just group the boxes relative to their length. Assume vertical is y. So yellow/blue/green has length 18. So area 18y = 128 + x + 40. Then same for blue/green/purple. 14y = 80 + 40 x. Then subtract one from the other and you get 4y =48 so y=12
Width=a so we can have 2 equations. Based on 14 cm part: 80+40+x=14a Based on 17 cm part: 128+40+x=18a To solve we can subtract 1st from 2nd: 48=4a ► a=12 Put quantity of a in one the equation (2): 128+40+x=(18)(12) ► 168+x=216 ► x=48
Thank you Sir. I have solved this using simultaneous equations by taking the breath as y 18y= 128+x+40 14y= x+40+80 Once solved you get y=12 and x=48 in very faster manner. Am I correct Sir.
Let height be 'h' From yellow, blue, and green rectangles, their total area is (18)(h) = 128 + 40 + x 18h = 168 + x....(Eq 1) From pink, green, and blue rectangles, their total area is (14)(h) = 80 + 40 + x 14h = 120 + x...(Eq 2) Subtracting Eq 2 from Eq 1, we get 4h = 168 -120 = 48 h = 48/4 = 12 Substituting value of h in Eq 2, (14)(12) = 80 + 40 + x 168 = 120 + x x = 168 - 120 x = 48
I am trying before watching I) area of Purple (P) eq1: ah=80 II) area of Yellow (Y) eq2: bh=128 III) note similar lines 18+a=14+b simplify to b=a+4 IV) substitute III into eq2 and simplify (a+4)h=128 ah+4h=128 V) subtract eq1 from the result of IV: ah is eliminated and 4h=48 h=12 VI) add BGP area of boxes and simplify, x+120 VII) calculate area 14h=x+120 substitute (h=12) and simplify x+120=168 x=48 cm2
Call the length of each rectangle e. call the width of yellow a. Call width of blue b.Call width of green c and call width of pink d. So a+b+c= 18. b+c+d=14 this means a=d+4. ae=128 so e(d+4)=128. That is ed+4e=128. Then 80+4e= 128 that is e=12. So (128/12)+ (x/12)+ (40/12)= 18. That is 168+x 216 so x=48.
Let height = h. Then total area of big rectangle = 14h +128. Also, total area of big rectangle = 18h + 80. Thus 14h + 128 = 18h + 80. 48 = 4h. 12 = h. Then area of blue + green + purple = 14 x 12 = 168. Area of blue = 168 - 40 - 80 = 48.
Let y be the height of the rectangle: 14y = x + 40 + 80 y = (x+120)/14 ---- (1) 18y = 128 + x + 40 y = (x+168)/18 ---- (2) Equate (1) and (2) and solve for x: (x+120)/14 = (x+168)/18 18(x+120) = 14(x+168) 18x + 2160 = 14x + 2352 4x = 192 x = 48 cm²
here is how i did it: let s be the value such that: 32 - s = width of the whole triangle (32 = 18+14) solve this system of equations: (18-s) * y = 128 (14-s) * y = 80 solution: y = 12 s = 22/3 x = 12*22/3 - 40 = 48
Let h be the height of the entire structure. Then: h = (168 + x)/18 = (120 + x)/14; now we don't need h any more; just cross-multiply on the right: (14)(168 + x) = (18)(120 + x); multiply everything out: 2352 + 14x = 2160 + 18x; collect terms and simplify: 192 = 4x; divide both sides by 4: x = 192/4 = 48; voila! Carpe Diem. 🤠
There is no need to think 'outside the box' for this problem as there is no need to partition the yellow rectangle. Height of each rectangle is same - let's say it is h. Then just add areas of yellow, blue and green rectangles 128+x+40 = 18h... equation 1 Add areas of blue, green and pink rectangles - x+40+80 = 14h...equation 2 Subtract equation 1 from equation 2 128+x+40 -(x+40+80) = 18h-14h => 48 = 4h => h = 12.. substitute this in any of equation 1 or 2. Let's substitute in equation 1 => 128+x+40 = 18*12 => 168 + x = 216 => 168 + x - 168 = 216 - 168 => x = 48 sq cm
48 Answer A different approach: Let the width of the yellow = r, then the length of yellow = the width of the rectangle= 128/r. Hence the width of blue =x/128 r = xr/128 Hence the width of green= 40r/ 128 Hence the width of purple = 80r /128 Since the width of the yellow= r, it can be expressed as 128r/128 which =r I focus on the width since information is given about it, such as yellow + blue + green = 18; hence 128r/128 + xr/128 + 40r/128= 18 168 r + xr = 18* 128 blue + green + purple = 14; hence xr/ 128 + 40r/128 + 80r/128= 14 120 r + xr = 14 *128 The simultaneous equations to be used A and B 168 r + xr = 18* 128 equation A 120 r + xr = 14* 128 equation B 48 r = 4 * 128 subtract equation B from A 48r =512 r = 10 and 2/3 Substituting the value of 'r' into equation A 168 * 32/3 + 32/3x = 18*128 gives 48 as the Answer Since the width =128/r = 128 *32/3 =12 So the width of the rectangle = 12
18 - 128/h = 14 - 80/h 18 - 14 = 128/h - 80/h 4h = 48 h = 12 cm Area X = A - A1 - A2 Area X = b.h - 80 - 40 Area X = 14 . 12 - 80 - 40 Area X = 48 cm² ( Solved√ )
thankyou v.much dear professor 😊
You are very welcome, dear
So nice of you.
Thank you! Cheers! 😀
The solution method is very interesting.
Thanks Sir
Thanks PreMath
just group the boxes relative to their length. Assume vertical is y. So yellow/blue/green has length 18. So area 18y = 128 + x + 40. Then same for blue/green/purple. 14y = 80 + 40 x. Then subtract one from the other and you get 4y =48 so y=12
Width=a so we can have 2 equations.
Based on 14 cm part: 80+40+x=14a
Based on 17 cm part: 128+40+x=18a
To solve we can subtract 1st from 2nd: 48=4a ► a=12
Put quantity of a in one the equation (2): 128+40+x=(18)(12) ► 168+x=216 ► x=48
Thank you Sir. I have solved this using simultaneous equations by taking the breath as y 18y= 128+x+40 14y= x+40+80 Once solved you get y=12 and x=48 in very faster manner. Am I correct Sir.
Let height be 'h'
From yellow, blue, and green rectangles, their total area is
(18)(h) = 128 + 40 + x
18h = 168 + x....(Eq 1)
From pink, green, and blue rectangles, their total area is
(14)(h) = 80 + 40 + x
14h = 120 + x...(Eq 2)
Subtracting Eq 2 from Eq 1, we get
4h = 168 -120 = 48
h = 48/4 = 12
Substituting value of h in Eq 2,
(14)(12) = 80 + 40 + x
168 = 120 + x
x = 168 - 120
x = 48
I tried to calculate height from both relations which is (168+x)/18=(120+x)/14 or 2352+14x = 2160+18x where x = 48
2352+14x = 2160+18x
Very well explained 👍
Thanks for sharing 😊
So nice of you
Thanks for video.Good luck sir!!!!!!!!!
Thank you too
Thank you Proffesor
Let B be the lenght of X & H be the common height. (14 - B)*H = 120, (18 - B)*H = 168. H = 12, B = 4. X = 48
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Thank you! Cheers! 😀
[(128+40)-(40+80)]/(18-14)= 48/4 =12 cm²/cm de base = Altura de todos los rectángulos → X=48
Gracias y saludos.
Excellent!
You are very welcome!
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Yay! I solved the problem.
Super!
Thanks, Professor, for a fun problem!❤
Glad you enjoyed it!
Thank you! Cheers! 😀
Partitioning at first is tricky but near the end, I found it easy
Excellent!
Thank you, Kevin! Cheers! 😀
These "outside the box" solutions are always amazing🤩
Yes they are!
Excellent!
Thank you! Cheers! 😀
Hey, I got one on my own!
Bravo!
I am trying before watching
I) area of Purple (P) eq1: ah=80
II) area of Yellow (Y) eq2: bh=128
III) note similar lines 18+a=14+b
simplify to b=a+4
IV) substitute III into eq2 and simplify
(a+4)h=128
ah+4h=128
V) subtract eq1 from the result of IV: ah is eliminated and 4h=48
h=12
VI) add BGP area of boxes and simplify, x+120
VII) calculate area 14h=x+120
substitute (h=12) and simplify
x+120=168
x=48 cm2
Call the length of each rectangle e. call the width of yellow a. Call width of blue b.Call width of green c and call width of pink d. So a+b+c= 18. b+c+d=14 this means a=d+4. ae=128 so e(d+4)=128. That is ed+4e=128. Then 80+4e= 128 that is e=12. So (128/12)+ (x/12)+ (40/12)= 18. That is 168+x 216 so x=48.
Let height = h.
Then total area of big rectangle = 14h +128.
Also, total area of big rectangle = 18h + 80.
Thus 14h + 128 = 18h + 80.
48 = 4h.
12 = h.
Then area of blue + green + purple = 14 x 12 = 168.
Area of blue = 168 - 40 - 80 = 48.
Let y be the height of the rectangle:
14y = x + 40 + 80
y = (x+120)/14 ---- (1)
18y = 128 + x + 40
y = (x+168)/18 ---- (2)
Equate (1) and (2) and solve for x:
(x+120)/14 = (x+168)/18
18(x+120) = 14(x+168)
18x + 2160 = 14x + 2352
4x = 192
x = 48 cm²
Please put some algebra questions
Sure, keep watching.
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(168+x)÷18 =(120+x)÷14 solve for x. A very simple solution.
Quite easy, thanks for thinking about us lesser morals 🤓
No worries!
Take care
Easy way:Let the height of the rectangle be y. Then we get2 equations 1)18y=128+40+x
2)14y=80+40+x. Solving these equations we get y=12 and x=48.
here is how i did it:
let s be the value such that:
32 - s = width of the whole triangle
(32 = 18+14)
solve this system of equations:
(18-s) * y = 128
(14-s) * y = 80
solution:
y = 12
s = 22/3
x = 12*22/3 - 40 = 48
Let h be the height of the entire structure. Then:
h = (168 + x)/18 = (120 + x)/14; now we don't need h any more; just cross-multiply on the right:
(14)(168 + x) = (18)(120 + x); multiply everything out:
2352 + 14x = 2160 + 18x; collect terms and simplify:
192 = 4x; divide both sides by 4:
x = 192/4 = 48; voila!
Carpe Diem. 🤠
x= 14.h - (40+80) = 14.h -120
x= 18.h - (128 +40) =18.h -168
18h-14h = 168-120 4h=48 h=12
x= 14 .12 - 120 = 168-120=48
Excellent!
Thank you! Cheers! 😀
。Simply (168+x)/(120+x)=18/14=9/7, 7x168+7x=9x120+9x, x=(7x168-9x120)/2=48.😊
Excellent!
Thank you! Cheers! 😀
There is no need to think 'outside the box' for this problem as there is no need to partition the yellow rectangle. Height of each rectangle is same - let's say it is h. Then just add areas of yellow, blue and green rectangles
128+x+40 = 18h... equation 1
Add areas of blue, green and pink rectangles -
x+40+80 = 14h...equation 2
Subtract equation 1 from equation 2
128+x+40 -(x+40+80) = 18h-14h
=> 48 = 4h
=> h = 12.. substitute this in any of equation 1 or 2. Let's substitute in equation 1
=> 128+x+40 = 18*12
=> 168 + x = 216
=> 168 + x - 168 = 216 - 168
=> x = 48 sq cm
48
Bravo!
(128+X+40)/18=(X+40+80)/14 is easier maybe.
Thank you! Cheers! 😀
👋👋👋👋
48 Answer
A different approach:
Let the width of the yellow = r, then the length of yellow = the width of the rectangle= 128/r.
Hence the width of blue =x/128 r = xr/128
Hence the width of green= 40r/ 128
Hence the width of purple = 80r /128
Since the width of the yellow= r, it can be expressed as 128r/128 which =r
I focus on the width since information is given about it, such as
yellow + blue + green = 18; hence 128r/128 + xr/128 + 40r/128= 18
168 r + xr = 18* 128
blue + green + purple = 14; hence xr/ 128 + 40r/128 + 80r/128= 14
120 r + xr = 14 *128
The simultaneous equations to be used A and B
168 r + xr = 18* 128 equation A
120 r + xr = 14* 128 equation B
48 r = 4 * 128 subtract equation B from A
48r =512
r = 10 and 2/3
Substituting the value of 'r' into equation A 168 * 32/3 + 32/3x = 18*128
gives 48 as the Answer
Since the width =128/r = 128 *32/3 =12
So the width of the rectangle = 12
18y=168+x
14y=120+x
4y=48
Y=12
14(12)=120+x
172=120+x
X= 52
Almost perfect !!!
14 . 12 = ??????
All others found X=48, except you
@@marioalb9726 12x12+2x14 according to my bad math.
18 - 128/h = 14 - 80/h
18 - 14 = 128/h - 80/h
4h = 48
h = 12 cm
Area X = A - A1 - A2
Area X = b.h - 80 - 40
Area X = 14 . 12 - 80 - 40
Area X = 48 cm² ( Solved√ )