Exercise 18 Page 784

The area of a parallelogram is the product of a base and its corresponding height.

We can consider the side whose length is $14$ milimeters as the base. However, we need to find the height. To do so, we will pay close attention to the right triangle formed by the height, a side, and a part of a nonparallel side.

We can see that the measure of two of the interior angles of the triangle are $60^{\circ }$ and $90^{\circ }.$ We can use the Triangle Angle Sum Theorem to find the measure of the third angle. The third angle measures $30^{\circ }$ and, therefore, we have a $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ triangle. In this type of triangle the length of the longer leg is $\sqrt{3}$ times the length of the shorter leg. With this information, and knowing that the shorter leg measures $7$ milimeters, we can find the length of the longer leg. Therefore, the height of the parallelogram is $7\sqrt{3}\text{ mm}.$ Now that we know that the base is $14\text{ mm}$ and that the height is $7\sqrt{3}\text{ mm},$ we can substitute these values in the formula for the area of a parallelogram. The area of the parallelogram to the nearest tenth is $169.7$ square millimeters.