Chapter 11.1, Problem 40PPE

To determine

Find the ratio of the shaded triangle to the area of the rectangle.

Answer to Problem 40PPE

Length of the Rectangle = $\text{x}\text{+}\text{8}$

Ratio of shaded Triangle and Rectangle: $\text{x}:2(\text{x}+8)$

Explanation of Solution

Given:

Find the ratio of the shaded triangle to the area of the rectangle.

Concept Used:

The area of the shaded region: $\text{Area of Triangle}=\frac{1}{2}·\text{base}·\text{height}$

  $\text{Area of Rectangle}=\text{length}·\text{height}$

Calculation:

Length of the Rectangle = $\text{x}\text{+}\text{8}$ and Height of the rectangle = x

Base of the Triangle = x and height of the Triangle = x

  $\begin{array}{l}\text{Area of Triangle}=\frac{1}{2}·\text{base}·\text{height}\\ \text{Area of Triangle}=\frac{1}{2}·\text{x}·\text{x}=\frac{1}{2}{\text{x}}^{\text{2}}\end{array}$

  $\begin{array}{l}\text{Area of Rectangle}=\text{length}·\text{height}\\ \text{Area of Rectangle}=\text{(x+8)}·\text{x}\text{=}{\text{x}}^{\text{2}}+8x\end{array}$

Ratio of shaded Triangle and Rectangle: $\frac{\frac{1}{2}{\text{x}}^{\text{2}}}{\text{x}·\text{(x+8)}}=\frac{1}{2}·\frac{\text{x}·\text{x}}{\text{x}·\text{(x+8)}}=\frac{\text{x}}{2(\text{x}+8)}$

Thus, Length of the Rectangle = $\text{x}\text{+}\text{8}$ and ratio of shaded Triangle and Rectangle: $\text{x}:2(\text{x}+8)$