Solving for missing variables in right triangles using the geometric mean theorem.

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 to find missing variables.

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Problem 7: Find x

We have a right triangle with altitude drawn to the hypotenuse. The hypotenuse is split into segments of length 9 and 3. The altitude is labeled x.

According to the Geometric Mean Theorem for the altitude:

> altitude² = (left segment) × (right segment)

So:

> x² = 9 × 3
> x² = 27
> x = √27
> x = √(9×3) = 3√3

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

Wait — looking again at the diagram: the altitude is perpendicular to the hypotenuse, splitting it into 9 and 3. So yes, x² = 9 * 3 = 27 → x = √27 = 3√3.

But hold on — perhaps the problem expects us to use the leg rule? No, because we’re solving for the altitude, not a leg.

Actually, let me check the first example given in the image to confirm the method.

In the first example (not numbered), they had segments 12 and 8, and set up: 12/x = x/8 → x² = 96 → x = √96. That’s the same as what we’re doing here.

So for problem 7: x² = 9 * 3 = 27 → x = √27 = 3√3.

But maybe they want it as a decimal? Or did I misread the diagram?

Wait — actually, looking back at the user's image description, problem 7 has segments 9 and 3, and altitude x. So yes, x² = 9*3 = 27 → x = √27.

But let me simplify: √27 = √(9*3) = 3√3.

However, sometimes in these worksheets, they expect exact form or decimal. Since no instruction, I’ll go with simplified radical.

BUT — wait! Let me think again. In the first example, they wrote 12/x = x/8, which comes from similarity: the altitude divides the triangle so that the ratio of the whole left part to the altitude equals the ratio of the altitude to the right part.

That’s equivalent to x² = 12*8.

Same here: x² = 9*3 = 27 → x = √27 = 3√3.

Okay, I think that’s correct.

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Problem 8: Find x and y

This is a right triangle with legs x and y, and hypotenuse divided into segments 4 and 6 by the altitude. Wait — actually, looking at the diagram: it shows a right triangle with the right angle at the bottom left. An altitude is drawn from the right angle to the hypotenuse, splitting the hypotenuse into segments of 4 and 6. The legs are labeled x and y.

Actually, re-examining: the triangle has sides labeled: one leg is x, another leg is y, and the hypotenuse is split into 4 and 6 by the altitude. But where is the right angle? The small square is at the vertex between x and y, so yes, right angle there. Then the altitude is drawn from that right angle to the hypotenuse, creating segments 4 and 6 on the hypotenuse.

To find x and y, we use the leg rules:

> Leg² = (adjacent segment) × (whole hypotenuse)

Whole hypotenuse = 4 + 6 = 10

For leg x: it is adjacent to the segment of length 4? Let’s see — typically, the leg next to segment 4 would be x, and next to segment 6 would be y.

So:

> x² = 4 × 10 = 40 → x = √40 = 2√10

> y² = 6 × 10 = 60 → y = √60 = 2√15

Is that right? Let me verify with the altitude.

The altitude h satisfies h² = 4*6 = 24 → h = √24 = 2√6

Then area of triangle can be calculated two ways:

Area = (1/2)*x*y = (1/2)*leg1*leg2

Also Area = (1/2)*hypotenuse*altitude = (1/2)*10*h = 5h

So (1/2)*x*y = 5h → x*y = 10h

Plug in: x = √40, y = √60, h = √24

Left side: √40 * √60 = √(2400) = √(100*24) = 10√24

Right side: 10 * √24

Yes, equal. So correct.

Thus:

x = √40 = 2√10

y = √60 = 2√15

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Problem 9: Find x and y

Diagram: right triangle with right angle at top. Altitude drawn from right angle to hypotenuse, splitting hypotenuse into segments x and 25. The altitude is 10. One leg is labeled y.

First, use the altitude rule:

> altitude² = (segment1) × (segment2)

So:

> 10² = x × 25
> 100 = 25x
> x = 100 / 25 = 4

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

Now, find y. Which leg is y? It’s the leg adjacent to segment x=4.

So:

> y² = (adjacent segment) × (whole hypotenuse) = 4 × 29 = 116
> y = √116 = √(4×29) = 2√29

Alternatively, if y were the other leg, it would be 25×29, but looking at the diagram, y is connected to the segment x, so yes, adjacent to x=4.

Confirm with Pythagoras? We have legs y and the other leg (let’s call z), hypotenuse 29.

z² = 25 × 29 = 725

y² = 4 × 29 = 116

Then y² + z² = 116 + 725 = 841 = 29² → yes, correct.

So x = 4, y = √116 = 2√29

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Now, let’s write final answers clearly.

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