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

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

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[SkinnerRobot]

5x = 8/√2  ⇒ x = 8/(5√2)

Yellow area = 2⋅1/2⋅x⋅5x/2
= 1/2⋅8/(5√2)⋅8/√2
= 32/10
= 3.2

Blue area = (8/√2)² - 3.2
= 32 - 3.2
= 28.8 square units

[misterenter-izrz]

The total area is 8^2/2=32, the blue shaded area is 32×(1-1/10)=32×(9/10)=28.8.😊

[yalchingedikgedik]

Very nice and enjoyable
Thanks Sir for your efforts
❤❤❤❤❤

[gaylespencer]

Calculate the value of x, * which allows you to calculate the area of the one of the yellow triangles, (x*(5x/2))/2, which you will then double. Subtract that total from the area of the square.

* Isolating the value of x was done via this equation, 8 (diagonal) = (5x)(2^.5)

[santiagoarosam]

Área amarilla =(1/5)*(1/2)=1/10 del área del cuadrado ---> Área azul =(9/10)(8²/2)=288/10=28, 8 ud².
Gracias y saludos

[marioalb]

Side of square:
s = 5x = 8/√2 = 4√2 cm
Yellow area:
x = s/5 = 0, 8√2 cm
A = ½b.h = ½ x s = 3, 2 cm²
Square area:
A = s² = 32 cm²
Blue shaded area:
A = A₁ -A₂ = 28, 8 cm² ( Solved √)

[DB-lgsq]

h=AT/2
S = 2[1/2(5x)^2 - 1/2(x . 5x/2)
2(5x)^2=8^2
S=32- 16/5
=28, 8
شكرا لكم على المجهودات .

[Waldlaeufer]

AB = 8 => AT = 8/√2 = 4√2 = 5x => x = 4/5 √2
A(yellow) = 1/2 * b * h = 1/2 * x * 5x = 1/2 * 4/5 √2 * 4√2 = 1/2 * 16/5 * 2 = 16/5
A(blue) = (5x)² - A(yellow) = (4√2)² - 16/5 = 32 - 16/5 = 160/5 - 16/5 = 144/5 = 28.8 units

[devondevon]

28.8
Area of square in terms of x = 25 x ^2
Area of yellow region in terms of x = x * 5x * 1/2 = 5x^2/2 or 2.5 x^2
Hence, yellow region is 1/10 the area of square
Hence, area of blue = area of square - 1/10 area of square
Since, the diagonal of square = 8,
then let 5x^2 + 5x^2= 64
50x^2 = 64
25 x^2 = 32
Area of square = 32
Hence area of blue = 32 - 3.2 = 28.8 a

[quigonkenny]

Triangle ∆APB:
AP² + PB² = AB²
(5x)² + (5x)² = 8²
25x² + 25x² = 64
50x² = 64
x² = 64/50 = 32/25
x = √(32/25) = 4√2/5

Let the center of square ATBP be O. As ∠DOA = ∠COB, as vertical angles, ∠aoAD = ∠OBC = 45°, and AO = BO = 4, ∆DOA and ∆COB are congruent. As PB is perpendicular to both BC and AD, the height of both triangles is equal to PB/2 = 5x/2. The blue area will be equal to rhe area of square ATBP minus the areas of the two triangles.

A = (5x)² - 2(x(5x/2)/2) = (5x)² - 5x²/2
A = (4√2)² - 5(4√2/5)²/2
A = 32 - 5(32/25)/2
A = 32 - 16/5 = 144/5 = 28.8 sq units

[rey-dqnx]

x=4√2/5
Area yellow= (4√2/5)(2√2)
Area yellow= 16/5
Area blue=32-16/5
144/5

[MrPaulc]

Square's full area is (8/sqrt(2))^2 = 64/2 = 32.
8/sqrt(2) is equivalent to 4*sqrt(2).
4*sqrt(2) is 5x, so x = (4/5)*sqrt(2) so x = 4*sqrt(2)/5
Take one yellow triangle: base 4*(sqrt(2))/5 and height of (4*sqrt(2)/2.
Multiply for (16*2)/10.
I won't take half the base because there are two equal triangles, so will take that answer as it is.
32/10 = 3.2 or, if preferred, 16/5.
Blue areas total 32 - 16/5 = 160/5 - 16/5 = 144/5 = 28.8 un^2
Now check the video: yes, we went in different directions but arrived at the same place. There appear to be several ways to solve this one.
Thanks again. You teach well.

[prossvay]

APBT Is a square
So AP=PB=BT=TA=5x
In ∆APB
AP^2+PB^2=AB^2
(5x)^2+(5x)^2=8^2
50x^2=64
x=8/5√2=4√2/5
So AP=PB=BT=TA=4√2
∆ADE Concurrent ∆ BCE
So BE=EA=4 and CE=ED
So E is middle AB and CD
By BC Parallel DP
So F is Middle BP
Area of triangle BCE and
Area of the square=(5x)^2=(4√2)^2=32
Blue shaded area=32-16/5=28.8 square units.❤❤❤

[marioalb]

Yellow Area:
A = ½b.h = ½ x 5x = 2, 5.x²
Square area:
A = s² = (5x)² = 25x² = ½8² = 32cm²
Segment "x" squared:
x² = 32/25 cm
Shaded Area:
A = A₁-A₂= 25x² - 2, 5x² = 22, 5 x²
A = 28, 8 cm² ( Solved √ )

[aljawad]

It’s easy to see that triangles AED & CED are identical (three internal angles are equal + one side), so directly the area of each is (1/2)*(x)*(h), where h = 1/2 the value of x.

[michaeldoerr]

The answer is 144/5 units square. I am glad that I have learned of another use of vertical angles!!!

[Khantia]

I got a result of 28.8 from the thumbnail and came to verify my answer.
First I calculated the sides and the area. Then I figured out that the yellow area is 1/10 of the whole thing (as it's 2 triangles, so half of the area a 1/5 would be), and then I just calculated 32 * 0.9 :P

[johnbrennan]

Triangles mbe and ane are congruent so 5x-h =h ie h=5x/2. Area triangle ade = 1/2 by x by 5x/2=(5x^2)/4. Total yellow area = (5x^2)/2 since yellow triangles congruent etc.

[Birol]

My way of solution ▶
Let's consider the right triangle ΔAPB
[AP]=[PB]= 5x
[BA]= 8
according to the Pythagorean theorem we can write:
[AP]²+[PB]²= [BA]²
⇒
(5x)²+(5x)²= 8²
25x²+25x²= 64
50x²= 64
x²= 64/50
x= 4√2/5

b) The intersection of the red line with the black line is O, so, both triangles with the yellow area are equal to each other: ΔADO = ΔCOB

c) A(ΔADO) = A(ΔCOB)
A(ΔADO)= x*h/2
h= 5x/2
Ayellow= A(ΔADO) + A(ΔCOB)
Ayellow= x*5x/2*(1/2)*2
Ayellow= 5x²/2
Ayellow= 5*(32/25)*(1/2)
Ayellow= 16/5 length units

d) Ablue= A(APBT) - Ayellow
A(APBT)= 5x*5x
A(APBT)= 25x²
A(APBT)= 25*(32/25)
A(APBT)= 32
⇒
Ablue= 32 - 16/5
Ablue= 144/5
Ablue= 28, 8 square units

[himo]

8*8/2=32 (5x)²+(5x)²=8² 50x²=64 x²=32/25

Blue shaded area = 32 - 3.2 = 28.8