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

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Discussion Overview

The discussion revolves around finding the length x in a geometric problem involving shaded areas, specifically focusing on the areas of a triangle and a rectangle. Participants explore the setup of equations based on given area conditions.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant proposes an equation for the area of a triangle as $A_{triangle}=(\frac{1}{2})({x-1})(x)$ and for a rectangle as $A_{rectangle}=x$, leading to the equation $A_{rectangle}+A_{triangle}=2400$.
  • Another participant questions the use of $x-1$ in the area of the triangle, prompting a reevaluation of the area formula.
  • A later reply suggests that the area of the triangle should be $A_{TRI}=\frac{1}{2}(x^2)$, leading to a revised equation $\frac{1}{2}(x^2)+x=2400$.
  • Further clarification is provided that the correct equation should be $\frac{1}{2}x^2+x=1200$, which simplifies to $x^2+2x-2400=0$.
  • Participants derive the roots of the equation, identifying $x=48$ cm as a solution while discarding the negative root.

Areas of Agreement / Disagreement

There is no consensus on the initial setup of the equations, as participants express differing views on the correct formulation of the area of the triangle. However, there is agreement on the final solution of $x=48$ cm after resolving the equation.

Contextual Notes

Participants have not fully clarified the assumptions behind the geometric setup, particularly regarding the dimensions involved in the triangle and rectangle. The discussion also reflects some uncertainty about the initial area expressions used.

paulmdrdo1
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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?
 

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

$$x=48\text{ cm}$$
 

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