Question

triangle abc is an equilateral triangle. segment ad measures 18 inches.
which statements about the diagram are correct?
check all that apply.
bd = 9 in.
ab = 36 in.
dc = 6√3 in.
ac = 12√3 in.
bc = 18√3 in.

Explanation:

Step1: Recall properties of equilateral triangle

In an equilateral triangle, the altitude (AD) also acts as a median and angle bisector. So, \(BD = DC\), and triangle \(ABD\) is a 30 - 60 - 90 triangle, where \(\angle ADB = 90^{\circ}\), \(\angle BAD=30^{\circ}\), \(\angle ABD = 60^{\circ}\). In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest one (\(BD\)), the side opposite \(60^{\circ}\) is \(AD\) (altitude), and the hypotenuse is \(AB\) (side of equilateral triangle).

Step2: Analyze each option

• For \(BD\): Let's use the 30 - 60 - 90 triangle ratios. In \(\triangle ABD\), \(\tan(60^{\circ})=\frac{AD}{BD}\), \(\tan(60^{\circ})=\sqrt{3}=\frac{18}{BD}\), so \(BD = \frac{18}{\sqrt{3}}=6\sqrt{3}\)? Wait, no, wait. Wait, in 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is \(BD\), side opposite \(60^{\circ}\) is \(AD = 18\). The ratio of sides: opposite \(30^{\circ}\): opposite \(60^{\circ}\): hypotenuse \(=x:x\sqrt{3}:2x\). So if \(AD\) (opposite \(60^{\circ}\)) is \(x\sqrt{3}=18\), then \(x=\frac{18}{\sqrt{3}} = 6\sqrt{3}\)? Wait, no, I made a mistake. Wait, in equilateral triangle, altitude \(h=\frac{\sqrt{3}}{2}a\), where \(a\) is the side length. So \(h = AD=18=\frac{\sqrt{3}}{2}a\), so \(a=\frac{36}{\sqrt{3}} = 12\sqrt{3}\). Then, since \(D\) is the mid - point of \(BC\), \(BD=\frac{a}{2}=\frac{12\sqrt{3}}{2}=6\sqrt{3}\)? Wait, no, wait, let's re - derive.

Wait, formula for altitude of equilateral triangle: \(h=\frac{\sqrt{3}}{2}s\), where \(s\) is the side length (\(AB = BC=AC = s\)). Given \(h = AD = 18\). So \(18=\frac{\sqrt{3}}{2}s\), solving for \(s\): \(s=\frac{36}{\sqrt{3}}=\frac{36\sqrt{3}}{3}=12\sqrt{3}\). So side length \(AB = AC=BC = 12\sqrt{3}\) inches. And since \(D\) is the mid - point of \(BC\), \(BD = DC=\frac{s}{2}=\frac{12\sqrt{3}}{2}=6\sqrt{3}\) inches.

Now let's check each option:

• Option 1: \(BD = 9\) in. From above, \(BD = 6\sqrt{3}\approx 10.392

eq9\), so this is incorrect.

• Option 2: \(AB = 36\) in. We found \(AB = 12\sqrt{3}\approx20.78

eq36\), so this is incorrect.

• Option 3: \(DC = 6\sqrt{3}\) in. Since \(D\) is the mid - point of \(BC\) and \(BC = 12\sqrt{3}\), \(DC=\frac{BC}{2}=6\sqrt{3}\), so this is correct.
• Option 4: \(AC = 12\sqrt{3}\) in. From the altitude formula, we found \(s = AC=12\sqrt{3}\), so this is correct.
• Option 5: \(BC = 18\sqrt{3}\) in. We found \(BC = 12\sqrt{3}

eq18\sqrt{3}\), so this is incorrect.

Answer:

The correct statements are:

• \(DC = 6\sqrt{3}\) in.
• \(AC = 12\sqrt{3}\) in.