Can you find the length X? | (Rectangles) | #math #maths | #geometry

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Can you find the length X? | (Rectangles) | #math #maths | #geometry

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[yalchingedikgedik]

Thanks Sir
That’s very useful and enjoyable
We are very interested from the exercises
❤❤❤❤❤❤

[quigonkenny]

I was able to eyeball 12 as the correct answer fairly quickly, but it was nice to see how it was proven.

[misterenter-izrz]

Rather difficult I suppose😢, (x-3)y=27, (5-y)z=8, (z+5)y=27, x=z+8, ....😅

[timeonly]

@ 8:20 Once you found that y=3, you can get the other side of the green rectangle by dividing y=3 into the area (27) and get 27/3 = 9... WITHOUT plugging y=3 into that ugly expression! Fun problem.

[Birol]

My way of solution ▶
20= 5*t
t= 4

15= 5*u
u= 3

a+b= 5
b+c= 4
⇒
a+b= 5
-b-c= -4
a-c= 1
a= c+1
we can write a= c+1

b*d=
a(5+d)=
a= c+1
⇒
(c+1)(5+d)=

c+1+b= 5
b+c= 4
b= 4-c
if we put this in the equation (I) above:
b*d= 8
(4-c)*d= 8
d= 8/(4-c)
⇒
(c+1)*(5+d)=
if we put d= 8/(4-c) we get:
(c+1)*[5+ 8/(4-c)] = 27
(c+1)*(20-5c+8)/(4-c)= 27
(c+1)(28-5c)= 27(4-c)
28c-5c²+28-5c= 108-27c
5c²-27c+108-28-23c=0
5c²-50c+80= 0
c²-10c+16=0
Δ= 100-4*1*16
Δ= 36
√Δ= 6

c₁= (10+6)/2
c₁= 8

c₂= (10-6)/2
c₂= 2

if c= 8
b= 4-c would be negative❗
⇒
c= 2

b= 4-c
b= 4-2
b= 2

d= 8/(4-c)
d= 8/(4-2)
d= 8/2
d= 4

x= 5+d+u
x= 5+4+3
x= 12 length units

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

Let's check these Solutions :

01) Sum of All Areas = 70
02) 20 / 5 = 4
03) 15 / 5 = 3
04) 27 = 3 * 9 (?)
05) 8 = 4 * 2 (?)
06) Left Side = 4 + 3 (?)
07) The only possible solution herein is x = 9 + 3 = 12

[santiagoarosam]

8=b*(8/b)→ 27=(5+b)[5-(8/b)]→ b=4→ X=(5+4)+(15/5)=9+3=12 ud.
Buen rompecabezas. Gracias y un saludo cordial.

[Nothingx]

0:10 blue, pink and yellow pre math is in my pillow 😊

[himo]

20/5=4 15/5=3
(x-3)y=27 (x-8)(5-y)=8 x-3=27/y
(27/y-5)(5-y)=8 135/y-27-25+5y=8 135/y-60+5y=0
5y²-60y+135=0 y²-12y+27=0 (y-3)(y-9)=0 y=9 is rejected, thus y=3
(x-3)*3=27 x-3=9 x=12

[Abdullah-okmw]

I can also find x from this 8/(x-8)+27/(x-3)=5
Wich 5 is right side of figure, x-8 is over side of S=8 figure, x-3 is over side of S=27. x=6 and 12. 6 is not suitable. Answer is x=12.

[ramnagarajan]

Extend the pink rectangle's into the blue rectangle to split it into two parts. Call the bottom area 5a.

Total area of the bigger rectangle formed by this is: 27+15+8+5a = 50+5a. Since the height is 5, the length x is (10+a)

That length can also be expressed as: 5 + (8/a) + 3 = 8 + (8/a)

So, 10 + a = 8 + 8/a
Solve to get a = 2
So, x = 10+a = 12

[adamrussell]

Just looking at the diagram it looks like pink is 2x4. Yellow is 5x3 so X is 3+4+5=12. And since the numbers all work out with no contradictions our original assumption for pink is correct and our answer is correct.

[timmcguire]

I always estimate a ballpark for the answer before I start - just to make sure my answer at the end makes sense. In this case I said x looks like it should be 11, 12, or 13ish. In my head 6 plugged the 12 in for x and bingo, saw it worked without picking up the pencil. Of course, that doesn't always work out that way :-)

[Pixel_Runner]

Did anyone else just work out 3 (for the bottom right rectangle). Then look at the bottom length rectangle and know that it must be 3*9.

So it must have been 12. Much easier and quicker!

[lilpapi]

Blue top side = 5
Yellow top side = 3 since area 15 and side of 5
Pink top = 4 since top is longer than side and area is 8, so it has to be 4 x 2 rectangle

So yeah x = 12

[frankmutuma]

Could solve for X directly by using the rectangle of area 8 and obtaining x2 - 18x + 72 = 0.

[singaporecambridgeolevelsm]

Noticed the light blue rectangle is redundant. That shows there's another way to do it . But the solution on the clip is good if it didn't even have to use anything linked to the blue triangle

[unknownidentity]

Let's find x:
.
..
...
....


Let's use the labels x and y for the horizontal and vertical side lengths of the rectangles, respectively. Then we obtain:

x(b) = 5 ; y(b) = A(b)/x(b) = 20/5 = 4
y(y) = 5 ; x(y) = A(y)/y(y) = 15/5 = 3

x(b) + x(p) = x(g)
y(g) + y(p) = y(y)
x(p)*y(p) = A(p)
x(g)*y(g) = A(g)

5 + x(p) = x(g)
y(g) + y(p) = 5
x(p)*y(p) = 8
x(g)*y(g) = 27

x(p)*y(p) = 8
(5 + x(p))*(5 − y(p)) = 27

x(p)*y(p) = 8
25 − 5*y(p) + 5*x(p) − x(p)*y(p) = 27

25 − 5*y(p) + 5*x(p) − 8 = 27
5*x(p) − 5*y(p) = 10
x(p) − y(p) = 2

x(p)*y(p) = 8
x(p) − y(p) = 2

(2 + y(p))*y(p) = 8
2*y(p) + y²(p) = 8
2*y(p) + y²(p) − 8 = 0

y(p) = −1 ± √[(−1)² + 8] = −1 ± √(1 + 8) = −1 ± √9 = −1 ± 3

Since y(p)>0, the only useful solution is:

y(p) = −1 + 3 = 2
x(p) = A(p)/y(p) = 8/4 = 2

y(g) = 5 − y(p) = 5 − 2 = 3
x(g) = A(g)/y(g) = 27/3 = 9

Now we are to calculate the value of x:

x = x(g) + x(y) = 9 + 3 = 12

Best regards from Germany

[michaelkouzmin]

Just another solution:
W - width; H - height;
W_yellow = 15/5=3;
W_green = x - W_yellow = x - 3;
H_green = 27/W_green = 27/(x-3);
W_red = W_green - W_blue = (x-3) - 5 = x-8;
H_red = 8/W_red = 8/(x-8);
H_yellow = H_green + H_red = 27/(x-3) + 8/(x-8) = 5;
2 roots: x1 = 12 (accepted); x2 = 6 (rejected as x>8 at least)))
x= 12.

[mracjesstark]

just by looking at it i had it 12. but its good to know the equation process.