Question

which three pairs of side lengths are possible measurements for the triangle? (image of right triangle ghi with right angle at h, angle at g is 30°, angle at i is 60°) options: gh = 2, hi = 2√3; gh = 8, gi = 16; gh = 8√2, gi = 16; gh = 2√3, hi = 2; gh = 8√3, gi = 16; hi = 7, gi = 14

Explanation:

Step1: Recall 30-60-90 triangle ratios

In a 30-60-90 right triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) (short leg) is \(x\), opposite \(60^\circ\) (long leg) is \(x\sqrt{3}\), and hypotenuse is \(2x\). Here, \(\angle I = 60^\circ\), \(\angle G = 30^\circ\), right angle at \(H\). So:
- \(HI\) (opposite \(30^\circ\)): short leg \(= x\)
- \(GH\) (opposite \(60^\circ\)): long leg \(= x\sqrt{3}\)
- \(GI\) (hypotenuse) \(= 2x\)

Step2: Analyze each option

1. **Option 1: \(GH = 2\), \(HI = 2\sqrt{3}\)**
If \(GH = x\sqrt{3}=2\), then \(x=\frac{2}{\sqrt{3}}\). Then \(HI = x=\frac{2}{\sqrt{3}} eq 2\sqrt{3}\). Incorrect.

2. **Option 2: \(GH = 8\), \(GI = 16\)**
\(GI = 2x = 16\Rightarrow x = 8\). Then \(GH = x\sqrt{3}=8\sqrt{3} eq 8\). Wait, no—wait, \(GH\) is adjacent to \(30^\circ\)? Wait, recheck angles: \(\angle G = 30^\circ\), so side opposite \(\angle G\) is \(HI\) (short leg), side adjacent to \(\angle G\) is \(GH\) (long leg), hypotenuse \(GI\). So:
\(\cos(30^\circ)=\frac{GH}{GI}\), \(\sin(30^\circ)=\frac{HI}{GI}\).
\(\cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{GH}{GI}\), \(\sin(30^\circ)=\frac{1}{2}=\frac{HI}{GI}\).
For \(GI = 16\), \(HI=\frac{1}{2}\times16 = 8\), \(GH=\frac{\sqrt{3}}{2}\times16 = 8\sqrt{3}\). Wait, no—wait, the triangle: right angle at \(H\), so \(GH\) and \(HI\) are legs, \(GI\) hypotenuse. \(\angle G = 30^\circ\), so \(\angle I = 60^\circ\). So:
- \(HI\) (opposite \(30^\circ\)): \(HI = \frac{GI}{2}\) (since \(\sin 30^\circ=\frac{HI}{GI}=\frac{1}{2}\))
- \(GH\) (opposite \(60^\circ\)): \(GH = \frac{GI\sqrt{3}}{2}\) (since \(\sin 60^\circ=\frac{GH}{GI}=\frac{\sqrt{3}}{2}\))

So for \(GI = 16\), \(HI = 8\), \(GH = 8\sqrt{3}\). Wait, but option 2 says \(GH = 8\), \(GI = 16\). No, that's wrong. Wait, maybe I mixed up legs. Wait, right angle at \(H\), so:
- \(H\) is right angle, so \(GH\) and \(HI\) are legs, \(GI\) hypotenuse.
- \(\angle G = 30^\circ\), so angle at \(G\) is \(30^\circ\), so the side opposite \(\angle G\) is \(HI\) (leg), side adjacent is \(GH\) (leg). So:
\(\tan(30^\circ)=\frac{HI}{GH}=\frac{1}{\sqrt{3}}\), \(\cos(30^\circ)=\frac{GH}{GI}=\frac{\sqrt{3}}{2}\), \(\sin(30^\circ)=\frac{HI}{GI}=\frac{1}{2}\).

So \(GI = 2HI\) (from \(\sin 30^\circ\)), and \(GH = HI\sqrt{3}\) (from \(\tan 30^\circ\)).

Let's redo:
- \(GI = 2HI\) (hypotenuse = 2×short leg)
- \(GH = HI\sqrt{3}\) (long leg = short leg×\(\sqrt{3}\))

Now check each option with this ratio (\(GH = HI\sqrt{3}\), \(GI = 2HI\)):

1. **Option 2: \(GH = 8\), \(GI = 16\)**
\(GI = 2HI\Rightarrow HI = 8\). Then \(GH = HI\sqrt{3}=8\sqrt{3} eq 8\). Wait, no—wait, maybe I had \(GH\) and \(HI\) reversed. If \(\angle I = 60^\circ\), then \(\tan(60^\circ)=\frac{GH}{HI}=\sqrt{3}\), so \(GH = HI\sqrt{3}\), and \(GI = 2HI\) (since \(\angle G = 30^\circ\), hypotenuse = 2×short leg (HI)).

So for \(GI = 16\), \(HI = 8\), \(GH = 8\sqrt{3}\). So option 2: \(GH = 8\), \(GI = 16\) is wrong. Wait, option 5: \(GH = 8\sqrt{3}\), \(GI = 16\). Let's check: \(GI = 16\), so \(HI = 8\) (since \(GI = 2HI\)), then \(GH = HI\sqrt{3}=8\sqrt{3}\). Yes! So option 5: \(GH = 8\sqrt{3}\), \(GI = 16\) is correct (since \(HI = 8\), \(GH = 8\sqrt{3}\), \(GI = 16\), which fits \(GH = HI\sqrt{3}\), \(GI = 2HI\)).

2. **Option 3: \(GH = 8\sqrt{2}\), \(GI = 16\)**
\(GI = 2HI\Rightarrow HI = 8\). Then \(GH = HI\sqrt{3}=8\sqrt{3} eq 8\sqrt{2}\). Incorrect.

3. **Option 4: \(GH = 2\sqrt{3}\), \(HI = 2\)**
Check ratio: \(GH = HI\sqrt{3}\Rightarrow 2\sqrt{3}=2\times\sqrt{3}\). Yes! \(HI = 2\) (short leg), \(GH = 2\sqrt{3}\) (long leg), then \(GI = 2HI = 4\) (hypotenuse). This fits \(GH = HI\sqrt{3}\), \(GI = 2HI\). Correct.

4. **Option 5: \(GH = 8\sqrt{3}\), \(GI = 16\)**
\(GI = 16\Rightarrow HI = 8\) (since \(GI = 2HI\)), then \(GH = HI\sqrt{3}=8\sqrt{3}\). Correct.

5. **Option 6: \(HI = 7\), \(GI = 14\)**
\(GI = 2HI\Rightarrow 14 = 2\times7\). Yes! Then \(GH = HI\sqrt{3}=7\sqrt{3}\). So \(HI = 7\), \(GI = 14\) (hypotenuse = 2×short leg), so this fits. Correct.

Wait, let's re-express the correct ratio:
- Short leg (opposite \(30^\circ\)): \(HI = x\)
- Long leg (opposite \(60^\circ\)): \(GH = x\sqrt{3}\)
- Hypotenuse: \(GI = 2x\)

So the three conditions are:
- \(GI = 2HI\) (hypotenuse = 2×short leg)
- \(GH = HI\sqrt{3}\) (long leg = short leg×\(\sqrt{3}\))

Now check each option:

1. \(GH = 2\), \(HI = 2\sqrt{3}\): \(GH = x\sqrt{3}=2\Rightarrow x=\frac{2}{\sqrt{3}}\); \(HI = x=\frac{2}{\sqrt{3}} eq 2\sqrt{3}\). Incorrect.

2. \(GH = 8\), \(GI = 16\): \(GI = 2x = 16\Rightarrow x = 8\); \(GH = x\sqrt{3}=8\sqrt{3} eq 8\). Incorrect.

3. \(GH = 8\sqrt{2}\), \(GI = 16\): \(GI = 2x = 16\Rightarrow x = 8\); \(GH = x\sqrt{3}=8\sqrt{3} eq 8\sqrt{2}\). Incorrect.

4. \(GH = 2\sqrt{3}\), \(HI = 2\): \(HI = x = 2\); \(GH = x\sqrt{3}=2\sqrt{3}\). Correct (since \(GI = 2x = 4\), which is consistent).

5. \(GH = 8\sqrt{3}\), \(GI = 16\): \(GI = 2x = 16\Rightarrow x = 8\); \(GH = x\sqrt{3}=8\sqrt{3}\). Correct.

6. \(HI = 7\), \(GI = 14\): \(GI = 2x = 14\Rightarrow x = 7\); \(HI = x = 7\), \(GH = x\sqrt{3}=7\sqrt{3}\) (not given, but \(GI = 2HI\) holds, and \(GH\) would be \(7\sqrt{3}\), but the option only checks \(HI\) and \(GI\), and \(GI = 2HI\) (14 = 2×7) is true. So this is correct.

Wait, let's recheck the angle labels. The triangle has right angle at \(H\), so:
- Vertex \(H\): right angle
- Vertex \(G\): \(30^\circ\)
- Vertex \(I\): \(60^\circ\)

So:
- Side opposite \(G\) (30°): \(HI\) (short leg, length \(x\))
- Side opposite \(I\) (60°): \(GH\) (long leg, length \(x\sqrt{3}\))
- Hypotenuse: \(GI\) (length \(2x\))

So the three valid pairs must satisfy:
1. \(GI = 2 \times HI\) (hypotenuse = 2×short leg)
2. \(GH = HI \times \sqrt{3}\) (long leg = short leg×\(\sqrt{3}\))

Now check each option:

- **Option 4: \(GH = 2\sqrt{3}\), \(HI = 2\)**
\(HI = 2\) (x = 2), so \(GH = 2 \times \sqrt{3}\) (x√3 = 2√3) ✔️, \(GI = 2 \times 2 = 4\) (consistent).

- **Option 5: \(GH = 8\sqrt{3}\), \(GI = 16\)**
\(GI = 16 = 2x \Rightarrow x = 8\) (HI = 8), so \(GH = 8 \times \sqrt{3} = 8\sqrt{3}\) ✔️.

- **Option 6: \(HI = 7\), \(GI = 14\)**
\(GI = 14 = 2x \Rightarrow x = 7\) (HI = 7), so \(GH = 7 \times \sqrt{3} = 7\sqrt{3}\) (not given, but the option only checks HI and GI, and \(GI = 2HI\) (14 = 2×7) is true, so this is valid (since HI is short leg, GI hypotenuse, ratio holds).

Wait, earlier I thought option 2 was wrong, and option 5 is correct. Let's confirm:

- Option 2: \(GH = 8\), \(GI = 16\)
\(GI = 16 = 2x \Rightarrow x = 8\), so \(GH should be 8\sqrt{3}\), but option 2 says \(GH = 8\). Incorrect.

- Option 5: \(GH = 8\sqrt{3}\), \(GI = 16\)
\(x = 8\) (HI = 8), \(GH = 8\sqrt{3}\), \(GI = 16\). Correct.

- Option 4: \(GH = 2\sqrt{3}\), \(HI = 2\)
\(x = 2\) (HI = 2), \(GH = 2\sqrt{3}\), \(GI = 4\). Correct.

- Option 6: \(HI = 7\), \(GI = 14\)
\(x = 7\) (HI = 7), \(GI = 14\) (2x = 14), so \(GH = 7\sqrt{3}\). The option only checks HI and GI, and \(GI = 2HI\) holds, so this is valid.

So the three correct options are:
- \(GH = 2\sqrt{3}, HI = 2\) (Option 4)
- \(GH = 8\sqrt{3}, GI = 16\) (Option 5)
- \(HI = 7, GI = 14\) (Option 6)

Wait, let's verify with \(GI = 2HI\) for each:

- Option 4: \(HI = 2\), so \(GI should be 4\) (not given, but the option gives \(GH\) and \(HI\), and \(GH = HI\sqrt{3}\) holds (2√3 = 2×√3). So valid.

- Option 5: \(GI = 16\), so \(HI = 8\) (since \(GI = 2HI\)), and \(GH = 8\sqrt{3}\) (HI×√3 = 8×√3). Valid.

- Option 6: \(HI = 7\), \(GI = 14\) (GI = 2×HI). Valid.

Yes, these three satisfy the 30-60-90 triangle ratios.

Answer:

• \(GH = 2\sqrt{3}, HI = 2\)
• \(GH = 8\sqrt{3}, GI = 16\)
• \(HI = 7, GI = 14\)

(In boxed form, listing the correct options as per the problem's checkboxes—assuming the options are labeled as:

1. \(GH = 2, HI = 2\sqrt{3}\)
2. \(GH = 8, GI = 16\)
3. \(GH = 8\sqrt{2}, GI = 16\)
4. \(GH = 2\sqrt{3}, HI = 2\)
5. \(GH = 8\sqrt{3}, GI = 16\)
6. \(HI = 7, GI = 14\)

So the correct options are 4, 5, 6:

1. \(GH = 2\sqrt{3}, HI = 2\)
2. \(GH = 8\sqrt{3}, GI = 16\)
3. \(HI = 7, GI = 14\)