Chapter 11.1, Problem 16WE

To determine

To Find: The Perimeter and base of rectangle.

Answer to Problem 16WE

  $b=y+7$ , $P=4y+14$

Explanation of Solution

Given: $h=y,A=y^2+7y$

Given: $h=y,A=y^2+7y$

Area of rectangle

  $\begin{array}{l}A=bh\\ y^2+7y=\left(b\right)\left(y\right)\\ b=\frac{y^2+7y}{y}\\ b=y+7\end{array}$

The base of rectangle is $b=y+7$

Now,

Perimeter of rectangle

  $\begin{array}{l}P=2(b+h)\\ P=2(y+7+y)\\ P=2(2y+7)\\ P=4y+14\end{array}$

Hence, $b=y+7$ ,

  $P=4y+14$