Geometric Mean Theorem: 8 Solved Exercises + Chart

The geometric mean theorems — known in many textbooks as Euclid’s theorems — describe what happens when you drop the altitude to the hypotenuse of a right triangle. There are two of them, and in symbols they fit on one line: a² = c · p (the leg rule) and h² = p · q (the altitude rule), where c is the hypotenuse, a is a leg, p is that leg’s projection onto the hypotenuse, q is the other projection and h is the altitude.

In words: each leg is the geometric mean of the hypotenuse and its own projection onto it, and the altitude is the geometric mean of the two projections. You need them in every problem that mentions the altitude to the hypotenuse or the projections of the legs, exactly where the Pythagorean theorem alone is not enough.

The five parts of the figure

Before the formulas, you have to read the diagram. In a right triangle ABC with the right angle at C, drawing the altitude from C to the hypotenuse AB creates five quantities.

Symbol	What it is	Where it is in the diagram
c	hypotenuse	the side opposite the right angle, the longest one
a, b	legs (catheti)	the two sides that form the right angle
h	altitude to the hypotenuse	from the right-angle vertex, perpendicular to the hypotenuse
p	projection of leg a	the piece of the hypotenuse on the side of a
q	projection of leg b	the piece of the hypotenuse on the side of b

Two relations always hold and give you the fastest possible check: p + q = c (the projections rebuild the hypotenuse) and h ≤ c / 2 (the altitude can never exceed half the hypotenuse).

All the formulas in one chart

Relation	Formula	Statement
Leg rule (first theorem)	a² = c · p and b² = c · q	each leg is the geometric mean of the hypotenuse and its projection
Altitude rule (second theorem)	h² = p · q	the altitude is the geometric mean of the two projections
Pythagorean theorem	a² + b² = c²	the square of the hypotenuse equals the sum of the squares of the legs
Area relation	a · b = c · h	the area can be computed from the legs or from hypotenuse and altitude

That last row is not one of Euclid’s theorems, but learn it alongside them: the area of the same triangle can be written as (a · b) / 2, using the legs as base and height, or as (c · h) / 2, using the hypotenuse and its altitude. It is the shortcut you need whenever the two legs are given and h is wanted.

The proportion form

Textbooks also state both rules as proportions, and in that form the name geometric mean explains itself:

c : a = a : p (leg rule) and p : h = h : q (altitude rule)

In each one the same quantity occupies both middle places. Since in any proportion the product of the means equals the product of the extremes, c : a = a : p immediately gives a² = c · p.

Which rule to use: a decision table

This is where marks get lost, not in the arithmetic. The rule is to look at which quantities the problem gives you.

What the problem gives	What to use	Formula
hypotenuse and one projection	leg rule	a = √(c · p)
a leg and its projection	leg rule, rearranged	c = a² / p
a leg and the hypotenuse	leg rule, rearranged	p = a² / c
both projections	altitude rule	h = √(p · q)
both legs	Pythagoras + area relation	c = √(a² + b²), then h = (a · b) / c
a leg and the altitude	Pythagoras in the small triangle	p = √(a² - h²)
hypotenuse and altitude	sum and product of the projections	x² - c·x + h² = 0

The last two rows are the ones that cause trouble: no geometric mean rule connects a leg directly to the altitude. When the problem hands you exactly that pair, the route is the Pythagorean theorem applied to one of the two small triangles, whose legs are h and a projection and whose hypotenuse is a leg of the big triangle.

A fully worked triangle: 15-20-25

It pays to keep in mind one triangle where every value is a whole number, so a wrong answer stands out instantly.

Quantity	Value	Where it comes from
legs	a = 15, b = 20	given
hypotenuse	c = 25	Pythagoras: √(225 + 400) = √625
projection of a	p = 9	leg rule: 225 / 25
projection of b	q = 16	leg rule: 400 / 25
altitude	h = 12	altitude rule: √(9 · 16) = √144
area	150	(15 · 20) / 2 and also (25 · 12) / 2

Every check works out: 9 + 16 = 25 and 15 · 20 = 25 · 12 = 300. Two more convenient all-integer triangles are 30-40-50 (p = 18, q = 32, h = 24) and 45-60-75 (p = 27, q = 48, h = 36). All of them are multiples of the 3-4-5 triple, which is exactly why the projections come out whole.

Why they work: similar triangles

The altitude splits the right triangle into two smaller right triangles. Each of them has a right angle and shares an acute angle with the original, so all three triangles are similar: same angles, proportional sides.

Triangle	Legs	Hypotenuse
the big one	a, b	c
small one, left	h, p	a
small one, right	h, q	b

Both rules fall out of that similarity. Comparing the big triangle with the left-hand one, the ratio of hypotenuse to adjacent leg is the same: c : a = a : p, which is the leg rule. Comparing the two small triangles with each other, the ratio of their legs gives p : h = h : q, the altitude rule. They are not two separate facts to memorise: they are one similarity read in two directions.

The limiting case: how tall can the altitude be?

Once the hypotenuse is fixed, the altitude cannot grow without bound. From p + q = c and p · q = h² it follows that p and q are the roots of x² - c·x + h² = 0, which exist only when the discriminant is non-negative:

c² - 4h² ≥ 0, that is h ≤ c / 2

The maximum happens when Δ = 0, so p = q = c / 2: that is the right isosceles triangle, the only one whose altitude is exactly half the hypotenuse. With a hypotenuse of 10 cm, for instance, the altitude cannot exceed 5 cm, and the legs of that limiting triangle measure √(10 · 5) = √50 ≈ 7.07 cm.

This is worth checking before you start computing: if a problem claims a hypotenuse of 10 cm and an altitude of 6 cm, the triangle does not exist, and saying so is the solution. Exercise 8 below is exactly that case.

The most common mistakes

• Pairing a leg with the wrong projection. a goes with p and b goes with q: each leg belongs with the projection starting at its own vertex. Writing a² = c · q is mistake number one, and it produces a plausible but wrong answer.
• Using the altitude rule with a leg. h² = p · q links the altitude to the projections, not to the legs. Writing h² = a · b is meaningless; the correct relation with the legs is a · b = c · h.
• Forgetting the square root. From a² = 225 the leg is 15, not 225. This is the single costliest slip in tests.
• Adding instead of subtracting for the second projection. If c = 25 and p = 9, then q = 25 - 9 = 16, not 25 + 9.
• Applying the rules to a non-right triangle, or to an altitude drawn to a leg instead of the hypotenuse. In both cases the formulas are simply false.
• Skipping the p + q = c check. It is the fastest verification available: if the projections do not rebuild the hypotenuse, something went wrong earlier.
• Rounding too early. With data like 5.4 and 9.6 the results stay exact all the way through; round only at the end, and only if the problem asks for it.

In the eight exercises below the difficulty really does climb, and no two are solved the same way. The first three apply the two rules forwards and backwards; the fourth and fifth need an outside step (Pythagoras in the small triangle, then the area relation); the sixth leads to a quadratic equation built from the sum and product of the projections; the seventh starts from a ratio instead of lengths; the eighth is a triangle that cannot exist, and you are meant to catch it from a negative discriminant.