Can you find the side length X? | (Step-by-step explanation) |

[@arnavkange1487]

Respect button for premath 👇

[@PreMath]

Thanks dear ❤️🌹

[@quigonkenny]

As all sections are y cm² and there are five of them, each one is one fifth of the total area. As the orange section takes up the entire bottom of the square and leaves an intact rectangle made up of the other four sections, that means the remaining rectangle is of the dimensions s × 4s/5, width by height, where s is the side length of the full square. Separating the orange section removes one fifth of the total area. As the blue section takes up the entire left side of the remaining rectangle and leaves an intact rectangle made up of the other three sections, that means the remaining rectangle is of the dimensions 3s/4 × 4s/5, width by height. Separating the blue section removes one fifth of the total area, or one fourth of the area minus the orange section. As one fourth of the wisth has been reloved, and that measures 8 units, we can determine that s = 4(8) = 32 cm. This means the remaining rectangle, minus blue and orange sections, is 3(32)/4 × 4(32)/5 = 24 × 128/5 [25.6], width by height. As the yellow section takes up the entire top side of the remaining rectangle and leaves an intact rectangle made up of the other two sections, that means the remaining rectangle is of the dimensions 24 × (2/3)(128/5) = 24 × 256/15 [17.0666], width by height. Separating the yellow section removes one fifth of the total area, or one third of the area minus the orange and blue sections. As x is the height of the green and pink sections: x = 256/15 cm ≈ 17.0666... cm. As the width of the green section is 24/2 = 12, the area is 12(256/15) = 1024/5. As 32² = 1024, this is 1/5 of rhe total area of the square.

[@PreMath]

Thanks ❤️

[@jimlocke9320]

At 6:30, we have found that the area of each rectangle is 1024/5 and the yellow rectangle has width 24. Combine the purple and green rectangles into one rectangle. It also has a width of 24 and its height is x. Its area is twice that of a single rectangle, or (2)(1024/5) = 2048/5. So, using the formula area of a rectangle equals base times height, 2048/5 = (24)(x) and x = 2048/((5)(24) = 2048/120 = 256/15 ~= 17.1, as PreMath also found.

[@PreMath]

Excellent! Thanks ❤️

[@cyruschang1904]

Area of the bottom rectangle is 1/5 of the square. Its length is DC, this means it's width is 1/5 of AD, which makes the length of the blue rectangle 4/5 of AD If (4/5 of AD)(8) = 1/5 of the square area (AD^2), then AD = 32 = the side length of the square y = (32)(32)/5 x = 32 - (y)/(32 - 8) - (32/5) = 32 - (32)(32)/(32 - 8)(5) - (32/5) = 32 (1 - 32/120 - 1/5) = 32 (1 - 4/15 - 1/5) = 32 (8/15) = 256/15 (cm)

[@PreMath]

Excellent! Thanks ❤️

[@cyruschang1904]

@@PreMath Thank you

[@devondevon4366]

17.067 Let , the side of the square = n, then area = n^2 and area of each rectangle = n^2/5 (since there are 5) The length of the blue = ( n^2/5 )/8 = n^2/40 Since the combined area of yellow, purple, green and blue = 4n^2/5 (4* n^2/5), and since the combined area of all four equals the length of blue, then the length of the square = (4n^2/5)/n^2/40 = 4n^2/5 * 40/n^2 4 /5 * 40 = 32. Hence, the length of the square = 32 Therefore, the length of the yellow is 24 ( 32-8) Therefore, the combined length of the purple and green is, also, 24 But since the combined area of the purple and the green is 2n^2/5 (n^2/5 *2), then X = 2n^2/5 divided by 24 = 2n^2/ 120 = n^2/60 = 32^2/60 = 1024/60 = 256/15 = 17.067 Answer

[@PreMath]

Excellent! Thanks ❤️

[@phungpham1725]

Let a be the side of the square and h the width of the rectangle at the bottom. Consider the blue and this rectangle. We have h.a = 8.(a-h) = 8a-8h---> h= 8a/(a+8) So a . 8a/(a+8) = sq a/5---> 8.sqa/(a+8)= sqa/5---> a=32 Therefore the width of the green rectangle= 24/2= 12 12x= sqa/5----> x= sqa/60= 1024/60 = 17,06 units

[@PreMath]

Excellent! Thanks ❤️

[@prossvay8744]

8+2y/x=3x/2+xy/8x+2y (1) (side of square) And area square=(8+2y/x)(3x/2+xy/8x+2y)=5y So y=12x (1): 8+24x/x=3x/2+x(12x)/8x+24x 8+24=3x/2+(12x^2)/32x 32=3x/2+3x/8 32×8=12x+3x 15x=256 So: x=256/15 or x=17.1 cm. ❤❤❤ Thanks sir.

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@tombufford136]

At a quick glance: y area is 1/5 of square area. square with side lengths, s. Then (s-8) * x = purple and green areas = 2/5 * s * s = s*x - 8*x. Yellow area has height = x/2 and brown area has height 1/5 * s. Then s= 1/5 * s + x + 0.5 *x . Then 5 *s = s + 7.5*x . Then 4 * s = 7.5 *x. Then x = 8/15 * s. Then 2/5 * S^2 = s*8/15 * (s- 8) = 8/15 * s^2 - 64/15 * s. Then 6/8 * s^2 = s^2 - 8 * s. Then 1/4 * s = 8 and s = 32 . Then x = .17.1 cm

[@PreMath]

Thanks ❤️

[@santiagoarosam430]

Amarillo+Púrpura+Verde=3*Azul AB=(1+3)8=32. Área ABCD=32*32=1024. Y=1024/5 Longitud amarillo=3*8=24. Ancho púrpura=Ancho verde=24/2=12 X=(1024/5)/12=1024/60=256/15 Bonito acertijo. Un saludo cordial.

[@PreMath]

Thanks ❤️🌹

[@MrPaulc222]

Call the square's sides 'a'. a^2=5y Yellow rectangle's length is a-8 y/(a-8)=width of yellow. Green and purple triangles together: x(a-8)=2y, so x is twice yellow's width. x=(2y)/(a-8). Also, 8((3/2)x)=y when checking the blue rectangle. Let's see if some rearranging etc can sort this out. Double the second equation 8(3x)=2y and in the first equation, x(a-8) = 2y Therefore, 24x = ax - 8x 32x = ax, so I suppose a (the square's sides) = 32. Therefore, each rectangle has an area of (32^2)/5 = 1024/5 Therefore, green and purple widths are each 12. 1024/5 divided by 12 = 1024/60, which I think must be x. Simplify: 512/30 -->256/15. I think that's as far as I can go before decimalising x = 17.067 (rounded). Your way looks cleaner than mine, but at least my answer was okay. Thank you again.

[@PreMath]

You are very welcome! Thanks ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@SkinnerRobot]

y=8a ab=3•8a ⇒ b=24 12x=(8+24)²/5 ⇒ x=256/15≈17.07

[@PreMath]

Thanks ❤️

[@aljawad]

I created a set of 6x6 simultaneous equations summarizing the what is known about the dimensions of the collection of rectangles, solved it to reach the same result.

[@marcgriselhubert3915]

The area of the suare is 5.y, so its length is sqrt(5.y). The length of the blue rectangle is y/8 and the width of the pink rectangle is y/sqrt(5.y) = sqrt(5.y)/5 AD = sqrt(5.y) = y/8 + sqrt(5.y)/5, so (4/5).sqrt(5y) = y/8, at the square we get: (16/25).(5.y) = (1/64). (y^2) which gives that y = (16/25).(5).(64) = 1024/5 The length of the yellow rectangle is sqrt(5.y) -8, so the area of the purple + green rectangles is (sqrt(5.y) -8). x = 2.y So x = (2.y) / (sqrt(5.y) -8) = (2048/5) / (sqrt(1024) -8) = (2048/5) / 24 = 256/15

[@PreMath]

Thanks ❤️

[@unknownidentity2846]

Let's find x: . .. ... .... ..... May h be the label of the horizontal side lengths and may v be the label of the vertical side lengths. Now we can conclude: A(purple) = A(green) and v(purple) = v(green) ⇒ h(purple) = h(green) ⇒ h(yellow) = 2*h(green) ⇒ v(yellow) = v(green)/2 = x/2 v(blue) = A(blue)/h(blue) = y/(8cm) v(blue) = v(yellow) + v(green) = x/2 + x = 3x/2 3x/2 = y/(8cm) ⇒ y = (12cm)*x ⇒ h(green) = 12cm ⇒ h(yellow) = 24cm ⇒ AB = h(blue) + h(yellow) = 8cm + 24cm = 32cm So the area of the square ABCD turns out to be (32cm)² = 1024cm². Now we can conclude: y = A(ABCD)/5 = (1024/5)cm² x = y/(12cm) = (256/15)cm Best regards from Germany

[@PreMath]

Excellent! Thanks ❤️

[@saptspt]

Brown area is 1/5 square,so By blue area -->>> Y=(8*4/5)*5(8*4/5)=204,8cm² So width of violet dan green area -->>> (32-8)/2=12cm X=204,8/12=17,067cm

[@PreMath]

Thanks ❤️

[@ybodoN]

In the form of an equation system: {xz = y, (x + w) 2z = 3y, (x + w) (2z + 8) = 4y, (2z + 8)² = 5y} The solutions are: w = 0, x = 0, y = 0, z = −4 ☒ or w = 128/15, x = 256/15, y = 1024/5, z = 12 🗹

[@PreMath]

Thanks ❤️

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال S=ABCDمساحة S=5y y=8a AB=32 y=32^2/5 2y=24x y=12x x=y/12= 256/15

[@PreMath]

Excellent! Thanks ❤️

[@allalwadi7063]

Total area of square = 5Y --> Square side = ✔️(5Y) length of blue rectangle = Y/8 length of yellow rectangle = ✔️(5Y) - 8 width of yellow rectangle = = Y/ [ ✔️(5Y) - 8 ] Area of ( pink + green ) triangles = 2Y = [ ✔️(5Y) - 8 ] X ---> X = 2Y / [ ✔️(5Y) - 8 ] Area of ( blu + yellow + green + pink ) rectangles = Y/8 • ✔️(5Y) = 4Y ✔️Y = 32 / ✔️5 Y = 32²/5 ---> X= 2Y / ( ✔️(5Y) -8) = 2(32²)/5 / (32 -8) = 2(32²)/5 / 24 = 32(8) / 15 = 2⁸ / 15 = 256/15

[@StephenRayWesley]

(8)^2=64° (8)^2=64° (64°+64°)=128° (360°-128°)=5 132° √13^√1 2^√1 12 (x (ABCDx+1ABCDy-2)

[@PreMath]

Thanks ❤️

[@BKNeifert]

Wow. That one had my head reeling. But I think I understand how it worked.

[@fearfulninja3450]

I have a question where can I show you it

[@Girlnextdoor-nala]

Hi! How did you get to 256/15?

[@LuisdeBritoCamacho]

Let's start giving names to the Rectangles: Yellow Rectangle = R(y) = y sq cm Blue Rectangle = R(b) = y sq cm Green Rectangle = R(g) = y sq cm Red Rectangle = R(r) = y sq cm Pink Rectangle = R(p) = y sq cm So the Total Area of the Square [ABCD] is 5y sq cm and the Side is sqrt(5y) cm Looking at the R(y) = y sq cm we can see that is twice the Area of R(r) + R(g) = 2y sq cm. So the height of the R(y) is x/2 cm The height of the R(b) = (8 * x/2) + 8x = 4x + 8x = 12x Now we have a connection between x and y; y = 12x And now the Total Area of the Square is (5 * 12x) = 60x Let's make a connection between x and 8! The Base of the R(r) + R(g) = 24x and the height is x. Dividing 24x by x we obtain 24. In this case the Side of the Square [ABCD] is 32 cm, and the Area is 32^2 sq cm = 1.024 sq cm Each Rectangle is equal to 1.024 / 5 = 204,8 sq cm Calculating x 24 * x = 2 * 204,8 24x = 409,6 x = 409,6 / 24 = 4.096 / 240 = 2.048 / 120 = 1.024 / 60 = 512 / 30 = 256 / 15 cm x ~ 17,07 cm My answer: The length x is equal to approx. 17,07 cm. That's all folks!!

[@JSSTyger]

3x/2=y/8 x=y/12

[@PreMath]

Thanks ❤️

[@shaozheang5528]

The answer should be 300/19 or 15.8

[@축복이-x6u]

asnwer=12 cm

[@축복이-x6u]

asnwer=17.5 cm her

[@arnavkange1487]

1st view

[@PreMath]

Super Thanks ❤️

[@wackojacko3962]

1024 256 128 64 32 16 8 4 2 1 1/2 1/4 1/8 1/16 1/32 1/64 ...The Eye of Horus! 🙂

[@PreMath]

Thanks ❤️

[@robertlynch7520]

Well … there certainly are other ways to solve this, right? I used an iterative approach, since the algebra in my formulation was getting too hairy. The all-rectangles-equal point was reached with [𝒔 = 32], with the [𝒙 = 17.066666…], which calculating backwards was 256 ÷ 15. Cheαting? Nah. If I were (and am not) an AI cyborg, I would generally use very-very high precision iterative solving of complex systems such as this, because it eliminates having to deal with all the algebra terms and expansions. It generally works for most everything - intersecting circles, triangles, rectangles, sines, tangents, parabolæ. Just saying. Here's the code: # PERL code my $s; my $ds = 0.1; my ( $A, $B, ¢, $D, € ); # the triangles; my ( ¥, $m, $k, ); # the itsy bits; for( $s = 2 × 8; $s < 55; $s += $ds ) { ¥ = 8 × $s / ( $s ⊕ 8 ); # napkin algebra; $A = ($s - $y) × 8; # more napkin algebra; € = $s • $y; $k = ( 8 × $s •• 2 ) / ( $s •• 2 - 8 × × 2 ); # even more napkin algebra; $B = ($s - 8) • $k; ¢ = ½ • (( $s - $k - ¥ ) • ($s - 8)); $D = ¢; my $x = ( $s - $k - ¥ ); next if (abs( $D - $B ) > 10⁻¹¹ ); # discard any but the iterative solution; printf "s %6.2f ", $s; printf "y %8.5f ", $y; printf "A %8.5f ", $A; printf "E %8.5f ", $E; printf "k %8.5f ", $k; printf "B %8.5f ", $B; printf "C %8.5f ", ¢; printf "D %8.5f ", $D; printf "x %8.5f ", $x; printf " "; }

[@PreMath]

Awesome! Thanks ❤️