Write a polynomial that represents the area of the shaded region.
Understand the Problem
The question asks to find the area of the shaded region in a given figure. The figure is a rectangle of sides (x+5) and (x+6) with a triangle inside. The area can be found by subtracting the area of the triangle from the area of the rectangle. We can consider the base of the triangle to be (x+6) and the height to be (x+5).
Answer
$\frac{1}{2}x^2 + \frac{11}{2}x + 15$
Answer for screen readers
$\frac{1}{2}x^2 + \frac{11}{2}x + 15$
Steps to Solve
- Calculate the area of the rectangle
The area of a rectangle is given by the formula $Area = length \times width$. In this case, the length is $(x+6)$ and the width is $(x+5)$. So, the area of the rectangle is $(x+6)(x+5)$.
- Expand the expression for the rectangle's area
Using the distributive property (also known as FOIL), we expand the expression: $(x+6)(x+5) = x(x) + x(5) + 6(x) + 6(5) = x^2 + 5x + 6x + 30 = x^2 + 11x + 30$.
- Calculate the area of the triangle
The area of a triangle is given by the formula $Area = \frac{1}{2} \times base \times height$. In this case, the base of the triangle is $(x+6)$ and the height is $(x+5)$. Therefore, the area of the triangle is $\frac{1}{2}(x+6)(x+5)$.
- Simplify the expression for the triangle's area
From step 2, we know that $(x+6)(x+5) = x^2 + 11x + 30$. Therefore, the area of the triangle is $\frac{1}{2}(x^2 + 11x + 30) = \frac{1}{2}x^2 + \frac{11}{2}x + 15$.
- Calculate the area of the shaded region
The area of the shaded region is the area of the rectangle minus the area of the triangle. So, $(x^2 + 11x + 30) - (\frac{1}{2}x^2 + \frac{11}{2}x + 15)$.
- Simplify the expression
Combine like terms: $x^2 - \frac{1}{2}x^2 + 11x - \frac{11}{2}x + 30 - 15 = \frac{1}{2}x^2 + \frac{11}{2}x + 15$.
$\frac{1}{2}x^2 + \frac{11}{2}x + 15$
More Information
The area of the shaded region is represented by the quadratic polynomial $\frac{1}{2}x^2 + \frac{11}{2}x + 15$.
Tips
A common mistake is not distributing the $\frac{1}{2}$ correctly when finding the area of the triangle. Another common mistake is incorrectly combining like terms when subtracting the triangle's area from the rectangle's area.
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