #maths | #geometry

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[bongiweunathi]

Math is like a puzzle to me, too trick but i love it😂

[MrPaulc]

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.

[alexniklas]

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

[Sam.Lord.Cambodia]

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.

[scottdort]

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.

[ОльгаСоломашенко-ьы]

DC= x, sin(<DAC)=x/10, cos(<DAC)=√(1-x^2/100). AK=2* cos(<DAC), AP=7* cos(<DAC), в квадрат правую и левую части уравнения. 25(1- x^2/100)=x^2, x^2=20.

[cyruschang]

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

[apooia]

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.

[marcgriselhubert]

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.

[santiagoarosam]

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.

[zdrastvutye]

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 x(2, 3), y(2, 3):goto 60
30
40
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<masy then mass=masx else mass=masy
160 goto 180
170 xbu=x*mass:ybu=y*mass:return
180 for a=0 to 2:gcol 9+a:x=x(a, 0):y=y(a, 0):gosub 170:xba=xbu:yba=ybu:for b=1 to 4
190 ib=b:if ib=4 then ib=0
200 x=x(a, ib):y=y(a, ib):gosub 170:xbn=xbu:ybn=ybu:goto 220
210 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
220 gosub 210:next b:next a:gcol 8:x=0:y=l4:gosub 170:xba=xbu:yba=ybu
230 x=x(2, 1):y=0:gosub 170:xbn=xbu:ybn=ybu:gosub 210
premath-can you find area of the green shaded square
4.47213596die gruene flaeche= 20
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[giuseppemalaguti]

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

[michaeldoerr]

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!!!

[prossvay]

∆ 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.❤❤❤

[devondevon]

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

[quigonkenny]

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

[himo]

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

[unknownidentity]

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

[shaozheang]

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.

[egillandersson]

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.