How to Find the Diagonalof a Parallelogram
A parallelogram is a four‑sided figure whose opposite sides are parallel and equal in length. Knowing how to calculate its diagonals is useful in geometry, engineering, and design because the diagonals reveal information about the shape’s symmetry, area, and internal angles. This guide walks you through the theory, formulas, and step‑by‑step examples you need to find the length of either diagonal in any parallelogram, using only side lengths and the included angle—or, alternatively, vector components.
1. Core Properties of a Parallelogram
Before diving into calculations, recall the defining traits:
- Opposite sides are equal: (AB = CD) and (BC = AD).
- Opposite angles are equal: (\angle A = \angle C) and (\angle B = \angle D).
- Consecutive angles are supplementary: (\angle A + \angle B = 180^\circ). - The diagonals bisect each other but are not necessarily equal unless the parallelogram is a rectangle or an isosceles trapezoid‑like shape.
These properties let us relate the sides and angles to the diagonals through trigonometry or vector algebra.
2. Deriving the Diagonal Formula (Law of Cosines)
Consider parallelogram (ABCD) with sides (AB = a) and (BC = b). Let the angle between these sides at vertex (B) be (\theta) (i.e., (\angle ABC = \theta)). The diagonal (AC) stretches from vertex (A) to vertex (C) and forms two triangles: (\triangle ABC) and (\triangle CDA). In (\triangle ABC), we know two sides ((a) and (b)) and the included angle (\theta). Applying the law of cosines gives the length of diagonal (d_1 = AC):
[ d_1^2 = a^2 + b^2 - 2ab\cos(\theta) ]
The other diagonal, (BD), connects vertices (B) and (D). In triangle (ABD), the known sides are again (a) and (b), but the included angle is the supplement of (\theta) because (\angle BAD = 180^\circ - \theta). Using the law of cosines:
[ d_2^2 = a^2 + b^2 - 2ab\cos(180^\circ - \theta) = a^2 + b^2 + 2ab\cos(\theta) ]
Thus, the two diagonals are:
[ \boxed{d_1 = \sqrt{a^2 + b^2 - 2ab\cos\theta}} \qquad \boxed{d_2 = \sqrt{a^2 + b^2 + 2ab\cos\theta}} ]
Key takeaway: If you know the side lengths and any interior angle, you can compute both diagonals directly.
3. Using Vectors to Find Diagonals
When the parallelogram is defined by coordinate points or vectors, the diagonal calculation becomes even more straightforward. Let vectors (\vec{u}) and (\vec{v}) represent two adjacent sides originating from the same vertex. Then:
- One diagonal is the sum of the vectors: (\vec{d}_1 = \vec{u} + \vec{v}).
- The other diagonal is the difference: (\vec{d}_2 = \vec{u} - \vec{v}) (or (\vec{v} - \vec{u}); magnitude is the same).
The length of a diagonal is the magnitude of the resulting vector:
[|\vec{d}_1| = \sqrt{(\vec{u}+\vec{v})\cdot(\vec{u}+\vec{v})} ] [ |\vec{d}_2| = \sqrt{(\vec{u}-\vec{v})\cdot(\vec{u}-\vec{v})} ]
Expanding the dot product reproduces the law‑of‑cosines formulas, confirming consistency between the geometric and algebraic approaches.
4. Special Cases
4.1 Rectangle
All angles are (90^\circ) ((\cos\theta = 0)). The formulas reduce to:
[ d_1 = d_2 = \sqrt{a^2 + b^2} ]
Both diagonals are equal, as expected.
4.2 Rhombus
All sides are equal: (a = b = s). The diagonals become:
[ d_1 = s\sqrt{2 - 2\cos\theta} = 2s\sin\frac{\theta}{2} ] [ d_2 = s\sqrt{2 + 2\cos\theta} = 2s\cos\frac{\theta}{2} ]
Notice that the diagonals are perpendicular ((d_1 \perp d_2)) only when (\theta = 90^\circ) (i.e., the rhombus is a square).
4.3 Square
Here (a = b = s) and (\theta = 90^\circ). Both diagonals equal:
[ d = s\sqrt{2} ]
5. Step‑by‑Step Example Problems### Example 1: Given Sides and Angle
Problem: A parallelogram has sides (a = 7\text{ cm}) and (b = 5\text{ cm}). The angle between them is (\theta = 60^\circ). Find both diagonals.
Solution:
- Compute (\cos 60^\circ = 0.5).
- First diagonal: [ d_1 = \sqrt{7^2 + 5^2 - 2\cdot7\cdot5\cdot0.5} = \sqrt{49 + 25 - 35} = \sqrt{39} \approx 6.24\text{ cm} ]
- Second diagonal: [ d_2 = \sqrt{7^2 + 5^2 + 2\cdot7\cdot5\cdot0.5} = \sqrt{49 + 25 + 35} = \sqrt{109} \approx 10.44\text{ cm} ]
Answer: (d_1 \approx 6.24\text{ cm}), (d_2 \approx 10.44\text{ cm}).
Example 2: Using Vectors Problem: Parallelogram vertices are (A(0,0)), (B(4,0)), (C(6,3)), (D(2,3)). Find the length of diagonal (AC).
Solution:
- Determine vectors for adjacent sides:
(\vec{AB} = (4,0)) and (\vec{AD} = (2,3)). - Diagonal (AC) equals (\vec{AB} + \vec{AD} = (4+2, 0+3) = (6,3)).
- Magnitude: [ |AC| = \sqrt{6^2 + 3^2} = \sqrt{36 +
