MHB 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:

$$\frac{1}{2}x^2+x=1200$$

or:

$$x^2+2x-2400=0$$

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

Discarding the negative root, we then find:

$$x=48\text{ cm}$$