Q47. - Algebra 2

Question

Parallelogram L has an area of $3x^2+10x+3$ square meters and a height of $3x+1$ meters. Parallelogram M has an area of $2x^2-13x+20$ square meters and a 

height of $x-4$ meters. Find the area of the rectangle N.

Step-by-Step Solution

Answer

The area of the rectangle N is $2x^2+x-15$ square meters.

Step 1. Given Information.

Given parallelogram L has an area of $3x^2+10x+3$ square meters and a height of $3x+1$ meters. Parallelogram M has an area of $2x^2-13x+20$ square meters and a height of $x-4$ meters.

The area of the rectangle N is to be determined.

Step 2. Explanation .

The area of a parallelogram is given by $A=base\times height$.

For parallelogram L:

Plugging the given values in the equation:

$\begin{array}{l}A=base\times height\\ 3x^2+10x+3=b\times \left(3x+1\right)\\ b=\frac{3x^2+10x+3}{3x+1}\\ b=\frac{3x^2+9x+1x+3}{3x+1}\\ b=\frac{3x\left(x+3\right)+1\left(x+3\right)}{3x+1}\\ b=\frac{\left(3x+1\right)\left(x+3\right)}{3x+1}\\ b=x+3\end{array}$

Hence the base of the parallelogram L is $x+3$ meters.

From the figure, base of the parallelogram L and the side $b_1$ of the rectangle N are same.

Hence, $b_1=x+3$ meters.

For parallelogram M:

Plugging the given values in the equation:

$\begin{array}{l}A=base\times height\\ 2x^2-13x+20=b\times \left(x-4\right)\\ b=\frac{2x^2-13x+20}{x-4}\\ b=\frac{2x^2-8x-5x+20}{x-4}\\ b=\frac{2x\left(x-4\right)-5\left(x-4\right)}{x-4}\\ b=\frac{\left(2x-5\right)\left(x-4\right)}{x-4}\\ b=2x-5\end{array}$

Hence the base of the parallelogram M is $2x-5$ meters.

From the figure, base of the parallelogram M and the side $b_2$ of the rectangle N are same.

Hence, $b_2=2x-5$ meters.

The area of a rectangle is given by $A=length\times width$.

Plugging the values in the equation:

$\begin{array}{l}A=length\times width\\ A=b_1\times b_2\\ A=\left(x+3\right)\left(2x-5\right)\\ A=2x^2-5x+6x-15\\ A=2x^2+x-15\end{array}$

Step 3. Conclusion .

Hence, the area of the rectangle N is $2x^2+x-15$ square meters.