Exercise 24 Page 784

We are told that the height of a parallelogram is $4\text{ mm}$ more than its base. Therefore, if we let $b$ be the base, the height can be expressed as $h=b+4.$ We are also told that the area is $221$ square millimeters. Let's draw a diagram to illustrate the situation.

The area of a parallelogram is the product of its base and its height. We can substitute $A=221$ and $h=b+4$ into this formula and solve for $b,$ the base of the parallelogram. Let's do it! Note it's a quadratic equation. To solve this equation we will use the Quadratic Formula. The variable in our equation is $b.$ Let's replace it with $x$ not to mistake it with the $b$ in the Quadratic Formula. Next, we need to identify the values of $a,$ $b,$ and $c.$ We see that $a=1,$ $b=4,$ and $c=\text{-}221.$ Let's substitute these values into the Quadratic Formula. Now, we can go back to the previous notation and replace $x$ with $b.$ The solutions of our equation are $b=\text{-}2\pm 15.$ Let's separate them into the positive and negative cases.

$b=\text{-}2\pm 15$
$b_1=\text{-}2+15$	$b_2=\text{-}2-15$
$b_1=13$	$b_2=\text{-}17$

Since $b$ is the length base of a parallelogram it must be positive. Therefore, the base of the parallelogram is $13\text{ mm}.$ The height is $4\text{ mm}$ more than its base, so we can add $4$ to $13$ to obtain the height.