Geometric Mean Worksheet #s 1-11 - Free Printable

Let’s solve each problem one by one using the Geometric Mean Theorem (also called the Right Triangle Altitude Theorem). This theorem says:

> In a right triangle, if you draw an altitude from the right angle to the hypotenuse, it creates two smaller triangles that are similar to each other and to the original triangle.
> Also:
> - The altitude is the geometric mean of the two segments of the hypotenuse.
> - Each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

We’ll use these relationships:

1. Altitude rule:
If the altitude splits the hypotenuse into parts `a` and `b`, then:
→ `altitude² = a × b`

2. Leg rule:
For a leg next to segment `a`:
→ `leg² = hypotenuse × a`

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Problem 7:

Triangle with altitude `x`, splitting hypotenuse into 9 and 3.

So total hypotenuse = 9 + 3 = 12

Use altitude rule:
→ `x² = 9 × 3 = 27`
→ `x = √27 = √(9×3) = 3√3`

But wait — let me double-check. Is this correct? Actually, in some textbooks, they might expect decimal or simplified radical. But since the first example used integers, maybe I made a mistake?

Wait — no, 9×3=27 is correct. So x = √27 = 3√3 ≈ 5.196... but perhaps we leave as exact value.

Actually, looking back at the first example on the sheet: they had 12 and 8, so x² = 12×8 = 96 → x = √96. They didn’t simplify, just wrote √96.

So for consistency, maybe we should write √27? But usually we simplify radicals.

Wait — let me check again: Is the altitude really between 9 and 3? Yes. So yes, x² = 9×3 = 27 → x = √27 = 3√3.

But let’s see what the problem expects. Since it's labeled “x=”, probably wants simplified radical or decimal? But in math class, simplified radical is preferred.

Actually, hold on — maybe I misread. Let me look again.

In problem 7: base is split into 9 and 3, altitude is x. So yes, x² = 9 * 3 = 27 → x = √27 = 3√3.

But let me verify with another method. Suppose we consider the whole triangle. Legs would be... actually, we don’t need to. The geometric mean theorem directly applies here.

✔ Final answer for #7: x = 3√3

Wait — but in the first example, they left it as √96. Maybe they want unsimplified? But √96 can be simplified too. Hmm.

Actually, looking at their work: they wrote “√x² = √96” which is weird — they meant x = √96. And they didn’t simplify. So perhaps for consistency, we should not simplify? But that’s bad practice.

I think better to simplify unless told otherwise. Let’s go with simplified form.

But let me check online or standard practice — actually, in most geometry classes, you simplify radicals. So 3√3 is correct.

Alternatively, maybe I made a mistake in applying the theorem?

No — altitude to hypotenuse: x² = product of the two segments → 9*3=27 → x=√27=3√3. Correct.

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Problem 8:

Right triangle, altitude drawn to hypotenuse, creating segments 4 and 6. We need to find legs x and y.

First, note: the altitude divides the hypotenuse into 4 and 6, so total hypotenuse = 4 + 6 = 10.

Now, use leg rule:

- Leg x is adjacent to segment 4 → so x² = hypotenuse × 4 = 10 × 4 = 40 → x = √40 = 2√10

- Leg y is adjacent to segment 6 → so y² = hypotenuse × 6 = 10 × 6 = 60 → y = √60 = 2√15

Wait — is that correct? Let me confirm.

Actually, in the diagram, x is the leg opposite the segment 6? Or adjacent?

Looking at the image description: it shows a right triangle, altitude to hypotenuse, left segment is 4, right segment is 6. Left leg is labeled x, right leg is labeled y.

Typically, the leg adjacent to a segment corresponds to that segment.

Standard rule:
If the hypotenuse is split into p and q, then:

- Leg adjacent to p: leg² = (p+q) * p
- Leg adjacent to q: leg² = (p+q) * q

Here, p=4, q=6, hypotenuse=10.

So:

Left leg (adjacent to 4): x² = 10 * 4 = 40 → x = √40 = 2√10

Right leg (adjacent to 6): y² = 10 * 6 = 60 → y = √60 = 2√15

Yes.

But let me make sure about labeling. In the image, x is on the left, next to segment 4; y is on the right, next to segment 6. So yes.

✔ So x = 2√10, y = 2√15

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Problem 9:

Right triangle, altitude = 10, splits hypotenuse into x and 25. Need to find x and y (where y is the left leg).

First, use altitude rule:

Altitude² = product of the two segments → 10² = x * 25 → 100 = 25x → x = 100 / 25 = 4

So x = 4

Now, total hypotenuse = x + 25 = 4 + 25 = 29

Now, find y — which is the leg adjacent to segment x=4.

So use leg rule: y² = hypotenuse × adjacent segment = 29 × 4 = 116 → y = √116 = √(4×29) = 2√29

Check: could also use Pythagoras later, but this is direct.

Verify: the other leg would be adjacent to 25: leg² = 29×25 = 725, and indeed, 10² + ? Wait no — the legs are y and the other one, and altitude is 10.

Actually, we can verify with area or something, but not necessary.

So:

x = 4
y = 2√29

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Now, let’s compile all answers:

Problem 7: x = 3√3
Problem 8: x = 2√10, y = 2√15
Problem 9: x = 4, y = 2√29

But wait — in problem 7, is there a possibility they expect decimal? Probably not. Radical is fine.

Also, in problem 9, y is the left leg, which is adjacent to x=4, so yes.

One more check for problem 9:

Altitude = 10, segments 4 and 25.

Check altitude rule: 10² = 4*25 → 100=100 ✓

Leg y: adjacent to 4, so y² = (4+25)*4 = 29*4=116 → y=√116=2√29 ✓

Other leg: adjacent to 25: z²=29*25=725 → z=√725=5√29

Then check Pythagoras: y² + z² = 116 + 725 = 841, and hypotenuse²=29²=841 ✓

Also, area: (1/2)*base*height = (1/2)*29*10 = 145
Also, (1/2)*y*z = (1/2)*√116 * √725 = (1/2)*√(116*725)

Calculate 116*725: 116*700=81200, 116*25=2900, total 84100 → √84100=290 → (1/2)*290=145 ✓

Perfect.

So all correct.

Final Answers:

7. x = 3√3
8. x = 2√10, y = 2√15
9. x = 4, y = 2√29

But in the worksheet, for problem 7, they might expect just the number, but since it's irrational, radical is fine.

Perhaps write as decimals? But no, exact values are better.

I think we’re good.

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Final Answer:
7. \( x = 3\sqrt{3} \)
8. \( x = 2\sqrt{10} \), \( y = 2\sqrt{15} \)
9. \( x = 4 \), \( y = 2\sqrt{29} \)