Interactive tool

Geometric Mean Altitude Theorem Calculator — h = √(p × q) with Proof

The Geometric Mean Altitude Theorem states that the altitude drawn from the right angle of a right triangle to its hypotenuse equals the geometric mean of the two hypotenuse segments it creates. Formula: h = √(p × q). Example: segments p = 4 and q = 16 → h = √(4 × 16) = √64 = 8. Hypotenuse = p + q = 20. Verify with the Pythagorean theorem on each sub-triangle: √(4² + 8²) = √80 = 4√5 and √(8² + 16²) = √320 = 8√5. This calculator solves for h, or rearranges to find a missing segment when h and one segment are known.

Client-side

Geometric mean of hypotenuse segments

Before you calculate

Best forCalculating the altitude to the hypotenuse in a right triangle when both hypotenuse segments are known. Also solves for a missing segment p or q when h is given. Useful for geometry homework, similarity proofs, and Euclidean mean visualizations. InputTwo positive hypotenuse segment lengths p and q (the lengths that the altitude creates on the hypotenuse, not the triangle’s legs). Both must be greater than zero. OutputAltitude h = √(p × q), full hypotenuse length (p + q), segment ratio p : q, and the geometric mean value alongside the arithmetic mean for comparison.

Three connections the altitude theorem reveals

h = √(pq) is more than a formula — it is the geometric mean definition made visible in a triangle, a proof of AM-GM inequality, and the key to unlocking all triangle measurements from just two numbers.

AA similarity proof

Why three triangles sharing two angles produce h = √(pq)

Dropping altitude CD from right angle C to hypotenuse AB creates triangle ACD and triangle CDB. Both sub-triangles share a right angle (at D) and one angle with the original triangle: ACD shares angle A with ABC; CDB shares angle B with ABC. Two equal angles (AA) force similarity. From ACD ~ CDB: AD/CD = CD/DB → p/h = h/q → h² = pq → h = √(pq). For p = 9, q = 4: h = 6. Both sub-triangles are similar to ABC (angles 90°, arctan(2/3), arctan(3/2)).

AM-GM connection

The altitude is always shorter than the average of the two segments

By the AM-GM inequality, (p+q)/2 ≥ √(pq), with equality only when p = q. The arithmetic mean of the hypotenuse segments is always at least as long as the altitude. For p = 4, q = 16: AM = 10, GM = h = 8 — the altitude is 20% shorter than the midpoint of the hypotenuse. For p = q = 9: AM = GM = h = 9 (isoceles right triangle — the altitude equals the midpoint).

Full triangle from two numbers

p and q determine everything about the triangle

Once p and q are known: altitude h = √(pq); hypotenuse = p+q; leg a = √(p(p+q)); leg b = √(q(p+q)). For p = 4, q = 16: h = 8; hypotenuse = 20; leg a = √80 = 4√5 ≈ 8.94; leg b = √320 = 8√5 ≈ 17.89. Pythagorean check: 80 + 320 = 400 = 20² ✓. Two input numbers fully define a right triangle.

Which version of the altitude theorem applies to your problem

Identify which value is unknown, then pick the correct form. All examples use the same triangle (p=4, q=16, h=8) for direct comparison.

Situation	Input	Best next move	Why
Find altitude h from both segments	p = 4, q = 16	h = √(4×16) = √64 = 8	Direct application of the geometric mean altitude theorem.
Find missing segment q from h and p	h = 8, p = 4	q = h²/p = 64/4 = 16 ✓	Rearrange h² = pq: divide both sides by p.
Find a leg of the original triangle	p = 4, q = 16, hypotenuse = 20	leg a = √(p × hyp) = √(4×20) = √80 ≈ 8.94	Use the geometric mean leg theorem — different from the altitude theorem.
Non-right triangle or altitude not to hypotenuse	Triangle with no 90° angle	Do not use this theorem	AA similarity requires a right angle at C. Without it, h/p ≠ q/h and h ≠ √(pq).

Step-by-step: how to use this calculator correctly

Identify the right angle in the triangle. The altitude must be drawn from this vertex perpendicular to the hypotenuse — not from any other vertex.

Label the two segments the altitude creates on the hypotenuse as p (shorter) and q (longer), or enter them in any order — h = √(p×q) is symmetric.

Enter p and q. Both must be positive real numbers. The formula is undefined for zero or negative segment lengths.

Read h as the altitude length. Verify: h² should equal p × q exactly. For p = 4, q = 9: h² = 36 and h = 6. Check: 4 × 9 = 36 ✓.

If you need a missing segment instead: rearrange h² = pq → p = h²/q or q = h²/p. For h = 6 and q = 9: p = 36/9 = 4 ✓.

Three ways to use the altitude theorem: solve for h, p, or q

Given

Right triangle ABC with right angle at C. Altitude CD meets hypotenuse AB at D.

Work

Case 1 (find h): AD = p = 4, DB = q = 16 → h = √(4×16) = √64 = 8. Case 2 (find q): h = 8, p = 4 → q = h²/p = 64/4 = 16 ✓. Case 3 (find p): h = 8, q = 16 → p = h²/q = 64/16 = 4 ✓.

Result

h = 8, hypotenuse = p + q = 20. Leg AC = √(p×(p+q)) = √(4×20) = √80 = 4√5 ≈ 8.94. Leg BC = √(q×(p+q)) = √(16×20) = √320 = 8√5 ≈ 17.89.

Takeaway

AM of p and q = (4+16)/2 = 10. GM = h = 8. AM > GM by 2 units (25% larger) — exactly as the AM-GM inequality predicts for unequal positive values.

Four ways to express the altitude theorem: formula, proportion, rearrangement, and leg theorem

The altitude theorem h = √(p×q) is one equation with four useful forms. Each form solves a different unknown. Use the table to identify which version matches your problem.

Method	Best for	Watch for	Example
Altitude formula: h = √(p×q)	Finding the altitude when both hypotenuse segments p and q are given.	Both segments must be the parts of the hypotenuse, not the legs of the triangle.	p = 4, q = 16 → h = √(4×16) = √64 = 8. Verify: h² = 64 = 4×16 ✓.
Proportion form: p/h = h/q	Understanding why h = √(pq) — it follows directly from AA-similar triangle ratios.	Cross-multiply correctly: p × q = h × h = h². Do not write p/q = h/h.	p = 4, h = 8, q = ?: 4/8 = 8/q → q = 8×8/4 = 16 ✓.
Segment rearrangement: p = h²/q or q = h²/p	Recovering a missing segment when the altitude and one segment are known.	Divide h² (not h) by the known segment. A common error is dividing h instead of h².	h = 6, q = 9 → p = 36/9 = 4. Check: h = √(4×9) = √36 = 6 ✓.
Geometric mean leg theorem: leg = √(adjacent segment × hypotenuse)	Finding a leg of the original triangle from hypotenuse segment data.	Different formula from the altitude theorem. Uses the full hypotenuse (p+q), not just p or q.	p = 4, q = 16, hypotenuse = 20 → leg a = √(4×20) = √80 ≈ 8.94; leg b = √(16×20) ≈ 17.89.

Triangle with p = 4, q = 16, h = 8: three similar triangles labeled

Point D on hypotenuse AB splits it into AD = 4 and DB = 16. Altitude CD = 8. Triangle ACD ~ Triangle CDB ~ Triangle ABC (AA similarity). AM of segments = (4+16)/2 = 10 > h = GM = 8.

Similar triangles h / p = q / h h^2 = p x q h = √(p × q)

Altitude theorem proof: 4 steps from right angle to geometric mean

For p = 4, q = 16: proportion 4/h = h/16 → h² = 64 → h = 8. Every step is a geometric fact, not just algebra.

01 Drop altitude

From right angle C, draw CD perpendicular to hypotenuse AB. D lies between A and B. Label AD = p = 4 and DB = q = 16.

02 Identify similar triangles

Triangle ACD: angles are A, 90°, (90°−A). Triangle CDB: angles are (90°−A), 90°, A. Both share the same two non-right angles as triangle ABC → AA similarity.

03 Write the proportion

From ACD ~ CDB: AD/CD = CD/DB → p/h = h/q → 4/h = h/16 → h² = 64.

04 Take square root and verify

h = √64 = 8. Verify: h² = 64 = p×q = 4×16 ✓. Also verify legs: AC = √(p×AB) = √80 ≈ 8.94; BC = √(q×AB) = √320 ≈ 17.89. 8.94² + 17.89² ≈ 400 = 20² ✓.

Altitude theorem diagram

The altitude h splits the hypotenuse into p and q. Similar triangles produce h / p = q / h, so h^2 = p x q.

Key facts before you calculate

How three similar triangles prove h = √(pq)

When altitude CD is dropped from right angle C to hypotenuse AB: Triangle ACD has angles A, 90°, and (90° − A). Triangle CDB has angles (90° − A), 90°, and A. Triangle ABC has angles A, 90°, and (90° − A). All three triangles share the same angle set, so they are AA-similar. From ACD ~ CDB: AD/CD = CD/DB → p/h = h/q → h² = pq. Taking the square root: h = √(pq), the geometric mean. For p = 3, q = 12: h = √36 = 6. Verify: leg AC = √(3×15) = √45 = 3√5; leg BC = √(12×15) = √180 = 6√5. Pythagorean check: (3√5)² + (6√5)² = 45 + 180 = 225 = 15² ✓.

Geometric mean altitude vs geometric mean leg theorem

There are two related theorems, often confused. (1) Altitude theorem: h = √(p×q) — the altitude equals the GM of the two hypotenuse segments. (2) Leg theorem (geometric mean leg): each leg equals the GM of the hypotenuse and the adjacent segment. For leg a (adjacent to p): a = √(p×(p+q)). For p = 4, q = 16, hypotenuse = 20: leg a = √(4×20) = √80 = 4√5 ≈ 8.94; leg b = √(16×20) = √320 = 8√5 ≈ 17.89. This calculator solves the altitude theorem (h). For leg calculations, use the rearranged Pythagorean theorem.

Three altitude theorem errors that produce wrong triangle measurements

Each mistake produces a specific wrong answer. Knowing the error helps you catch it during a self-check.

Watch for

Using leg lengths as p and q instead of hypotenuse segments

If you enter the legs of the triangle (e.g., a = 8.94 and b = 17.89) as p and q, you get h = √(8.94×17.89) = √159.9 ≈ 12.65 — wrong. The correct altitude is h = 8. Always use the two pieces of the hypotenuse (AD and DB), not the original legs.

Watch for

Computing p×q without taking the square root

4×16 = 64 is the value of h² (variance units), not h (length units). Forgetting √ gives 64 instead of 8 — a factor of 8 error. Always complete the formula: h = √(p×q).

Watch for

Using the altitude theorem on an obtuse or acute triangle

The theorem requires the 90° angle to be inside the triangle. For an obtuse triangle, the altitude from the largest angle falls outside the triangle. The proportion p/h = h/q no longer holds. Confirm the right angle exists before applying the theorem.

Keep going

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Frequently asked questions

Do p and q have to be in a specific order?

No. h = √(p×q) is symmetric: √(4×16) = √(16×4) = 8. Enter either segment as p and the other as q. The altitude length is the same regardless of order.

How do I find a leg of the triangle using this theorem?

Use the geometric mean leg theorem: leg a = √(p×(p+q)) and leg b = √(q×(p+q)), where p is the segment adjacent to leg a. For p = 4, q = 16: leg a = √(4×20) = √80 ≈ 8.94; leg b = √(16×20) = √320 ≈ 17.89. Verify: 8.94² + 17.89² = 80 + 320 = 400 = 20² ✓.

Do p and q have to be entered in a specific order?

No. h = √(p×q) is symmetric: √(4×16) = √(16×4) = 8. Enter either segment as p and the other as q — the altitude is the same. However, if you are also computing the legs, you must track which segment is adjacent to which leg: leg a = √(p_adjacent × hypotenuse).

What is the point D in the diagram?

D is the foot of the altitude — the point where the perpendicular from right angle C meets the hypotenuse AB. It divides AB into segments AD = p and DB = q. For p = 4 and q = 16, D is 4 units from A and 16 units from B on a 20-unit hypotenuse.

Can I find a leg of the triangle from p and q?

Yes, using the geometric mean leg theorem: leg a (adjacent to p) = √(p × (p+q)); leg b (adjacent to q) = √(q × (p+q)). For p = 4, q = 16: leg a = √(4×20) = √80 ≈ 8.94; leg b = √(16×20) = √320 ≈ 17.89. Pythagorean check: 80 + 320 = 400 = 20² ✓.

Why does the altitude equal the geometric mean, not the arithmetic mean?

The proportion from similar triangles gives p/h = h/q, which is the definition of geometric mean: h is the value between p and q such that the ratio from p to h equals the ratio from h to q. AM = (p+q)/2 would be the midpoint of AB, which is a different point. For p = 4, q = 16: AM = 10 (midpoint of hypotenuse), GM = h = 8 (foot of altitude). They are 2 units apart on the hypotenuse.