A Diagonal Is Just the Law of Cosines in Disguise
Most students meet the parallelogram in Class 8 — opposite sides equal, opposite angles equal, diagonals bisect each other. Then in Class 10 or 11 they're asked to compute the diagonal length, and the formula looks intimidating: $p = \sqrt{x^2 + y^2 - 2xy \cos A}$. Square root, cosine, two squared sides — it reads like an equation pulled from advanced geometry.
The diagonal of parallelogram formula says:
$$p = \sqrt{x^2 + y^2 - 2xy \cos A},$$
where $x$ and $y$ are adjacent sides and $A$ is the angle between them.
It looks like a calculus-style formula. It's not — it's the law of cosines applied to the triangle formed by the diagonal. The whole formula is one law-of-cosines step.
The Diagonal of Parallelogram Formula
For a parallelogram with adjacent sides $x$ and $y$ and the angle $A$ between them, the two diagonals are:
$$\boxed{;p = \sqrt{x^2 + y^2 - 2xy \cos A} \quad\text{and}\quad q = \sqrt{x^2 + y^2 + 2xy \cos A};}$$
Notice the sign — the shorter diagonal carries the minus sign (opposite vertex angle is acute), the longer diagonal carries the plus. The two diagonals split the included angle differently and that splits the cosine sign.
The companion identity — the parallelogram law:
$$p^2 + q^2 = 2(x^2 + y^2).$$
The sum of the squares of the two diagonals equals twice the sum of the squares of the two sides. The cosine terms cancel when you add $p^2$ and $q^2$, leaving a clean side-length identity that has nothing to do with the angle.
Quick facts.
Type: A geometric length formula derived from the law of cosines.
Inputs needed: Two adjacent sides + one angle, or two adjacent sides + one diagonal (the parallelogram law gives the other).
Special cases: Square — $x = y$ and $A = 90°$, so $p = q = x\sqrt{2}$. Rhombus — $x = y$, diagonals perpendicular, $p^2 + q^2 = 4x^2$.
Grade introduced: NCERT Class 9 Chapter 8 — Quadrilaterals (properties); diagonal-length computation appears in Class 10 trigonometry chapters paired with CCSS-M HSG-SRT.D.10 (law of sines/cosines).
Related identity: Apollonius's theorem (a special case for the median of a triangle).
How the Diagonal of Parallelogram Formula Is Derived — One Application of the Law of Cosines
Take a parallelogram $ABCD$ with $AB = x$ along the base, $AD = y$ rising at angle $A$, and diagonal $AC = p$ from vertex $A$ to opposite vertex $C$.
The diagonal $AC$ closes a triangle with two of the parallelogram's sides — but not directly with $x$ and $y$. The triangle is $A$, $B$, $C$, where:
$AB = x$
$BC = y$ (since opposite sides of a parallelogram are equal, $BC = AD = y$)
The angle at $B$ — let's call it $B$ — is supplementary to angle $A$, since consecutive angles of a parallelogram sum to $180°$. So $B = 180° - A$.
The law of cosines applied to triangle $ABC$ gives:
$$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B$$
$$p^2 = x^2 + y^2 - 2xy \cos(180° - A).$$
Using the identity $\cos(180° - A) = -\cos A$:
$$p^2 = x^2 + y^2 - 2xy \cdot (-\cos A) = x^2 + y^2 + 2xy \cos A.$$
Wait — that gives the plus version. Let me reconsider which diagonal is which.
Actually, the convention depends on which diagonal is which. The diagonal from the acute-angle vertex picks up the minus sign (the formula in the boxed statement); the diagonal from the obtuse-angle vertex picks up the plus sign. Let me redo this more carefully.
Take the diagonal that originates from vertex $A$ — that's $AC = p$, going from $A$ across to the opposite vertex $C$. The triangle that contains this diagonal and the included angle $A$ is triangle $ABD$ — no wait, that's not right either. The two triangles $ABD$ and $ABC$ each contain one diagonal.
The cleanest version: in triangle $ABD$ (vertices $A$, $B$, $D$), the sides are $AB = x$, $AD = y$, and diagonal $BD = q$ is opposite the angle $A$. Law of cosines:
$$q^2 = x^2 + y^2 - 2xy \cos A.$$
So $q$ — the diagonal opposite vertex $A$ — uses the minus sign. By the parallelogram-law identity $p^2 + q^2 = 2(x^2 + y^2)$, the other diagonal $p = AC$ satisfies $p^2 = 2(x^2 + y^2) - q^2 = x^2 + y^2 + 2xy \cos A$. So $p$ uses the plus sign.
The diagonal whose endpoints are across the angle $A$ uses $\cos A$ with a minus; the other diagonal uses $\cos A$ with a plus. Naming convention varies by textbook — what matters is which triangle the diagonal belongs to.
I went through that derivation more carefully than necessary because the sign convention is exactly where most students slip, and the formula card alone doesn't make it obvious.
Three Worked Examples of Diagonal of Parallelogram Formula
Quick. A parallelogram has adjacent sides 6 cm and 8 cm with the included angle $60°$. Find the shorter diagonal.
Use $p = \sqrt{x^2 + y^2 - 2xy \cos A}$ with $x = 6$, $y = 8$, $A = 60°$, $\cos 60° = 0.5$:
$$p = \sqrt{36 + 64 - 2 \cdot 6 \cdot 8 \cdot 0.5} = \sqrt{100 - 48} = \sqrt{52} = 2\sqrt{13} \approx 7.21 \text{ cm}.$$
Final answer: $2\sqrt{13} \approx 7.21 \text{ cm}$.
Standard (Wrong Path First — Three Habits That Lose Marks). A parallelogram has sides 5 cm and 12 cm with one angle $120°$. Find both diagonals.
The wrong path. A student computes $\cos 120° = -0.5$ but plugs in the positive value 0.5:
$$p = \sqrt{25 + 144 - 2 \cdot 5 \cdot 12 \cdot 0.5} = \sqrt{169 - 60} = \sqrt{109} \approx 10.44 \text{ cm}.$$
The flaw: dropping the negative sign on $\cos 120°$. The cosine of an obtuse angle is negative; substituting only the magnitude gives the wrong diagonal length.
The rescue. Use $\cos 120° = -0.5$:
$$p = \sqrt{25 + 144 - 2 \cdot 5 \cdot 12 \cdot (-0.5)} = \sqrt{169 + 60} = \sqrt{229} \approx 15.13 \text{ cm}.$$
For the second diagonal, flip the cosine sign:
$$q = \sqrt{25 + 144 + 2 \cdot 5 \cdot 12 \cdot (-0.5)} = \sqrt{169 - 60} = \sqrt{109} \approx 10.44 \text{ cm}.$$
Final answer: Long diagonal $\approx 15.13$ cm; short diagonal $\approx 10.44$ cm.
Check with the parallelogram law: $p^2 + q^2 = 229 + 109 = 338 = 2(25 + 144) = 2 \cdot 169 = 338$ ✓.
Stretch. A parallelogram has adjacent sides $a$ and $b$ with one diagonal of length $d_1$. Find the other diagonal in terms of $a$, $b$, $d_1$ — no angle needed.
Apply the parallelogram law:
$$d_1^2 + d_2^2 = 2(a^2 + b^2)$$
$$d_2 = \sqrt{2(a^2 + b^2) - d_1^2}.$$
Final answer: $d_2 = \sqrt{2(a^2 + b^2) - d_1^2}$.
This is the side-and-one-diagonal version of the problem — common in JEE-Main coordinate-geometry questions where one diagonal is known from the vertices' coordinates and the other must be derived. The parallelogram law bypasses the angle calculation entirely.
Where the Diagonal of Parallelogram Formula Shows Up
The formula isn't a textbook curiosity. It appears wherever rigid two-dimensional structures meet at non-right angles.
Truss-bridge engineering. A parallelogram-shaped truss panel has diagonal braces whose length is computed exactly from the parallelogram-diagonal formula — the brace is one of the diagonals.
CAD software and 3D modelling. When a designer skews a rectangle to make a parallelogram, the software internally computes the new diagonal lengths using this formula to update grid snaps and dimension labels.
Crystallography. Two-dimensional crystal lattices are catalogued by the parallelogram of their unit cell. The diagonal lengths characterise the lattice symmetry and determine the diffraction pattern.
Vector addition (the parallelogram law of forces). When two forces $\vec{F_1}$ and $\vec{F_2}$ act on a body at an angle $\theta$, the resultant force's magnitude is $|\vec{F_1} + \vec{F_2}| = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}$ — the same formula. The resultant is the diagonal of the parallelogram formed by the two force vectors.
Hinged mechanisms in furniture. A folding table or scissor jack whose linkage forms a parallelogram has a diagonal that determines maximum extension and load-bearing capacity. Mechanical engineers compute it from the parallelogram diagonal formula.
For a Class 10 / Class 11 student, the most-encountered context is the JEE coordinate-geometry vector chapter — but the formula reaches into engineering mechanics, materials science, and any field where two vectors combine to produce a resultant. The parallelogram diagonal is the vector sum.
How the Diagonal Formula Reached Its Modern Form
The result was implicit in classical Greek geometry but didn't have its modern algebraic form until the development of trigonometry.
Euclid (c. 300 BCE, Alexandria) in Elements Book I proved the diagonal of a parallelogram bisects it into two congruent triangles, and established the law of cosines geometrically in Proposition II.13 — though without the trigonometric notation.
Apollonius of Perga (c. 240–190 BCE, Greece) proved the special case for the median of a triangle — Apollonius's theorem — which is geometrically equivalent to the parallelogram law $p^2 + q^2 = 2(x^2 + y^2)$ applied to a degenerate parallelogram.
François Viète (1540–1603, France) developed the trigonometric notation that lets us write the law of cosines as $c^2 = a^2 + b^2 - 2ab \cos C$. Before Viète, the same fact required pages of geometric proportion arguments.
Callout — Apollonius's theorem and the parallelogram law.
Apollonius of Perga earned the title "the Great Geometer" by extending Euclid's results to the conic sections — ellipses, parabolas, hyperbolas. Buried inside his work, almost as a side result, is the statement: in any triangle $ABC$ with median $AM$ to the midpoint $M$ of $BC$, we have $AB^2 + AC^2 = 2(AM^2 + BM^2)$. Fold a parallelogram along one diagonal, and the two resulting triangles share that median — which is exactly the parallelogram law in disguise. A theorem from 230 BCE about medians of triangles is, by an algebraic step, the parallelogram law a Class 11 student uses today.
Slip-Ups That Cost Marks on the Diagonal Formula
Mistake 1: Using $\cos$ of the wrong angle.
Where it slips in: A parallelogram has angles $A$ and $B = 180° - A$. A student computes the diagonal opposite vertex $A$ but plugs in $\cos B$ instead of $\cos A$ (or vice versa).
Don't do this: Apply the formula without identifying which diagonal originates from which vertex angle.
The correct way: Identify the triangle formed by the diagonal and the two adjacent sides. The angle in the formula is the angle of that triangle at the vertex opposite the diagonal. Or — use the cross-check: $\cos(180° - A) = -\cos A$, so the two diagonal formulas are sign-flipped versions of each other.
Mistake 2: Dropping the negative sign on $\cos$ of an obtuse angle.
Where it slips in: Computing $\cos 120°$, $\cos 135°$, or $\cos 150°$, a student takes the magnitude and forgets the negative sign.
Don't do this: Treat $\cos$ as always positive. For obtuse angles ($90° < A < 180°$), $\cos A$ is negative.
The correct way: $\cos 120° = -0.5$, $\cos 135° = -\sqrt{2}/2$, $\cos 150° = -\sqrt{3}/2$. The negative sign means the $-2xy \cos A$ term adds to the length (subtracting a negative). The diagonal across an obtuse angle is the longer of the two.
Mistake 3: Confusing the diagonal formula with the area formula.
Where it slips in: A student computes the diagonal as $xy \sin A$ — which is actually the area of the parallelogram, not the diagonal.
Don't do this: Memorise the formula by surface pattern (sides + angle) without distinguishing which output it produces.
The correct way: Area of a parallelogram = $xy \sin A$. Diagonal of a parallelogram = $\sqrt{x^2 + y^2 \pm 2xy \cos A}$. The units are different — area is in square units; diagonal is in linear units. The unit-check catches the mistake immediately.
A real-world version of "wrong-formula-for-the-output" mistakes. In 1986, NASA's Space Shuttle Challenger disintegrated 73 seconds after launch because an O-ring seal failed at low temperature. Engineers had data showing the O-ring's performance versus temperature — but they read the chart as showing the probability of failure when it actually plotted the number of past incidents. Same shape of data, wrong interpretation of axis. The structural mistake is identical to a student using a formula whose inputs match the problem but whose output answers a different question entirely. Always check what the formula returns, not just what it accepts.
Bottom Line
The diagonal of parallelogram formula is $p = \sqrt{x^2 + y^2 - 2xy \cos A}$ for one diagonal; the other diagonal uses $+2xy \cos A$.
The formula is the law of cosines applied to the triangle formed by the diagonal and two adjacent sides — no new geometry needed.
The parallelogram law, $p^2 + q^2 = 2(x^2 + y^2)$, is the cosine-free cross-check.
The single most common error is dropping the negative sign on $\cos$ of an obtuse angle — which flips one diagonal into the wrong length entirely.
The same formula governs vector addition: the resultant of two forces meeting at angle $\theta$ is the diagonal of the parallelogram they form.
Take These for a Test Drive — Three Problems
A parallelogram has adjacent sides 7 cm and 10 cm with the included angle $45°$. Find both diagonals.
A parallelogram has sides 8 cm and 6 cm, and one diagonal of length 12 cm. Find the other diagonal using the parallelogram law.
A square has diagonal $5\sqrt{2}$ cm. Find its side length, then verify the parallelogram law.
If Problem 1 gave you the same number for both diagonals, you forgot the sign flip on $\cos A$. Return to the cosine-sign table in Mistake 2.
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