Question

a rectangle has a length of x - 2 units and a width of y + 5 units. a triangle has a base of y + 5 units and a height of x - 4 units. jenna mistakenly says that the areas of the rectangle and triangle are equal. what is jenna likely error? click on the icon to view area formulas for several shapes.

Explanation:

Step1: Recall area formulas

The area formula for a rectangle is $A_{r}=length\times width$. Given length $x^{2}$ and width $y$, so $A_{r}=x^{2}y$. The area formula for a triangle is $A_{t}=\frac{1}{2}\times base\times height$. Given base $y$ and height $x^{4}$, so $A_{t}=\frac{1}{2}yx^{4}$.

Step2: Set up the equation

Jenna says $A_{r} = A_{t}$, so $x^{2}y=\frac{1}{2}yx^{4}$.

Step3: Solve for when the areas are equal

Assume $y
eq0$ (if $y = 0$, both shapes have area 0). Divide both sides of the equation $x^{2}y=\frac{1}{2}yx^{4}$ by $y$ (since $y
eq0$), we get $x^{2}=\frac{1}{2}x^{4}$. Rearrange to $\frac{1}{2}x^{4}-x^{2}=0$. Let $u = x^{2}$, then the equation becomes $\frac{1}{2}u^{2}-u = 0$. Factor out $u$: $u(\frac{1}{2}u - 1)=0$. So $u = 0$ or $\frac{1}{2}u-1=0$. If $u = 0$, then $x^{2}=0$, so $x = 0$. If $\frac{1}{2}u - 1=0$, then $u = 2$, and since $u=x^{2}$, $x=\pm\sqrt{2}$. But in the context of length, $x>0$, so $x=\sqrt{2}$.

Answer:

$x = \sqrt{2}$