Exercise 7 Page 562

Let's begin with recalling the $30^{\circ }\text{- }60^{\circ }\text{- }90^{\circ }$ Triangle Theorem. This theorem tells us that the length of the hypotenuse of this right triangle is $2$ times the length of the shorter leg $s,$ and the length of the longer leg is $\sqrt{3}$ times the length of the shorter leg.

Now let's take a look at the given picture. We will draw an altitude of an equilateral triangle base. Let $h$ represents this altitude.

Notice that after drawing an altitude we got the $30^{\circ }\text{- }60^{\circ }\text{- }90^{\circ }$ triangle. Therefore, the shorter leg of this triangle has a length two times less than $4.$

Next, according to the recalled theorem, we know that the longer leg, $h,$ is $\sqrt{3}$ times the length of the shorter leg. This means that the height of the opening of the mailer is approximately $3.46$ inches. To determine whether an engraved plaque will fit the opening of the mailer, let's compare the height of this plaque, which is $3\frac{1}{4}$ inches, with the height of the opening of the mailer. Since the height of the engraved plaque is less than the height of the opening of the mailer, it will fit the opening the of the mailer.