Chapter 6.4, Problem 6STP

Rectangle EFGH has a perimeter of 24 inches, and triangle BCD has a perimeter of 18 inches.

a. What is a system of equations for the perimeters of the figures?

b. Without solving, what method you would use to solve the system? Explain.

To determine

To calculate: Rectangle $EFGH$ has a perimeter of 24 inches, and triangle $BCD$ has a perimeter of 18 inches. What is the system of equations for the perimeters of the figures?

  

Answer to Problem 6STP

The system of equations is $\left(x-3\right)+\left(x-3\right)+2y=18$ and $2x+2y=24$

Explanation of Solution

Given:

Perimeter of triangle $BCD$ is 18 inches and perimeter of rectangle $EFGH$ has a perimeter of 24 inches.

Calculation:

We know that perimeter of any figure is sum of length of all sides of that figure

Perimeter of rectangle $EFGH$ is given as 24 inches, so

  $\begin{array}{l}2x+2y=24\\ 2\left(x+y\right)=24\\ or\left(x+y\right)=12\end{array}$

Perimeter of triangle $BCD$ is given as 18 inches, so

  $\begin{array}{l}\left(x-3\right)+\left(x-3\right)+2y=18\\ 2x-6+2y=18\\ 2x+2y=24\\ 2\left(x+y\right)=24\\ or\left(x+y\right)=12\end{array}$

Conclusion:

The system of equations is $\left(x-3\right)+\left(x-3\right)+2y=18$ and $2x+2y=24$

To determine

To explain: Rectangle $EFGH$ has a perimeter of 24 inches, and triangle $BCD$ has a perimeter of 18 inches. Without solving, what method you would use to solve the system? Explain.

  

Explanation of Solution

Given:

From above part we know the system of equations is $\left(x-3\right)+\left(x-3\right)+2y=18$ and $2x+2y=24$

To solve the above system of equations I would use elimination method because after simplifying the equation we get the $y-$ coefficients are same.