Find the Area of the Green Rectangle | Step-by-Step Explanation

[MathZoneKH]

I’ll to get what you teach me by this video😊

[JohnAlcaraz]

I solved the problem based on the thumbnail alone, so I did not know or need the assumption that all measurements were whole numbers. Let h be the height of the green rectangle, and let x be the base of the left rectangle. Then the height of the left rectangle is 12/x. It is also h-5, so h-5 = 12/x or

h = (12/x) + 5 (Equation1)

The base of the right rectangle is 9-x, and its height is 20/(9-x). The height of the right rectangle is also h-4, so h-4 = 20/(9-x) or

h = 20/(9-x) + 4 (Equation 2)

Combining equations 1 and 2 and manipulating yields the quadratic equation x^2 + 23x - 108 = 0, with solutions x = -27 and x = 4. Since x must be positive, x=4. Setting x=4 in equation 1 yields h=8, and therefore the area of the green rectangle is (base)(height)=(3)(8)=24.

[philipkudrna]

Simply by looking at it and having the info that the solution is whole numbers: the left rectangle must be 4*3=12 (base*height) and the left 5*4=20. Check base: 4+5+3=12; Heights: 3+5=4+4=8. Area: 8*3=24.

[ebich]

I substituted b=(12/a) from ab=12 into the equation 9a-b=23 to get 9a-(12/a)=23, and multiplied both sides by a to get 9a^2-23a-12=0. Factorizing the left side, (a-3)(9a-4)=0, so A=3or(4/9). When a=3 which satisfies the condition If a=3, then b=4, and 3+5=4+4, which satisfies the condition, but if a=(4/9), then b=27, which does not satisfy the condition, so a=3.

[hongningsuen]

Let a, b be the values as you stated.
Find b to get height of green rectangle = 5 + b hence its area = 3 x (5 + b).
Simultaneous equations are formed as ab = 12 and (9 - a) x (b + 1) = 20.
Substituting a as 12/b into 2nd equation get (9 - 12/b)(b + 1) = 20.
Transform it into a quadratic equation 9b^2 - 23b - 12 = 0.
Solving the quadratic equation get the roots b = 3 and b = 0.44.
With b = 0.44, a = 12/0.44 = 27.2 which is > 12 and is not possible. Hence b = 0.44 is rejected.
With b = 3, area of green rectangle = 3 x (5 + 3) = 24.

[math]

you are getting very creative with the questions, great job solving

[procash]

Solved mentally within a minute

But the method used will be great to solve problems having bigger & fraction numbers. I like

[williamwingo]

Perhaps a more pedestrian method, but it works:
Left-hand rectangle:
ab = 12; so
a = 12/b;

Now right-hand rectangle:
(9 - a)(b + 1) = 20; multiply out:
9b + 9 - ab - a = 20; substitute the values of ab and a above:
9b + 9 - 12 - 12/b = 20; collect terms:
9b - 3 - 12/b = 20; subtract 20 from both sides:
9b - 23 - 12/b = 0; multiply both sides by b:
9b^2 - 23b - 12 = 0; gee, that form looks familiar....

Now invoking the quadratic formula: Shazam!!

b = (+23 +/- sqrt(23^2 - 4*9*(-12))/(2*9); [23^2 = 529; and 4*9*12 = 4*3*3*12 = 3*3*4*12 = 3*12*12 = 3*144 = 432]
= (+23 +/- sqrt(529 + 432))/18; [529 + 432 = 961]
= (+23 +/- sqrt(961))/18; [sqrt(961) = 31; see note below]
= (23 +/- 31)/18; reject the 23 - 31 possibility since b must be positive: 23 + 31 = 54;
= 54/18; divide top and bottom by 9;
= 6/2;
= 3.

So b = 3; b + 5 = 8; and 3(b + 5) = 3*8 = 24.

whew!

Note: Calculators are not specifically prohibited in this problem; but if they were, it could be done by hand; or we might remember that 30^2 = 900, and try 31.

[sarojahebbar]

After watching your video s, I solved this without watching the video. Thanks a lot.

[drpkmath]

Hey my friend! Very cool thumbnail and also interesting question as always! Great job haha

[mustafizrahman]

I shouldn't want to see your rule because I want to try at first. (If I fail, I see your rule). And many many thanks for giving this technical problem.

[montynorth]

ab=12
Therefore b = 12/a
(9-a) x (b+1) = 20

Substituting for b in terms of a.
(9-a) x [ (12/a) +1] = 20
108/a - a -12a/a -a +9 = 20
108/a -a -12 +9 =20
108/a -a -3 = 20
multiply both sides by "a"
108 -a^2 -3a =20a
rearranging:-
a^2 = 108 -23a
Then a^ + 23a - 108 = 0
This is a quadratic equation.
Using the formula, answer 4.

[Viesto]

We have 3 rectangles. The area of the first rectangle is 12, and sides of this rectangle are a(horizontal) and b (vertical). Sides of second rectangle are 3 and x. The area of third rectangle is 20 and sides of this rectangle are c(horizontal) and d (vertical).

a*b=12 => a=12/b
c*d=20 => c=20/d
b+d+3=12 => b=9-d => a=12/(9-d)
x-c=4 => x=4+c
x-a=5 => x=5+a
4+c=5+a => a=c-1

12/(9-d)=(20/d)-1
12/(9-d)=(20-d)/d
12d=180-9d-20d+d^2
d^2-41d+180=0

We can factor dis equation: (d-36)*(d-5)=0 sp d=36 or d=5. d cannot be equal to 36 because b+d+3=12, so d=5.

c=20/d=20/5=4.
x=4+c=8

The area of the green rectangle is x*3=8*3=24.

[stephenremo]

Easy to do in your head...
The length of 12 is the key..
Only a few combos with 3 to make 12.
And since you know that the boxes are 1 different it can only be 3 and 4 for height

[tosuchino]

Since all measures are whole numbers, think of the area of the left rectangle as 2*2*3, and the area of the right rectangle as 2*2*5. The heigh of the left one is 1 less than that of the right one, which leaves 2 possibilities: (the height of the left one, the height of the right one) is either (3, 4) or (4, 5). In the former case, (the width of the left one, the width of the right one) = (4, 5). In the latter case, the widths are (3, 4). The widths of the 3 rectangles add up to 12 in the former case. Thus, the height of the green rectangle is 8 (= 3 + 5 = 4 + 4). Hence, the area of the greem one is 8*3 = 24.

[kaliprasadguru]

Sir, Pranam
Is it necessary that 'a' and or 'b' will have to be whole number

[mr.]

Interesting...I chose my “a” as the right side base, with the left side to be (9-a)....This results in 9y^2 -23y-12 = 0 which factors into (9y+4) (y-3)...Great question!

[rayhanantor]

Which whiteboard software do you use to make videos?

[David__U]

At 5:00 it is unnecessary to plug in integer values of "a" to check if b is an integer. Furthermore, your method only shows that there is _at_ _least_ one solution, not that it is the _only_ solution. Furthermore there's no reason to restrict "a" and"b" to be integers; the problem only has had an integer solution even if it is not required..

To solve this problem properly, you could have solved equation (1) for a or b and then plugged that into "9b-a=23". Equation (1) says ab=12, so a=12/b, so 9b - 12/b = 23. That is an equation with one variable, which can be solved with the quadratic formula (or completing the square, etc.).

Solving this quadratic will give two solutions but one is negative, which doesn't make sense for a length. Therefore the only solution (integer or otherwise) is the positive solution. The other length can then be found using equation (1) again.

[randaya]

Another solution simple, although I did the long method of quadratic. Nonetheless, the answer is still the same and I checked it already. I also sent you something as well. Look at the mail and see what's in it.