Geometric Mean Worksheet.pdf - Keller ISD Schools - Free Printable

Let’s solve each problem step by step. We’ll use the definition of geometric mean and properties of right triangles with altitudes.

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Geometric Mean Definition:

The geometric mean of two numbers *a* and *b* is the square root of their product:
→ √(a × b)

Also, in a right triangle, if you draw an altitude from the right angle to the hypotenuse, it creates two smaller similar triangles. The length of the altitude is the geometric mean of the two segments it divides the hypotenuse into.

That means:
If hypotenuse is split into parts *p* and *q*, then altitude *x* = √(p × q)

Also, each leg of the original triangle is the geometric mean of the whole hypotenuse and the adjacent segment.

So for example, if hypotenuse = p + q, and one leg is next to segment p, then:
leg = √[(p + q) × p]

We’ll apply these rules as needed.

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Problem 1: Find the geometric mean of 8 and 18.

Geometric mean = √(8 × 18)
= √(144)
= 12.0

✔ Rounded to tenths place → still 12.0

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Problem 2: Find the geometric mean of 20 and 25.

√(20 × 25) = √(500)
≈ 22.3607…
Rounded to tenths → 22.4

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Problem 3: 15 is the geometric mean of 25 and what other number?

Let the unknown number be *x*.
Then:
15 = √(25 × x)
Square both sides:
225 = 25x
Divide both sides by 25:
x = 9

✔ Answer: 9

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Problem 4: Find the geometric mean of 3 and 7.

√(3 × 7) = √21 ≈ 4.5826…
Rounded to tenths → 4.6

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Problem 5: 32 is the geometric mean of 16 and what other number?

Let unknown = x
32 = √(16 × x)
Square both sides:
1024 = 16x
x = 1024 ÷ 16 = 64

✔ Answer: 64

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Now let’s do the triangle problems (6–11). These involve right triangles with altitudes drawn to the hypotenuse.

In such cases:

- Altitude squared = product of the two segments of the hypotenuse.
- Each leg squared = whole hypotenuse × adjacent segment.

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Problem 6: Triangle with hypotenuse split into 12 and 8, altitude = x

Altitude x = √(12 × 8) = √96 ≈ 9.7979…
Rounded to tenths → 9.8

✔ x = 9.8

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Problem 7: Hypotenuse split into 9 and 3, altitude = x

x = √(9 × 3) = √27 ≈ 5.196…
Rounded to tenths → 5.2

✔ x = 5.2

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Problem 8: Right triangle with legs labeled x and y, altitude splits hypotenuse into 4 and 6.

First, find the altitude? Wait — actually, looking at diagram description:

It says “x” and “y” are the legs, and there’s an altitude drawn to the hypotenuse, splitting it into 4 and 6.

But wait — in standard notation, when altitude is drawn to hypotenuse, we usually label the segments and the altitude.

Actually, re-examining: In problem 8, the diagram shows a right triangle with legs x and y, and an altitude drawn to the hypotenuse, which is split into segments 4 and 6.

Wait — that doesn’t make sense unless we know which side is which.

Actually, more likely: The altitude is drawn from the right angle to the hypotenuse, dividing it into segments of 4 and 6. Then:

- The altitude h = √(4×6) = √24 ≈ 4.9 — but that’s not asked.

Wait — the question asks for x and y — which are probably the legs.

In this case:

Whole hypotenuse = 4 + 6 = 10

Leg adjacent to segment 4: x = √(hypotenuse × segment) = √(10 × 4) = √40 ≈ 6.3

Leg adjacent to segment 6: y = √(10 × 6) = √60 ≈ 7.7

Wait — but which is which? Since the diagram isn't visible, we assume:

Typically, if the segment next to leg x is 4, then x = √(10×4), etc.

But since no labels are given on which segment goes with which leg, perhaps we can assign arbitrarily — but let’s check common convention.

Alternatively, maybe x and y are the segments? No — problem says “solve for missing variable”, and diagram has x and y on the legs.

Looking back at problem 8 description: “x=______ y=______” and diagram shows a right triangle with legs labeled x and y, and altitude to hypotenuse splitting it into 4 and 6.

So yes — legs are x and y.

Rule: Each leg is geometric mean of whole hypotenuse and its adjacent segment.

Assume:

- Leg x is adjacent to segment 4 → x = √(10 × 4) = √40 ≈ 6.3
- Leg y is adjacent to segment 6 → y = √(10 × 6) = √60 ≈ 7.7

But which is which? Actually, without diagram, we might have to guess — but often in textbooks, they list them in order.

Alternatively, maybe x is the shorter leg? Let’s compute both:

√40 ≈ 6.324 → 6.3
√60 ≈ 7.746 → 7.7

Since 4 < 6, the leg adjacent to 4 should be shorter → so x = 6.3, y = 7.7

But problem doesn’t specify which is which — however, since it lists “x=___ y=___”, and in diagram likely x is left or first, we’ll go with:

x = √(10×4) = √40 ≈ 6.3
y = √(10×6) = √60 ≈ 7.7

✔ x = 6.3, y = 7.7

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Problem 9: Diagram shows right triangle with altitude 10, hypotenuse split into x and 25. Need to find x and y (where y is probably a leg).

Wait — description: “x=___ y=___” and diagram has altitude = 10, hypotenuse split into x and 25, and y is one leg.

Standard setup:

Altitude h = 10, segments = x and 25.

Then:
h² = x × 25
→ 100 = 25x
→ x = 4

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

Now, leg y — which leg? Probably the one adjacent to segment x=4.

So y = √(whole hypotenuse × adjacent segment) = √(29 × 4) = √116 ≈ 10.770 → 10.8

Or if y is adjacent to 25, then y = √(29×25) = √725 ≈ 26.9 — but that seems too big.

Looking at typical diagrams, y is often the leg corresponding to the smaller segment.

But let’s see: altitude is 10, which is fairly large, so legs should be larger than 10.

Actually, both legs will be longer than altitude.

But let’s calculate both possibilities.

If y is adjacent to segment x=4: y = √(29×4) = √116 ≈ 10.8

If y is adjacent to segment 25: y = √(29×25) = √725 ≈ 26.9

Which one makes sense? In most diagrams, y is labeled on the side opposite or adjacent — but since not specified, perhaps we need to infer.

Wait — in problem 9, it says “y=” and in diagram, likely y is the leg that forms the triangle with segment x and altitude.

Actually, another way: the leg can also be found using Pythagoras with the small triangle.

For example, the small triangle with base x=4 and height 10, then leg y = √(4² + 10²) = √(16+100)=√116≈10.8

Similarly, the other leg would be √(25² + 10²)=√(625+100)=√725≈26.9

But since the problem asks for “y=”, and typically y is the one associated with the unknown segment, probably y is the leg adjacent to x.

Moreover, in many worksheets, they ask for the leg corresponding to the segment you just found.

So I think:

x = 4
y = √(4² + 10²) = √116 ≈ 10.8

But wait — is that correct? Actually, in the small right triangle formed by the altitude, the leg of the big triangle is the hypotenuse of the small triangle.

Yes! So if we have a small right triangle with legs 4 and 10, then the hypotenuse of that small triangle is part of the big triangle’s leg.

Actually, no — the altitude is perpendicular, so the small triangle has:

- One leg = segment (say 4)
- Other leg = altitude (10)
- Hypotenuse = the leg of the big triangle (which is y)

So yes: y = √(4² + 10²) = √(16 + 100) = √116 ≈ 10.770 → 10.8

And x = 4 (from earlier)

✔ x = 4.0, y = 10.8

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Problem 10: Hypotenuse split into 7 and 28, altitude = x, and y is a leg.

First, altitude x = √(7 × 28) = √196 = 14.0

Now, whole hypotenuse = 7 + 28 = 35

Now, y is probably one leg. Which one? Likely the one adjacent to 28 or 7.

Again, if y is adjacent to 28, then y = √(35 × 28) = √980 ≈ 31.3

If adjacent to 7, y = √(35 × 7) = √245 ≈ 15.7

But also, we can use the small triangle: for example, if y is the leg forming triangle with segment 28 and altitude 14, then y = √(28² + 14²) = √(784 + 196) = √980 ≈ 31.3

Similarly, other leg = √(7² + 14²) = √(49 + 196) = √245 ≈ 15.7

Now, which is y? In diagram, likely y is the longer leg, since 28 > 7.

But problem says “y=”, and in many cases, y is labeled on the side opposite or something.

To be safe, let’s see common practice: often y is the leg corresponding to the larger segment.

But let’s check the answer format — it asks for x and y.

x is altitude = 14.0

y — perhaps it's specified in diagram as the leg adjacent to 28? Or maybe we can leave it as is.

Another thought: in some diagrams, y is the entire leg, and we can compute it as geometric mean.

I think it's safer to assume y is the leg adjacent to the segment 28, since 28 is larger and often labeled first or something.

But let’s calculate both and see.

Perhaps from the diagram description: "y" is written on the side that includes the 28 segment.

I recall that in such problems, sometimes y is the leg that is not the one we're solving for directly.

Wait — problem 10: “x=___ y=___” and diagram has altitude x, and y is one leg.

Given that, and since we have two possible legs, but typically in such worksheets, they expect you to find the leg using the geometric mean formula with the whole hypotenuse and the adjacent segment.

Moreover, since 28 is given, and 7 is given, and y is likely associated with 28.

Let me compute y = √(35 * 28) = √980

Calculate √980: 31.30495... → 31.3

And x = 14.0

But let’s verify with Pythagoras: the two legs should satisfy a² + b² = c²

c = 35

If legs are √245 and √980, then:

(√245)^2 + (√980)^2 = 245 + 980 = 1225

35^2 = 1225 → yes, correct.

So both are valid, but which is y?

Perhaps in the diagram, y is labeled on the side that has the 28 segment adjacent.

I think it's reasonable to take y as the leg adjacent to 28, so y = √(35*28) = √980 ≈ 31.3

Some might argue y is the other one, but I'll go with this.

To confirm, let's see problem 11 for pattern.

Problem 11: similar, but only x is asked.

For now, I'll set:

x = 14.0
y = 31.3

But let's double-check the calculation:

√(35 * 28) = √(980)

980 = 100 * 9.8, sqrt(100*9.8)=10*sqrt(9.8)≈10*3.13=31.3, yes.

√245 = √(49*5)=7√5≈7*2.236=15.652→15.7

But since the problem likely expects y to be the larger leg, I'll go with 31.3.

✔ x = 14.0, y = 31.3

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Problem 11: Right triangle with legs 5 and 15, altitude to hypotenuse is x. Find x.

First, find hypotenuse: √(5² + 15²) = √(25 + 225) = √250 = 5√10 ≈ 15.811

But we don't need hypotenuse directly.

Area of triangle = (1/2)*leg1*leg2 = (1/2)*5*15 = 37.5

Also, area = (1/2)*hypotenuse*altitude

So: (1/2)*h*x = 37.5, where h is hypotenuse.

h = √(5² + 15²) = √(25+225)=√250=5√10

So: (1/2)*(5√10)*x = 37.5

Multiply both sides by 2: 5√10 * x = 75

x = 75 / (5√10) = 15 / √10

Rationalize: (15√10)/10 = (3√10)/2

Numerically: √10 ≈ 3.162, so 3*3.162/2 = 9.486/2 = 4.743 → 4.7

Alternatively, use the property: in right triangle, altitude to hypotenuse h = (leg1 * leg2) / hypotenuse

So x = (5 * 15) / √(5² + 15²) = 75 / √250

Simplify √250 = √(25*10) = 5√10

So x = 75 / (5√10) = 15 / √10 = (15√10)/10 = (3√10)/2 ≈ (3*3.162)/2 = 9.486/2 = 4.743 → rounded to tenths: 4.7

✔ x = 4.7

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Now, compiling all answers:

1. 12.0
2. 22.4
3. 9
4. 4.6
5. 64
6. 9.8
7. 5.2
8. x=6.3, y=7.7
9. x=4.0, y=10.8
10. x=14.0, y=31.3
11. 4.7

Note: For problem 8,9,10, we assumed which leg is y based on typical diagram labeling. If the actual diagram differs, answers might swap, but based on standard interpretation, this should be correct.

Final Answer:
1. 12.0
2. 22.4
3. 9
4. 4.6
5. 64
6. 9.8
7. 5.2
8. x = 6.3, y = 7.7
9. x = 4.0, y = 10.8
10. x = 14.0, y = 31.3
11. 4.7