How Do You Find the Length x in a Geometric Problem with Shaded Areas?

[paulmdrdo1]

Find the length x if the shaded area is 1200 cm^2

I tried to solve this is what I get

since $A_{triangle}=(\frac{1}{2})({x-1})(x)$

and $A_{rectangle}=x$

$A_{rectangle}+A_{triangle}=2400$

Is the set-up of my equation correct?

 

[MarkFL]

Why are you using $x-1$ in the area of the triangle?

 

[paulmdrdo1]

Why are you using $x-1$ in the area of the triangle?

I see it. $A_{TRI}=\frac{1}{2}(x^2)$

now I will have

$\frac{1}{2}(x^2)+x=2400$

solving for x $(x-48)(x+50)=0$

$x=48 in.$ :D

 

[MarkFL]

Well, you actually have:

\(\displaystyle \frac{1}{2}x^2+x=1200\)

or:

\(\displaystyle x^2+2x-2400=0\)

\(\displaystyle (x+50)(x-48)=0\)

Discarding the negative root, we then find:

\(\displaystyle x=48\text{ cm}\)

 

[CellCoree]

Yes, your set-up is correct. To solve for x, you can simplify the equation to:
$x + \frac{1}{2}x^2 - \frac{1}{2}x = 2400$
Then, you can use algebraic methods to solve for x, which in this case is approximately 55.56 cm.