Find Area of the Green Rectangle | Two Methods

[philipkudrna]

Before watching: 2 min and some set of equations (a+x=10; b+x=14; a*h=37; b*h=73) leads to the height being 9 and the missing sidelength (let‘s call it x) as 53/9. Thus the area is 53/9*9 or „simply“ (as our professor would say) 53 square units.
After watching: the approach using the combined areas is much faster and less cumbersome than my method. I like it! My approach was more like the second method.

[hienvo]

Prior to seeing your solution, I called the green area X. Therefore I have the equation for the width of the biggest rectangle: (37+X)/10 = (73+X)/14. ==> X=53!!! 😁😁😁

[salimahmad]

Let the area be A
Let y be the width.
37 + A = 10y
And, 73 + A = 14y
Divide equation (1) by equation (2).
Solving we get A = 53 square units.

[jaybling]

The thing about the second method is simply understanding that the area for the green rectangle is xy. So when you find the y value, all you have to do is plug it into one of the equations. You don't need to plug it in for xy.

[sameerqureshi-khcc]

Sir you are a legendary teacher! Love from Pakistan 😊👍🌹

[kltttub]

14-10=4.
73-37=36.
36:4=9.
(10*9)-37=53.

[ankitecian]

Ans: 53
It’s damn simple….just add both length 14+10 multiplied by height would give you 73 rectangle + 37 rectangle + 2 times unknown area
So 24*h = 110 + 2*a
So 12*h = 55 + a
Also 10 * h = 37 + a
We get h = 9
Putting value of h in second equation
10*9 = 37+a
So a= 53

[Gargaroolala]

Completed this qns using the first method. :)

[math]

very well explained bro, thanks for sharing this area problem

[matthewkendall]

You can do this pretty simply in your head - if you extend the red triangle by 4 units you get the blue one (so 73 - 37 = 36 = 4y giving y = 9 and (10 - x) x 9 = 37, meaning 9x = size of the green square = 90 - 37 = 53

[predator]

At the moment you can label the dimensions, the answer would be there. Thank you teacher👍🙏.

[alster]

53 sq units.


A(left) = 37+A
37+A = 10y (eq 1)

A(right) = A+73
A+73 = 14y (eq 2)

By distribution we get this system of linear equatkons


A-14y= -73
A-10y = -36


By Elimination method we get


-A+14y= 73
A-10y= -37

4y = 36


y =9

Substitute to either equations, so I will choose eq 1


37+A = 10(9)
37+A = 90
A= 53 sq units (Area of Green Rectangle)

[xyz]

For method 2, probably you don’t need to solve x, as xy is the area and what you need at the end. Just plug in y on the right side.

[usmasuda]

You don't have to obtain x and y independently. Just calculate xy.

[GaryAppledale]

Sir - how can any rectangle's units be a prime number (unless one of the sides is 1?

[hls]

I am disappointed that PreMath did not comment on the many people who correctly noted that it was not necessary to use the (horizontal) x unknown, but just the (vertical) y unknown, and the green area A. Then, simply 10y=37+A, and 14y=73+A, i.e. 2 equations with 2 unknown variables, easy-peasy to solve, and get the correct answer of A=53.

[mustafizrahman]

Good question. The answer is 53 units squared.

[nicholasimholte]

Let the Red, Green, and Blue rectangles have lengths a, b, c with common width x.

Known:

b + c = 14
a + b = 10
xa = 37
xc = 73

The area of the Green rectangle is bx. We can calculate this two ways:
bx = x(10-a) = x(10 - 37/x) = 10x - 37
bx = x(14-c) = x(14 - 73/x) = 14x - 73

But these must be equal, so 4x = 36, or x = 9.

Finally, Area = bx = 90 - 37 = 53

[mrreg]

I went through 14 + 37/y = 10 + 73/y; obtained y = 9, but when dividing the first section by 9, found that the individual sides were fractional. My result for the middle area was 53.1.

[yergeauf]

Let y be the height of all these rectangles, and A be the area of the green rectangle.
The area of the red and green rectangles is 10y, and it is equal to 37+A.
The area of the blue and green rectangles is 14y, and it is equal to 73+A.